More from Ravindra Shinde

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1, , Gravitation, CONCEPT MAP, KEPLER’S LAWS, Kepler’s law, , Kepler's, first law, of orbit, , Kepler's, second, law of area, , Helped in, formulating, , Escape velocity, Value of, it on Earth, , Explains, , the inverse, square Law, , Kepler's third law, of time period, T2 α r3, , vesc = 11.2 km/s, , Newton’s Universal Law of Gravitation, , vesc =, , 2gR, , Gravitational Force, Velocity to, escape the, , influence of, Gravitational force, , F=, , GMm, d2, , Force gives, rise to, , Acceleration due to gravity (g), Weight, W = mg, , gives, rise to, , Shape, Altitude, 2, Pole - 9.832 m/s, Altitude increases, Equator - 9.78 m/s2 value of `g' decreases, , g = GM, R2, , Variation in the, value of g, Depth, Depth increases, value of `g' decreases, , Points to Remember:, zz Gravitation was discovered by Sir Isaac Newton., zz Gravitational Force is an universal force which is acting not only on any two bodies on the surface of the Earth, , but also on any two celestial objects., , zz Gravitational force is an attractive force., zz Every object in the universe attracts every other object with a force, which is directly proportional to the product, , of their masses and inversely proportional to the square of the distance between their centres F =, , G m1m2, , ., d2, zz If two objects are spherical, the force between them is along the straight line joining their centres and the segment, joining their centres is considered the distance between them., zz Out of the two objects, if mass of one of the objects is doubled, then according to law, the gravitational force, , between the two objects gets doubled. If the distance between the two objects is doubled, the gravitational force, between them becomes one fourth., , zz Value of G is 6.67 × 10-11 Nm2/kg2, zz The centripetal force is centre-seeking or the force which is acting towards the centre of the circle., , (5)

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Master Key Science and Technology Part-1 (Std. X), , 6, , zz The centripetal force between the Sun and the planet Earth is responsible for the revolution of the Earth. It is, , inversely proportional to the square of the distance between them., , zz The force that the Earth exerts on any object at or around the surface of the Earth is called the Earth’s gravitational, , force. According to Newton’s second law, the acceleration is produced in the body due to applied force., , zz The acceleration produced by the Earth’s gravitational force is called acceleration due to gravity or gravitational, , acceleration. It is denoted by g =, , GM, , ., R2, zz The acceleration due to gravity depends on mass of the Earth (M) and radius of the Earth (R) for an object on, the surface or height (h) at which the object is placed from the surface of the Earth. It does not depend upon the, mass of the falling object. Thus, acceleration due to gravity is same for all freely falling objects., zz The unit of ‘g’ in SI system is m/s2. The mass of Earth is 6 × 1024 kg and its radius is 6.4 × 106 m., zz The Earth does not have perfectly spherical shape. It is slightly flattened at the poles and bulged at the equator., , Therefore, the radius of the Earth at the poles is less than that at the equator., , zz The value of ‘g’ is maximum at poles and it is 9.83 m/s2 or 983 cm/s2., zz The value of ‘g’ is minimum at the equator and it is 9.78 m/s2 or 978 cm/s2., zz As altitude increases, value of g decreases., zz As depth increases, value of g decreases., zz The gravitational acceleration is considered 9.77 m/s2 for all the practical purposes., zz The acceleration on the Moon is, , 1, , th, , of that on the surface of the Earth., 6, zz W,  hen a body falls towards the Earth under the influence of the Earth’s gravity alone, its motion is called ‘free fall’., zz In ‘free fall’, body comes to the surface of the Earth with uniform acceleration. Initial velocity may be zero but, , it increases at every time interval, due to acceleration due to gravity., , zz The mass of an object is the quantity of matter contained in it. The SI unit of mass is (kg) and CGS unit is, , gram (g)., , zz From Newton’s first law, we know that mass is the measure of inertia of a body., zz Mass is a scalar quantity and it does not change from place to place on the surface of the Earth and even it does, , not change on the different planets., , zz The weight of the body is defined as the force with which the Earth attracts it towards its centre W = mg., zz Since the weight is a ‘force’, its SI unit is newton and CGS unit is dyne. (1N = 105 dyne), zz Weight is a vector quantity. It changes from place to place on the surface of the Earth and even changes with the, , planet., , zz Since the value of ‘g’ is not same everywhere, the weight of a body keeps on changing with place., zz The weight of an object on the Moon is 1 th of that on the Earth. We can write it as, wm = m × gm., , 6, zz Spring balance is used for measuring weight and two scale balance is used to compare masses., zz Johannes Kepler (1571 – 1630) was a German astronomer and a Mathematician., zz Kepler’s first law: The orbit of a planet is an ellipse with the sun at one of the foci., , zz Kepler’s second law: The line joining the planet and the sun sweeps equal areas in equal intervals of time., zz Kepler’s third law: The square of time period of revolution around the sun is directly proportional to the cube, , of the mean distance of a planet from the Sun., , zz A body can have energy because of its shape or position or configuration. This is known as potential energy., , –GMm ., R+h, zz E,  scape velocity is that initial velocity of an object by which it overcomes the Earth’s gravitational attraction., zz The gravitational potential energy at the height (h) from the ground is, , zz Escape velocity (vesc) can be calculated by the law of conservation of energy and is equal to vesc = √2gR .

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Gravitation, , 7, , MASTER KEY QUESTION SET - 1, Q.1. (A) 1. Fill in the blanks:, (1), , Gravitation was discovered by, , (2), ,  very object in the universe attracts every other, E, object with, force., m1m2, Fα, means F =, ., d2, If the distance between the two objects is doubled,, then the gravitational force between them will be, times., , (3), (4), , ., , (5), , If mass increases, force, , (6), , If mass triples, value of G, , (7), , I f Earth attracts a body with a force of 10 N, then, the body attracts the Earth with , N., , (8), , S.I. unit of G is, , (9), ,  laces on Earth exactly 90° to the direction of the, P, moon experiences, ., , (10), , , is a constant known as universal, gravitational constant., , (11), ,  he Earth’s gravitational force is always in the, T, direction of the, of the Earth., , (12), ,  he Earth’s gravitational acceleration is denoted, T, by letter, ., , (13), , The relation between g and G is, , (14), ,  he value of acceleration due to gravity at poles, T, is, m/s2., , (15), ,  he value of acceleration due to gravity at the, T, equator is, cm/s2., If altitude increases, value of g, ., When a spaceship is two Earth radii distance from, the centre of Earth, its g becomes, ., During free fall, the object comes vertically, downward with uniform, ., , (16), (17), (18), , (28), , he energy possessed because of position or, T, configuration is called, ., , (29), , vesc =, , (30), , Gravitational waves are detected by, , (31), , During ascent, is zero and during, descent, is zero., (32) Formula for centripetal force is, ., Gm1m2, 1, Ans. (1) Isaac Newton (2) gravitational (3), (4) 4, 2, d, (5) increases (6) remains constant (7) 10 (8) Nm2/kg2, GM, R2, 1 th, (14) 9.83 (15) 978 (16) decreases (17), 4, 24, (18) acceleration (19) inertia (20) 6 × 10 (21) 6.4 × 106, (9) low tide (10) G (11) centre (12) ‘g’ (13) g =, , ., ., , (22) Zero (23) newton (N) (24) weight (25) six, , ., , (26) poles (27) ellipse (28) potential energy (29) 2gR, (30) LIGO (31) Final velocity (v), Initial velocity (u), mv2, (32) F =, r, Q.1. (A) 2. Find the odd word out:, (1), , Acceleration, mass, force, weight, , Ans. Mass - It is scalar, while others are vectors., (2), ,  hange in value of ‘g’ at surface, change in value, C, of ‘g’ at height, change in value of ‘g’ at depth,, change in value of ‘g’ on thickness, , Ans. C,  hange in value of ‘g’ on thickness. Others are, related factors to the value of ‘g’., *(3), , Light, sound, heat, laws of planetary motion, , Ans. L,  aw of planetary motion. Others are related to, Newton., (4), , Mass, potential energy, radius, weight, , Ans. Weight. It is vector, while others are scalars., (5), , 9.83 m/s2, 9.8 m/s2, 980 cm/s2, 9.77 m/s2, , (19), , Mass is also a measure of, , of an object., , Ans. 980 cm/s2. Others are values of g in MKS., , (20), , The mass of the Earth is, , kg., , (6), , (21), , The radius of the Earth is, , (22), , Value of g at the centre of the Earth is, , (23), , The SI unit of weight is, , (24), ,  he gravitational force acting on any object on, T, the Earth is called it's, ., , Ans. N,  ewton’s law of gravity. Other are related to law, of motion., , (25), , The weight of an object on the Earth is, , times its weight on the Moon., , Ans. Ohm. Others are concerned with laws of gravity., , (26), , The weight of an object is maximum at, , on the surface of the Earth., , (9), , (27), ,  he orbit of a planet is an, T, Sun at one of the foci., , (10), , m., , Weight, Thrust, Force, Pressure, , Ans. Pressure. Others are vectors., ., , ., , with the, , (7), , (8), ,  ewton’s first law, Newton’s law of gravity,, N, Newton’s third law, Newton’s second law, , Newton, Ohm, Kepler, Galileo, 983 m/s2, 977 m/s2, 980 m/s2, 9.83 m/s2, , Ans. 9.83 m/s2. Others are in CGS system., 9.83 m/s2, 98.3 m/s2, 983 m/s, 98.03 m2s, , Ans. 9.83 m/s2. Others are not values of ‘g’.

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Master Key Science and Technology Part-1 (Std. X), , 8, , Q.1. (A) 3. Complete the analogy:, (1), , (2), , 6 × 10 kg : Mass of the Earth : : 6.4 × 10 m :, ., 24, , Column A, , Ans. Radius of the Earth - Mass is measured in kg and, distance is measured in metres., (2), ,  eight of a weather satellite : 8.7 m/s2 : : Height, H, of Communication satellite :, ., , Ans. 0.225 m/s2 - The value of ‘g’ changes with the, change in the height of the satellites., (3), , Match the columns:, , 6, , Mass : Scalar quantity : : Weight :, , ., , (1) Mass, , (a) m/s, , (2) Weight, , (b) m/s2, , (3) Acceleration, , (c) kg, , (4) Velocity, , (d) N, , Ans. (1 – c), (2 – d), (3 – b), (4 – a), (3), , Complete the chart:, , Ans. Vector quantity - Weight being a force is a vector, quantity and its direction is towards the centre of, the earth., , (1), , (4), , ., , (2), , 30 × 10, , Ans. 9.78 m/s2 - Earth’s radius is the largest at the, equator and the smallest at the poles., , (3), , 16 × 109, , (4), , 250 × 10-7, , (5), , At poles : 9.83 m/s2 : : At equator :, ,  hape of the Earth at equator : Bulged : : Shape of, S, the Earth at poles :, , Ans. F,  lattened - The shape of the Earth is not exactly, spherical. Due to its rotation, the earth bulges at, the equator and is flattened at the poles., 1, (6), Kinetic energy : mv2 : : Gravitational Potential, 2, energy :, , Column B, , F (N), , 27, , m1 (kg), , m2 (kg), , 500, , 84, , 15 × 10, , d (m), 02, 03, , 5, , 17, 45, , 34, , 47, , Ans. (1) 7 × 10-7 (2) 2.69 × 1033kg (3) 1.63 × 1022kg, (4) 7.5 × 10-2m, Q.1. (A) 5. State whether the following statements, are True or False. Correct the false statement., (1), , Force = mass × velocity, , (2), , ‘G’ is called gravitational acceleration., , –GMm, Ans. , - For small distances, i.e. heights, the, R+h, potential energy is less than zero, i.e. it is negative., , (3), , Acceleration is a scalar quantity., , (4), ,  ravitational force at the Moon is double than, G, the Earth’s gravitational force., , (7), , (5), , 1 N = 1 kg × 1 m/s2., , (6), , 1 dyne = 105 N., , (7), ,  he force towards the centre of the circular orbit, T, is called centripetal force., , (8), ,  he gravitational acceleration does not become, T, zero at the centre of the Earth., , (9), , Q.1. (A) 4. With the information in three columns,, match up and complete the chart:, ,  t the poles, the acceleration due to gravity is, A, 9.77 m/s2., , (10), , ‘g’ is called universal constant., , *(1), , (11), , Mass is a scalar quantity., , Force : ma : : Gravitational force :, Ans. Gm1m2 - Mathematically, the gravitational force of, d2, attraction between two bodies is given by Gm1m2, d2, (8), Force : Vector : : Weight :, Ans. Vector - Weight is also a force and its direction is, towards the centre of the earth. Weight = m × g, , I, , II, , III, , (1) Mass, , (a) m/s2, , (i) Zero at the, centre, , (12), , Beyond the surface of the Earth, g α, , (13), , Weight is a vector quantity., , (2) Weight, , (b) kg, , (ii) Measure of, Inertia, , (14), , The mass of the Earth is 6.4 × 106 kg., , (3) Acceleration, due to, gravity, , (c) Nm2/, kg2, , (iii) Same in, the entire, universe, , (4) Gravitational, constant, , (d) N, , (iv) Depends, on height, , Ans. (1-b-ii), (2-d-iv), (3-a-i), (4-c-iii),, , (15), , 1, (R + h)2, , ., , At a height of ‘h’ from the ground, the gravitational, –GMm, potential energy is, ., R+h, Ans. (1) False. Force = mass × acceleration (2) False., ‘G’ is called Universal gravitational constant, (3) False. Acceleration is a vector quantity, 1 th, (4) False. Gravitational force at the moon is, of, 6

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Gravitation, , 9, , the Earth’s gravitational force (5) True (6) True, (7) True (8) False. The gravitational acceleration, becomes zero at the centre of the Earth (9) False., The acceleration due to gravity is 9.83 m/s2 at, the poles (10) False. ‘g’ is called acceleration due, to gravity (11) True (12) True (13) True (14), False. Mass of the earth is 6 × 1024 kg (15) True., , (9), , (a) 9.72 m/s, (c) 9.83 m/s, (10), ,  he gravitational force of attraction between two, T, objects is given by, ., (a) F α, (c) F α, , (2), , m1m2, , (b) F α, , d2, , m1m2, , (d) F α, , d2, , (3), , (4), , (b) one fifth, (d) six, , The gravitational force causes, , (c) None of these, , (a) less than 1020 N, , (b) Circular motion of, moon, (d) Both a and b, ,  he SI unit of gravitational constant is, T, (a) Nm2 / kg2, (b) Nkg2 / m2, (c) m/s2, (d) N cm2 / g2, , (8), ,  he value of acceleration due to gravity at the, T, height ‘h’ from the ground is, ., , (c) g =, , (b) g =, , R+h, GM, (R + h), , 2, , GM, R+h, , (d) g = GM (R + h)2, , as the depth from, (b) fluctuates, (d) varies, , (b) less, (d) can’t say, ,  he mass of objects, T, the surface on the Earth., (a) remains constant, (c) changes, , at any place on, , (b) is non-uniform, (d) increases, ,  ccording to Newton’s first law, if mass is more,, A, then the inertia of the body is, ., (b) very less, (d) can’t say, , (16), , The weight of body gradually decreases from, , ., (a) equator to poles, (b) poles to equator, (c) pole to pole, (d) height to surface, , (17), ,  body of mass 1 kg is attracted by the Earth with, A, a force which is equal to, ., -11, (a) 9.8 N, (b) 6.67 × 10, (c) 1 N, (d) 9.8 m/s, , (18), , , The, gravitational potential energy at the height, of ‘h’ from the ground is, ., , (b) 1020 N, , (7), , (b) is in the same, direction as that of, motion, (d) increase as it rise, up, ,  s the height of the object from the surface of the, A, Earth increases, value of ‘g’ becomes, ., (a) more, (c) equal, , (14), , (b) Poles, (d) Both a and c, ,  he value of ‘g’, T, surface increases., , (a) less, (c) more, , (c) greater than 1020 N (d) 10-20 N, , GM, , (13), , (15), , ., ,  he Earth attracts moon with a force of 1020 N. The, T, moon attracts Earth with a force of, ., , (a) g =, , (12), , (a) increases, (c) decreases, , If the distance between two objects increases, 5 times, the gravitational force becomes, times., 1, (a) 5, (b) 15, (c), (d) 25, 25, The gravitational force on the surface of the Moon, is, times than that on the surface of, the Earth., , (a) Tides, , (6), , d3, , ., ,  hen an object is thrown upward, the force of, W, gravity, ., (a) is opposite to, the direction of, motion, (c) becomes zero at, higher point, , m1m2, m1m2, , (b) one forth, (d) 2 times, , (a) five, (c) one sixth, (5), , d2, , I f the distance between two bodies becomes half,, the gravitational force between them becomes, ., (a) half, (c) 4 times, , (11), , (b) 9.83 m/s2, (d) 9.72 m/s2, , The value of ‘g’ on Earth is zero at, (a) Centre of Earth, (c) Infinite distance, , Q.1. (B) Choose and write the correct option:, (1), , The value of ‘g’ is maximum at poles and it is, ., , ., , (a), (c), , –GMm, R+h, GMm1, R +h, 2, , 2, , (b), (d), , –GMm1, R2 + h, GMm, R2 + h

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Master Key Science and Technology Part-1 (Std. X), , 10, , (19), , (20), ,  he orbit of a planet is an ellipse with the Sun at, T, one of the, ., (a) foci, (b) centre (c) middle (d) surface, , Ans. Given:, , r=1m, , , , m1 = 75 kg, ,  he straight line joining the planet and the Sun, T, sweeps equal, in equal interval of time., , , , (a) volume, (c) density, (21), ,  he square of time period of revolution around the, T, Sun is directly proportional to the, of the planet from the Sun., (a), (b), (c), (d), , (22), , (b) angle, (d) area, , mean distance, square of the distance, cube to the distance, cube of the mean distance, ,  hich of the following is not an example of free, W, fall?, (a), (b), (c), (d), , Moon revolving around the Earth, Earth revolving around the Sun, Parachute jumping, Artificial satellites revolving around the Earth, , (23), , The centre of mass of an object having uniform, density is at its, ., (a) centre, (b) Geometrical centre, (c) centroid, (d) Circumference, mm, Ans. (1) (a) F α 1 2 (2) (c) 4 times (3) (c) 1 (4) (c) one, d2, 25, sixth (5) (d) both a and b (6) (b) 1020 N (7) (a) Nm2/kg2, GM, (8) (c) g =, (9) (b) 9.83 m/s2 (10) (d) Both a, (R + h)2, and c (11) (a) Is opposite to the direction of motion, (12) (c) decreases (13) (b) less (14) (a) remains, constant (15) (c) more (16) (b) poles to equator, –GMm, (17) (a) 9.8 N (18) (a), (19) (a) foci, R+h, (20) (d) area (21) (d) cube of the mean distance, (22) (c) Parachute jumping (23) centroid, Q.2.1. Solve the following:, Type: A, Gm1m2, F=, , v = u + at, r2, 1 2, F = ma , s = ut +, at, 2, v2 = u2 + 2as, , , *(1) Mahendra and Virat are sitting at a distance of, 1 metre from each other. Their masses are 75 kg, and 80 kg respectively. What is the gravitational, force between them?, , , , m2 = 80 kg, , G = 6.67 × 10-11 Nm2/kg2, , To find: Force (F) = ?, Gm1m2, Formula: F =, r2, Solution:, F=, , 6.67 × 10–11 × 75 × 80, , 12, F = 6.67 × 10 × (6 × 103), –11, , = 40.02 × 10-8, = 4.002 × 10-7 N, The gravitational force between Mahendra, and Virat is 4.002 × 10–7 N, *(2) The mass of the Earth and Moon are 6 × 1024, kg and 7.4 × 1022 kg respectively. The distance, between them is 3.84 × 105 km. Calculate the, gravitational force of attraction between the, two? G = 6.7 × 10-11 Nm2/kg2., Ans. Given:, , , , , m1 = 6 × 1024 kg, , m2 = 7.4 × 1022 kg, , r = 3.84 × 105 km = 3.84 × 108 m, G = 6.7 × 10-11 Nm2/kg2, , To find: F = ?, Formula: F =, Solution: F =, , Gm1m2, r2, 6.7 × 10–11 × 6 × 1024 × 7.4 × 1022, , (3.84 × 108)2, 297.48 × 10-11 × 1024 × 1022, =, 14.74 × 1016, 300 × 1019, , ≈, 15, = 20 × 1019, F = 2 × 1020 N, The gravitational force of attraction between, Earth and Moon is 2 × 1020 N, *(3) The mass of the Earth is 6 × 1024 kg. The distance, between the Earth and the Sun is 1.5 × 1011 m. If, the gravitational force between the two is 3.5 ×, 1022 N, what is the mass of the sun? G = 6.7 × 10-11, Nm2/kg2., Ans. Given:, , m1 = 6 × 1024 kg, , , , r = 1.5 × 1011 m, , , , F = 3.5 × 1022 N, , , , G = 6.7 × 10-11 Nm2/kg2, , To find: m2 = ?

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Gravitation, , 11, , Formula: F =, Solution:, m2 =, =, , Gm1m2, , Fr2, Gm1, , Formula: s = ut +, , r2, , 3.5 × 1022 × (1.5 × 1011)2, , 6.7 × 10-11 × 6 × 1024, 7.875 × 1044, , 1, , = t2, 100 × 2.5 × 10-9, 1 × 109, , = t2, 250, 1000 × 106, , = t2, 250, , = 1.96 × 1030 kg, In the previous example, assuming that, the bench on which Mahendra is sitting is, frictionless, starting with zero velocity. What, will be Mahendra’s velocity after 1s and how, will it change with time? Mass of Mahendra (75, kg) and force (4.002 × 10-7 N)., , Ans. Given:, , Initial velocity (u) = 0 m/s, , , , Time (t) = 1 s, , , , Force (F) = 4.002 × 10-7 N, , , , mass (m) = 75 kg, , -7, , , , Time (t) = 20 s, , , , Mass of truck (m) = 7000kg, Force (F) = ?, , (ii) F = ma, , 400 = 0 × 20 +, , *(5) , Assuming that acceleration remains constant, (5.34 × 10-9 m/s2), How long will Mahendra take, to move 1 cm towards Virat if he starts from, rest? (Use your brain power; Textbook Page no. 6), Ans. Given: Acceleration (a) = 5.34 × 10 m/s, -9, , 2, , ≈ 5 × 10 m/s, 1, Displacement (s) = 1 cm =, m, 100, Initial velocity (u) = 0 m/s, , To find: time (t) = ?, , Distance travelled (s) = 400 m, , Solution: (i) s = ut +, , As distance decreases, force increases thereby, increasing acceleration. Hence velocity increases, , , , , , , , Velocity will increase with time, , , , (6) A truck starts from rest and rolls down a hill, with a constant acceleration. It travels a distance, of 400 m in 20s. Find its acceleration. Also find, the force acting on it if its mass is 7000 kg., , Formulae: (i) s = ut +, , v = u + at, , , , Mahendra will take 2 × 103 s to move towards, Virat., , , , a = 5.34 × 10-9 m/s2, , = 5.34 × 10-9 m/s, , t = 2 × 103 s, , , , To find: Acceleration (a) = ?, , Solution: a = 4.002 × 10, 75, -7, a = 0.0534 × 10 m/s2, , , , t = √ 4 × 106 s, , , , Ans. Given: Initial velocity (u) = 0 ms-1, , To find: Final velocity (v) = ?, F, Formulae: v = u + at , a =, m, , = 0 + 5.34 × 10-9 × 1, , 4 × 106 = t2, , , , The mass of the Sun is 1.96 × 1030 kg, , , , 1, = 2.5 × 10-9 × t2, 100, , , , 40.2 × 1013, 7.875, =, × 1031, 40, = 0.196 × 1031 kg, , *(4), , 1, 1, = 0×t+, × 5 × 10-9 × t2, 100, 2, , Solution:, , = m2, , 1, at2, 2, , -9, , 2, , 1, at2, 2, , 1, at2, 2, , 1, a × (202), 2, , 400 a, 2, 400 × 2, ∴a=, = 2 m/s2, 400, , 400 =, , (ii) F = ma, ∴ F = 7000 × 2, ∴ F = 14000N, The truck moves with an acceleration of 2 m/s2, and the force acting on it is 14000N, (7) Karan and Arjun are two friends of mass m1 and, m2 respectively, separated by a distance d.

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Master Key Science and Technology Part-1 (Std. X), , 12, , What would happen to the force between them if, , Type: B, , (i) Mass of Arjun is doubled., (iii) Distance between them is doubled., (iv) Value of G doubled., Ans. The force between Karan and Arjun is, G m1 m2, F=, ...(i), d2, (i) If ‘m2’ is doubled., , , F1 =, , G m1 (2m2), d2, , =, , d2, , , , , F = 2F, Force becomes double., , , , (ii) If ‘m1’ and ‘m2’ are doubled., , , , 1, , F1 =, , G (2m1) (2m2), d2, , =, , d2, , F1 = 4F, Force becomes 4 times., , , , , , (iii) If ‘d’ is doubled., G m1m2, 1 (G m1m2), F1 =, =, 2, 4, d2, (2d), 1, F1 =, F, 4, Force becomes one forth., , , , (iv) If ‘G’ is doubled., , , , , , , F1 =, , 2G m1m2, , = 2F, d2, Force is doubled., , ...[from(i)], , 4(G m1m2), , , , , , , , 2(G m1m2), , ...[from(i)], , *(1) , Calculate the gravitational force due to, the Earth on Mahendra, if mass of Earth is, 6 × 1024 kg, Radius is 6.4 × 106 m , g = 9.77 m/s2 and, mass of Mahendra is 75 kg., Ans. Given:, , Mass of Earth (M) = 6 × 1024 kg, , , , Radius of Earth (R) = 6.4 × 106 m, , , , Mass of object (m) = 75 kg, , , , Gravitational acceleration (g) = 9.77 m/s2, , To find: Force (F) = ?, Formula: F =, , GMm, R2, , Solution:, ...[from(i)], , ...[from(i)], , NUMERICALS FOR PRACTICE, (1), , GM, , v = u + at, R2, 1 2, W = F = mg , s = ut +, at, 2, GM m, W=F=, v2 = u2 + 2as, R2, Note:, For Earth g = 9.77 m/s2, For Moon gm = 1.63 m/s2, g=, , (ii) Mass of both Karan and Arjun is doubled., ,  wo boys are sitting very close to each other at, T, a distance of 0.5 m from each other. If the mass, of one boy is 40 kg and other is 50 kg, find the, gravitational force between them. , , (5.336 × 10–7 N), , (2), ,  the force of gravitation between the Earth and, If, an object of mass ‘m’ is 9 × 107 N, find the mass, of an object if the mass of the Earth is 6 × 1024 kg, and its radius is 6.4 × 106 m., (9.2 × 106 kg), , (3), , If two objects of masses 500kg and 84kg, respectively are at a distance of 2m apart from, each other. Find gravitational force between, them? , (7 × 10-7 N), , (4), , I f two objects of 45 kg and 47 kg respectively are, attracted towards each other by a gravitational, force of 250 × 10-7 N, find the distance between, their centres. , (7 × 10-7 N), , F=, , 6.67 × 10–11 × 6 × 1024 × 75, , 40.96 × 1012, 9.77 × 75, =, 1, = 732.75 N, F ≈ 733 N, The gravitational force is 733 N, *(2) Starting from rest, due to the gravitational force, of the Earth i.e. 733 N, What is the speed of, Mahendra after 1 second? If his mass is 75 kg., Ans. Given:, , Initial velocity (u) = 0 m/s, , , , Force (F) = 733 N, , , , Mass (m) = 75 kg, , , , Time (t) = 1 s, , To find: Final velocity (v)= ?, F, Formulae: v = u + at, a =, M, Solution:, 733, a =, 75, ∴ a = 9.77 m/s2, v = u + at, = 0 + 9.77 × 1, = 9.77 m/s

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Gravitation, , 13, , The speed of Mahendra after 1 second is 9.77, m/s, *(3) If a person weighs 750 N on Earth, how much, would be his weight on the Moon given that, 1, Moon’s mass is, of that of the Earth and its, 81, 1, radius is, of that of Earth., 3.7, Ans. Given: Weight on Earth (WE) = 750 N, , , 1, , Mm =, , , , 81, 1, , Rm =, , ( M )=, R, ∴(, =, R ), , ME ∴, RE, , 3.7, , Mm, , 1, , m, , 81, 1, , E, , 3.7, , E, , To find: Weight on Moon (Wm) = ?, GMm, R2, , Formula: W = F =, Solution:, For Earth , WE =, , G ME m, , Wm =, , RE, Dividing equation (ii) by (i),, WM, WE, , WM, WE, , WM, WE, , Wm, 750, Wm, 750, , =, =, =, =, =, , G Mm m, Rm2, , G Mm m, Rm2, , Mm, ME, , RE, , ×, , G Mm m, Rm2, , ...(ii), , RE2, , RE2, , ×, , G ME m, 2, , × (3.7)2 . . . from ‘given’, 81, 13.69, 81, , . . . (i), , GMB, , . . . (ii), RB2, Dividing equation (ii) by (i),, gB GMB GMA, =, ÷, gA, RB2, RA2, gB GMB RA2, =, ×, gA, RB2 GMA, gB MB, R, =, × A, gA M, RB, A, , 2, , 4 2 MB, =, ∴ 2MA = MB, 2 1 MA, MB = 2MA, The mass of B is twice mass of A, *(5) The mass and weight of an object on Earth is 5 kg, and 49 N respectively. What will be their values, on the Moon? Assume that the acceleration due, 1, to gravity on the Moon is, th of that on the Earth., 6, Ans. Given: me = 5 kg, We = 49 N, 1, 1, , gm =, ge =, × 9.8 = 1.633 m/s2, 6, 6, To find: Mass on Moon mm = ?, , , Weight on Moon Wm = ?, , Formula: W = F = mg, , 750, , Solution: Mass remains same : mm = 5 kg, , 81, , Wm = mm × gm, , Wm = 126.8 N, , , , The weight on the Moon is 126.8 N, *(4) The radius of the planet A is half the radius of, planet B. If the mass of A is MA, What must be, the mass of B so that the value of g on B is half, that of its value on A?, 1, RA, 1, Ans. Given: RA =, RB, ∴, =, RB, 2, 2, , , RA2, , , , Rm, , 1, , Wm = 13.69 ×, , gB =, , GMA, , 1, M, 1, = B ×, . . . from ‘given’, MA, 4, 2, , G ME m, , ÷, , gA =, , For Moon, , ...(i), , 2, , GM, R2, Solution: For planet A, , Formula: g =, , gB =, , 1, 2, , ga, , gB, 1, ∴, =, gA, 2, , To find: Mass of planet B, MB = ?, , = 5 × 1.633, , Wm = 8.17 N, Mass on Moon is 5 kg and weight is 8.17 N, *(6) Suppose you are standing on a tall ladder. If, your distance from the centre of the Earth is 2R,, what will be your weight? (Use your brain power;, Textbook Page no. 10), Ans. Given: R1 = 2R, To find: W1 = ?, Formula: W = F =, , GMm, R2

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Master Key Science and Technology Part-1 (Std. X), , 14, , GMm, , Solution: W =, , R2, , , , W1 =, , , , =, , W1 =, W1 =, , 1, 4, 1, 4, , ×, , . . . (i), , GMm, (2R)2, GMm, 4R2, , . . . (ii), , GMm, R2, , W, , ...[from(i)], , Weight will be one fourth the original weight, *(7) What would be the value of g on the surface, of the earth if its mass was twice as large and, its radius half of what it is now? (Can you tell?;, Textbook Page no. 8), Ans. Given: M' = 2M, R, , R' =, 2, ', To find: g = ?, Formula: g =, Solution: g' =, , R2, GM', R'2, G × 2M, , =, , , , = 4 × 2 × g ................ ( g =, , , , 2, , =, , G × 2M, , , , ( ), ,  alculate the value of ‘g’ on the Moon, if its, C, mass is 7.4 × 1022 kg and radius is 1740 km., , (1.63 m/s2), , (6), , If the weight of a body on the surface of the, Moon is 100 N, what is its mass?, (61.34 kg), , (7), , If the acceleration due to gravity on the, surface of the Earth is 9.8 m/s2, what will be, the acceleration due to gravity on the surface, of the planet whose mass and radius both are, two times the corresponding quantities for the, Earth., (4.9 m/s2), vesc = √ 2gR, T2, =K, R3, , Type: C, , *(1) Calculate the escape velocity on the surface of, the Moon given the mass and radius of the Moon, to be 7.34 × 1022 kg and 1.74 × 106 m respectively., Ans. Given:, , Mass (M) = 7.34 × 1022 kg, , , , Radius (R) = 1.74 × 106 m, , Gravitational acceleration on Moon(gm), 1, 9.8, , =, ge =, m/s2 = 1.63 m/s2, 6, 6, To find: vesc = ?, , GM, , R, 2, , (5), , R2, 4, , =, , 4 × 2 × GM, R2, GM, R2, , Formula: vesc = √ 2gm R, Solution: vesc = √ 2 × 1.63 × 1.74 × 106, , ), , g' = 8 g, , The value of g will become 8 times., , NUMERICALS FOR PRACTICE, , = √ 5.67 × 106, , , , = 2.38 × 103 m/s, , vesc = 2.38 km/s, , The escape velocity on Moon is 2.37 km/s, *(2) Let the period of revolution of a planet at a, distance R from a star be T. Prove that if it was, at a distance of 2R from the star, its period of, revolution will be √8 T., , (1), , , Find the weight of a man whose mass is 50 kg., (490 N), , (2), , , Find, the gravitational force between man of, mass 60 kg and the Earth., (586 N), , Ans. Given:, , Distance from Sun = R, , , , Time of Rotation = T, ,  stone of mass 2 kg is falling from a certain, A, height. Find the force of attraction between the, Earth and the stone. Also, find the acceleration., , (19.6 N, 9.8 m/s2), , , , New distance = 2R, , (3), , (4), , , The, planet in space has mass twice as that of the, Earth and a radius thrice as that of the Earth. If, the weight of a book is 90 N on the Earth, what, would be the weight on that planet?, , (20 N), , To find: New time TN= ?, T2, Formula:, =k, R3, Solution: Case (i), T2, , = k . . . (i), R3, Case (ii)

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Gravitation, , 15, , TN2, = k OR, (2R)3, From (i) and (ii),, , , TN2, 8R3, , = k . . . (ii), , T 2, T2, N =, 8R3, R3, , , , Gravitational acceleration (g) = 10 m/s2, , , , Initial velocity (u) = 0 m/s, , To find: Final velocity (v) = ?, , , Displacement (s) = ?, , TN2 = 8T2, , Formulae: v = u + gt, s = ut +, , TN = √ 8T2, , Solution: v = u + gt, , (2), , s=0+, ,  he escape velocity for mass is 5.02 km/s. If its, T, radius is 3390 km, What is the value of g on its, surface., (g = 3.71 m/s2), , s=5m, ,  planet orbits the Sun in time T at a distance, A, of R from it. Another planet orbits the Sun in a, time of 8T. What is its distance R' from the Sun., , (R' = 4R), , v = u + gt, 1, gt2, 2, v2 = u2 + 2gs, s = ut +, , , , , NUMERICALS FOR PRACTICE, , Type: D, , 2, , gt2, , = 0 + 10 × 1, , TN = √ 8 T, , (1), , 1, , Downward/Dropped/Falling, g (positive) = 10 m/s2, , 1, 2, , v = 10 m/s, × 10 × 12, , Speed on reaching ground is 10 m/s and height, of the table is 5 m., *(3) An iron ball of mass of 3 kg is released from, height of 125 m and falls freely to the ground., Assuming that the value of g is 10 m/s2, calculate, (i) time taken by the ball to reach the ground, (ii) velocity of the ball on reaching the ground, (iii) the height of the ball at half the time it takes, to reach the ground., , Upward/Thrown up, g (negative) = -10 m/s2, 3 kg, , *(1) An object takes 5s to reach the ground from a, height of 5m on a planet. What is the value of g, on the planet?, Ans. Given:, , Displacement (s) = 5 m, , , , Time (t) = 5 s, , , , Initial velocity (u) = 0 m/s, , 125 m, , ht/2 = ?, , To find: Gravitational acceleration (g) = ?, 1, gt2, 2, 1, Solution: 5 = 0 +, × g × 52, 2, 25, 5=g×, 2, 5×2, =g, 25, 2, g=, 5, g = 0.4 m/s2, Formula: s = ut +, , Value of g on the planet is 0.4 m/s2, , t=?, v=?, , (Diagram is only for reference), Ans. Given:, , Mass (m) = 3 kg, , , , Displacement (s) = 125 m, , , , , Initial velocity (u) = 0 m/s, , To find: Time (t) = ?, , , Final velocity (v) = ?, , , , Height (ht/2) = ?, , *(2) A ball falls off a table and reaches the ground in, 1 s. Assuming g = 10 m/s2, calculate its speed on, reaching the ground and the height of the table., , Formulae: s = ut +, , Ans. Given:, , s = ut +, , Time (t) = 1 s, , g = 10 m/s2, , Solution: Case (i), 1, gt2, 2, , 1, gt2, v = u + gt, 2

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Master Key Science and Technology Part-1 (Std. X), , 16, , , , Final velocity (v) = 0 m/s, , To find:, , s, , Initial velocity (u) = ?, , , 125 m, , h = 125 – 5, , ttotal = ?, , Formulae:, v2 = u2 + 2gs, v = u + gt, , , (Diagram is only for reference), 1, 125 = 0 × t +, × 10 × t2, 2, 125 = 5t2, , t2 = 25, t = 5 s ...(i), Case (ii), v = u + gt, = 0 + 10 × 5, v = 50 m/s...(ii), Case (iii), , , , 5, 2, , ...(from (i)), , = 2.5s, , Height of the ball at 2.5s = s, ∴ Using Newton's second equation,, s = ut +, , 1, , to find u, , , , , , , 0, = 9+(-10)×t (from(i)), 10 t = 9, t, = 0.9 s, , Time of ascent = Time of descent, ttotal = 2t, , , = 2 × 0.9, , , , = 1.8 s, , 2, , *(5) An object thrown vertically upwards reaches a, height of 500 m. What was its initial velocity?, How long will the object take to come back to, the Earth? Assume g = 10 m/s2., , s = 31.25 m, ∴ Height of the ball at half time, = 125 - 31.25, = 93.75 m, (i) Time taken by the ball to reach the ground, is 5s., (ii) Velocity of ball on reaching ground is 50 m/s., (iii) Height of ball at half time is 93.75 m., *(4) A tennis ball is thrown up and reaches a height, of 4.05 m before coming down. What was its, initial velocity? How much total time will it, take to come down? Assume g = 10 m/s2, Displacement (s) = 4.05 m, , , , (Diagram is only, for reference), , The initial velocity was 9 m/s and total time, taken to come down, is 1.8 s., , gt, 2, 1, s=0+, × 10 × (2.5)2, 2, , Ans. Given:, , ∴ v, g and s are given,, , (a) v2 = u2 + 2gs, 02 = u2 + 2×(-10)×4.05, 0 = u2 - 81, u2 = 81, u = 9 m/s ...(i), ∴ we have v, u and g to find t., we use 1st equation, (b) v, = u + gt, , t, =, 2, , ttotal = ?, , Solution: , , 25, , =, , u=?, , we use 2nd equation, , 125, = t2, 5 1, , Half time, , ttotal = 2 × t, , S = 4.05 m, , g = -10 m/s2, , Ans. Given: Displacement (s)= 500 m, , , Gravitational acceleration (g) = -10 m/s2, , , , Final velocity (v) = 0 m/s, , To find: Initial velocity (u) = ?, , , ttotal = ?, , Formulae: v2 = u2 + 2gs, ttotal = 2t, , , v = u + gt, , ∴ v, g and s is given,, to find u, we use 3rd equation, Solution: v2 = u2 + 2gs, , , , , 0 = u2 - 10000, 10000 = u2, u = 100 m/s

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Gravitation, , 17, , Since, we have v, u and g to find t,, we use 1st equation., v = u + gt, 0 = 100 + (-10) × t, 10 t = 100, t = 10s, Time of ascent = Time of descent, ∴ ttotal = 2 × 10, , NUMERICALS FOR PRACTICE, , ∴ ttotal = 20s, , (1), ,  ball thrown up vertically returns to the, A, person after 6s. Find the velocity with which it, was thrown up., (29.4 m/s), , (2), ,  boy drops a coin from the top of a building, A, which is 49m high. Find the velocity with which, the coin strikes the ground., (v = 31 m/s), , (3), ,  ball is thrown vertically upwards with velocity, A, of 49 m/s. Calculate (i) Maximum height to, which it rises (ii) total time (t) it takes to return, to the surface of Earth., (s = 122.5 m, t = 10 s), , (4), ,  stone is thrown vertically upwards with, A, initial velocity of 40 m/s. Taking g = 10 m/s2 find, the maximum height and total distance covered, by stone., (h = 80 m, d = 160 m), , Initial velocity was 100 m/s and time taken, to come back to Earth is 20s., (6) Find a formula for maximum height attained by, object., Ans. Solution: From 3 equation of motion, v2 = u2 + 2as, v = 0, a = -g, ∴, 02 = u2 + 2(-g)s, 2gs = u2, u2, s=, 2g, rd, , *(7) A stone thrown vertically upwards with initial, velocity u reaches a height ‘h’ before coming, down. Show that the time taken to go up is, same as the time taken to come down., Ans. When object is thrown up., Initial velocity = u m/s, Final velocity = 0 m/s, Acceleration = –g m/s2, Time taken to go up = t1, According to 1st eq. of motion, v = u + gt, 0 = u - gt1, gt1 = u, u, t1 =, (s) . . . (i), g, , (During Ascent), B, , u = 0 m/s, , t2, , A, , v = u m/s, + g m/s2, , When object falls down. (During Descent), Initial velocity = 0 m/s, Final velocity (v) = u m/s, , B, , v = 0 m/s, , Time taken to go down = t2, , t1, , According to 1st eq. of motion, v = u + gt, . . . (ii), , From (i) and (ii),, t1 = t2 , , (1), Force, Ans. The force is that physical quantity which changes, or tends to change the state of rest or of uniform, motion in a straight line., (2), Newton’s universal law of gravitation, Ans. , According to the Newton’s universal law of, gravitation, every object in the universe attracts, every other object with a force, which is directly, proportional to the product of their masses, and inversely proportional to the square of the, distance between their centres., (3), Universal constant of gravitation (G), Ans. Universal constant of gravitation is the force of, attraction between two unit masses placed at unit, distance apart from each other., (4), Centre of mass, Ans. It is the point inside or outside the object at which, the total mass of the object can be assumed to be, concentrated., *(5), , Acceleration = g m/s2, , u = 0 + gt2, u, t2 =, (s), g, , Q.2.2. Define/write the laws:, , A, , u m/s, – g m/s2, , (Diagram is only, for reference), , Gravitational acceleration (g) OR Acceleration, due to gravity, Ans. , The gravitational force due to earth on a, body results in its acceleration. This is called, acceleration due to gravity., *(6) Free fall, Ans. When a body falls towards the Earth under the, influence of the Earth’s gravity alone, its motion, is called a ‘free fall’.

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Master Key Science and Technology Part-1 (Std. X), , 18, , (7), Mass (m), Ans. The mass of an object is the quantity of matter, contained in it. Mass is the measure of the inertia, of a body., (8), Weight (W), Ans. The weight of a body is defined as the force with, which the Earth attracts the object., *(9) Centripetal force, Ans. The force that acts on any object moving along a, circle, directed towards the centre of the circle is, called centripetal force., (10) Uniform circular motion, Ans. When an object moves in a circular path with, uniform speed, its motion is uniform circular, motion., (11) Potential energy, Ans. A body can have energy because of its shape, or position or configuration. This is known as, potential energy., (12) Escape velocity, Ans. , The minimum initial velocity needed by an, object projected upwards to overcome Earths, gravitational force and not fall back on Earth is, called escape velocity., Q.2.3. Answer the following in one or two sentences:, *(1) What do you know about gravitational force?, (Can you recall; Textbook Page no. 1), Ans. E,  very object in the universe attracts every, other object with a force. This force is called, gravitational force., *(2) , Will the direction of the gravitational force, change as we go inside the earth? (Think about it;, Textbook Page no. 9), Ans. , The direction of earth’s gravitational force is, towards the center of earth so it will not change., *(3) What would happen if there were no gravity?, (Think about it; Textbook Page no. 8), Ans. If gravitational force of the Earth would not have, existed, then all the object would have floated in, the air. Life would be unstable., *(4) What would happen if the value of G was twice, as large? (Think about it; Textbook Page no. 8), Ans. I f the value of ‘G’ is doubled, then the gravitational, force (F) also will get doubled., *(5), , What types of forces are we familiar with?, (Can you recall; Textbook Page no. 1), , Ans. Types of forces are:, , , (i) Gravitational force, , , , (ii) Electro-magnetic force, , , , (iii) Nuclear force, , , , (iv) Balanced force, , , , (v) Unbalanced force, , *(6) What is the value of ‘g’ at the centre of the Earth?, (Think about it; Textbook Page no. 9), Ans. At the centre of the Earth the value of ‘g’ is zero., *(7) Will the mass and weight of an object on the, earth be same as their values on Mars? Why?, Ans., (i),  he mass of the object will remain same but its, T, weight will differ on Mars., (ii) Since mass of an object is the amount of matter, present in the body, its value remains the same, on Earth and Mars., (iii) Weight is the force with which the Earth attracts, the object. It is given as W = mg. As g changes,, weight (W) also changes., Q.2.4. Write short notes:, (1), Earth’s gravitational force, Ans., (i), The Earth attracts all the objects towards the, surface of the Earth., (ii) The attraction is towards the centre of the Earth., Hence, the gravitational force of the Earth acts, towards the centre of the Earth., (iii) Due to this force, all objects fall vertically, downwards on the earth., (2), Earth’s gravitational acceleration, Ans., (i), The Earth applies a force of gravitation on any, object at or around its surface., (ii) The uniform acceleration produced in a freely, falling body due to the gravitational force of, the Earth is called gravitational acceleration or, acceleration due to gravity., (iii) Gravitational acceleration is denoted by ‘g’ and is, a vector quantity. Its SI unit is m/s2., (3) Variation in the value of g., Ans. The value of g varies based on the following, (a), , Along the surface of the earth., - The earth is not perfectly spherical., - It is bulged at the equator and flattened at the, poles, due to its rotation., - The radius is highest at the equator and least at, the poles., - Value of g is highest at poles(9.832 m/s2)

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Gravitation, , 19, , - Value of g is least at equator (9.78 m/s2), (b), , Q.2.5. Complete the following table:, , Change with height., - As height increases, value of g decreases., , (c), , Place, , Altitude from, the Earth’s, surface (km), , - The change is negligible for height smaller than, the radius of the earth., , (i), , Surface of Earth, , 0, , Change with depth., , (ii), , Mount Everest, , 8.8, , - As depth increases, value of g decreases., , (iii) Altitude attained by, Man-made balloon, , 36.6, , (iv) Orbit of space shuttle, , 400, , - This is because, as we go inside the earth, the, mass under consideration decreases., , (v), , - Hence, value of g decreases., , Communication, satellite, , Ans. (i) 9.83, , (ii) 9.8, , g, (m/s2), , 35700, (iii) 9.77 (iv) 8.7, , (v) 0.225, , Q.2.6. Distinguish between:, (1) Gravitational constant and Gravitational acceleration., Ans., , Gravitational constant (G), (i), , Gravitational constant is the force of, attraction between two unit masses placed at, unit distance apart from each other., , (i), , The uniform acceleration produced in a freely, falling body due to the gravitational force of the, Earth is called gravitational acceleration., , (ii), , It has SI unit Nm2/kg2, while CGS unit is, dyne.cm2/g2., , (ii), , It has SI unit m/s2, while CGS unit is, cm/s2., , (iii) It is always denoted by ‘G’., , (iii) It is always denoted by ‘g’., , (iv) Its S.I. value is 6.673 × 10–11 Nm2/kg2, while, in CGS, it is 6.673 × 10–8 dyne cm2/g2., , (iv) Its S.I. value is 9.77 m/s2, while in CGS, it is 977, cm/s2., , (v), *(2), , Gravitational acceleration (g), , Its value is fixed and does not change with, (v), conditions, hence it is called universal constant., , The value of ‘g’ depends on various factors like, altitude, depth, shape etc., , Weight and Mass, , Ans., , Mass (m), , Weight (W), , (i), , Mass is the amount of matter present in the (i), object., , Weight is the force with which the Earth attracts, the object., , (ii), , Mass does not change with place., , (ii), , Weight(W) = mg. As g changes, W also changes., , (iii), , SI unit is kg. CGS unit is gram (g)., , (iii) SI unit is newton. CGS unit is dyne., , (iv), , Mass is a scalar quantity., , (iv) Weight is a vector quantity., , Q.2.7. Give scientific reasons:, (1) High and low tides are regular phenomena., Ans., (i), The sea level on the Earth is directly influenced, by the gravitational force of the moon., (ii) Due to the gravitational force, the water on the, Earth experiences pull towards the direction of, the Moon, hence there occurs high tide., (iii) At the same time, the place which is at right, angles or at 90°, the water level gets reduced and, experiences low tide., , (iv), ,  his situation occurs two times a day on regular, T, basis, water level rises two times in a day and, reduces two times in a day., Low tide, , (v), , High, tide, , Earth, Low tide, , The Moon’s, High gravitational, tide, force, Moon

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Master Key Science and Technology Part-1 (Std. X), , 20, , (2) In the spacecraft, travellers and objects appear, floating., Ans., (i), Though the spacecraft is taken to a huge height, from the Earth, the value of acceleration due to, gravity does not become zero., (ii) The weightlessness is because the spacecraft is in, a state of free fall., (iii) Any object in a state of free fall will feel weight, lessness., (3) Weight of an object changes from place to place, on the surface of the Earth., Ans., (i), The shape of the Earth is not exactly spherical. It, is flattened at poles, while bulged at equator., (ii), ,  eight is the Earth’s gravitational force acting on, W, the object, which is generally W = mg., , (iii), ,  ass of the object remains the same but value of, M, ‘g’ keeps on changing from place to place., , (iv), ,  he value of ‘g’ is maximum at poles i.e. 9.83 m/s2, T, while it is the least at equator i.e. 9.78 m/s2., , (v), ,  ue to this the weight of an object changes from, D, place to place on the surface of the Earth., , (vi), , It is maximum at poles but gets reduced at, equator., , Q.3.1. Explain the following:, (1) Explain the terms:, (a) Free fall, (b) Acceleration due to gravity, (c) Escape velocity (d) Centripetal force, (e) Potential energy, Ans. , *(a) Free fall:, , (e), , Potential energy:, , , The energy possessed due to position or, configuration is called potential energy., (2) Define: The Universal law of gravitation and, derive mathematically., Ans. T,  he Universal law of gravitation: Every object, in the universe attracts every other object with a, force which is directly proportional to the product, of their masses and inversely proportional to the, square of the distance between their centres., m1, , F1, , F2, , m2, , d, Fig 1.1 Gravitational force between two objects., Let m1 and m2 be the masses of two objects, the, distance between their centres be ‘d’., , , ∴, ,  he gravitational attraction between the two will, T, be given by,, m1m2, Fα, d2, F=, , Gm1m2, d2, , , , (G is constant), , G is called gravitational constant; it is also known, as Universal constant of gravitation., *(3) What are Newton’s laws of motion?, , (Can you recall; Textbook Page no. 1), Ans. , (i) Newton’s first law of motion : Every inanimate, object continues to be in a state of rest or of, uniform motion in a straight line unless it is acted, upon by an external unbalanced force., , *(b) Acceleration due to gravity:, ,  (ii) Newton’s second law of motion : The rate of, change of momentum is directly proportional, to the applied force and it takes place in the, direction of force., , , The gravitational force due to earth on a, body results in its acceleration. This is called, acceleration due to gravity., ,  (iii) Newton’s third law of motion : To every action, there is always instantaneous, equal and opposite, reaction., , *(c), , *(4), , When a body falls towards the Earth under the, influence of the Earth’s gravity alone, its motion, is called free fall., , Escape velocity:, , , The minimum initial velocity needed by an, object projected upwards to overcome Earths, gravitational force and not fall back on Earth is, called escape velocity., *(d) Centripetal force:, The force that acts on any object moving along a, circle, directed towards the centre of the circle is, called centripetal force., , What are the effects of force acting on an object?, (Can you recall; Textbook Page no. 1), , Ans. The applied force can:, (i), change the state of rest., (ii) change the motion of an object, (iii) change the direction of motion, (iv) change the speed of a moving object, (v) change the shape of the object temporarily.

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Gravitation, , 21, , (5) Define: Acceleration due to gravity and derive, mathematically., Ans. , The gravitational force due to the Earth on, an object results in its acceleration. It is called, acceleration due to gravity., We know,, GMm, . . . (i), R2, F = mg . . . (ii), F=, , 2, Centripetal force F = mv, ...(i), r, Distance travelled Circumference, by planet in one revolution = of orbit, , = 2πr , Time period of revolution = (T), distance, ∴, speed of planet v =, time, , , , From (i) and (ii),, , , , GMm, mg =, R2, , T, Substituting eq. (ii) in (i), , (, , m, ∴, , GM, g= 2, R, For Earth, g = 9.77 m/s2, , Ans. The acceleration due to gravity is given as, GM, g= 2, R, At the centre of the Earth, the mass under, consideration is zero., , F =, , ∴, , F =, , ∴, , F =, , , , Hence, acceleration due to gravity, at the centre, of the earth, is zero., *(7) Write the three laws given by Kepler. How did, they help Newton to arrive at the inverse square, law of gravity?, Ans. Kepler’s laws:, , 4mπ2r, , m4π2r2, , =, , rT2, , ×, , r2, , r2, T2, 4mπ2r3, , r2T2, According to Kepler's 3rd law, r, , 3, , r3, , = k or, , T, , 2, , =, , 1, , Substituting (iv) in (iii), , , , F =, , , , Rearranging eq. (v), , 4mπ2r3, r2T2, , 4mπ2, , (ii) , Kepler’s second law : The line joining the planet, and the Sun sweeps equal areas in equal intervals, of time., , , , Since, , k, k, , ∴, , F = constant ×, , ∴, , Fα, , (8), , 1, k, , =, , 4mπ2, r2k, , ...(v), , × 1, r2, , 4mπ2, , 1, , ×, , ...(iii), , ...(iv), , k, , , , (i) , Kepler’s first law : The orbit of a planet is an, ellipse with the Sun at one of the foci., , Sun, , ), , r, 4mπ r, , F=, , r, , T, , 2, , 2, , , , (iii) , Kepler’s third law: The square of its period of, revolution around the Sun is directly proportional, to the cube of the mean distance of a planet from, the Sun., m, v, , 2πr, , ...(ii), , , T2, Multiplying numerator and denominator with ‘r2’., , T2, , i.e. M = O, ∴ g=0, , F =, , ∴, , *(6) The value of ‘g’ at the centre of the Earth is zero., Explain?, , 2πr, , =, , = constant., 1, r2, , , r2, State Kepler’s third law and derive mathematically, to obtain constant., , Ans. , Kepler’s third law: The square of its period of, revolution around the Sun is directly proportional, to the cube of the mean distance of a planet from, the Sun., If ‘T’ is a periodic time and the average distance, of a planet from the Sun is ‘r’, then,, , , T2 α r3

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Master Key Science and Technology Part-1 (Std. X), , 22, , ∴, , T2 = kr3, , ∴, , T, r3, , k =, , E1 = E2, , ... k is constant, k ≠ 0, , 2, , 1, GMm, mv2esc –, =0, 2, R, 1, GMm, mv2esc =, 2, R, , *(9) If the value of g suddenly becomes twice its, value, it will become two times more difficult to, pull a heavy object along the floor. Why?, Ans., (i), If the value of g becomes double,the force with, which the earth pulls the object i.e weight of, object becomes double., (ii) As weight increases, frictional force also increases, while pulling the object, ( iii) Hence, it will become two times more difficult to, pull a heavy object along the floor., (10) Explain centripetal force with suitable example., Ans. The force on an object revolving in a circular path, towards the centre is called centripetal force., , Fig 1.2 Centripetal force, e.g. A stone tied to the string moves in a circular, path and its velocity is tangential., (11) Define:, Escape, velocity, and, derive, mathematically., Ans. , The velocity required to overcome the Earth’s, gravitational force to move into the space is, called escape velocity., On surface of Earth, 1, A, K.E. = mv2esc, 2, B , C , , 1, GMm, E1 = mv2esc –, 2, R, , B , C , , , K.E. = 0, GMm, P.E. = –, =0, ∞, T.E. = P.E. + K.E., , vesc =, , 2GM × R, R ×R, , vesc =, , 2GM × R ... g = GM, R2, R2, , vesc =, , 2gR, , For Earth, vesc = 11.2 km/s., Q.3.2. Open Ended Question:, *(1) According to Newton’s law of gravitation, every, object attracts every other object means if the, Earth attracts an apple towards it, then an apple, also attracts the Earth with the same force, then, why an apple falls down but the Earth does not, move towards the apple? (Use your brain power;, Textbook Page no. 7), Ans. Y,  es, the apple attracts the Earth with equal, force but the mass of the apple is much less as, compared to that of the Earth., F, , We know, F = ma , or a =, m, , Hence, greater the mass, lesser the acceleration., Therefore, due to greater mass, the Earth’s, acceleration is almost zero., *(2) , Is there a gravitational force between two, objects kept on a table or between you and your, friend sitting next to you? If yes, why don’t the, two move towards each other? (Use your brain, power; Textbook Page no. 5), , , , We know,, , , , F α m1m2, , Since the mass of me and my friend is very less as, compared to the Earth, the force is also very less., Hence, we don't move towards each other., , At infinity, A, , 2GM, R, , Ans. Y,  es, there is a force of attraction between the two, objects as well as me and my friend., , GMm, P.E. = –, R, T.E. = P.E. + K.E., , , , v2esc =, , R+h=∞, , =0+0=0, , According to the law of conservation of energy,, , *(3) , Will your weight remain constant as you go, above the surface of the earth? (Use your brain, power; Textbook Page no. 10), Ans., (i) Weight of a body depends on acceleration due to, gravity. W = m × g

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Gravitation, , 23, , (ii) As we go above the surface of the earth height, increases and hence acceleration decreases., (iii) , As acceleration due to gravity decreases, consequently weight decreases., , (iii), , *(4) , According to Newton’s law of gravitation,, earth’s gravitational force is higher on an object, of larger mass. Why doesn’t that object fall, down with higher velocity as compared to an, object with lower mass? (Use your brain power;, Textbook Page no. 12), Ans., (i) The acceleration due to gravity (g) on an object, only depends on mass (M) and radius (R) of the, , (iv), , earth i.e. g =, , GM, ., R2, , (ii) It does not depend on mass (m) of the object., (iii) The acceleration produced at a given point is the, same for all objects., (iv) Hence, object of larger mass does not fall down, with higher velocity as compared to an object, with lower mass., , The mass of m2 was reduced to 50% and the, force exerted by m1 on m2 was found to be 20 N,, what will be the force exerted by m2 on m1?, , Ans. The force exerted by m2 on m1 is 20 N., ,  hy gravitational constant is called universal, W, constant?, , Ans. , The value of gravitational constant does not, depend upon the nature and size of the bodies., It also does not depend upon the nature of the, medium between two bodies; hence it is called, universal constant., (v), ,  hat will happen to gravitational force if mass, W, of one of the objects is doubled?, , Ans. I f the mass of one of the objects is doubled, then, the gravitational force between them also gets, doubled., (vi), , What is the value of universal constant in SI?, , Ans. I n SI system, the value of ‘G’ is 6.67 × 10–11, Nm2/kg2., *(2) Observe the following diagram and answer the, questions:, , Q.3.3. Application based Questions / Questions based, , on figures:, , D, E, F, , *(1) Observe the figure and answer the following, questions: (Textbook Page no. 4), m1, , m2, , F2, , F1, d, , (i), , The force of Gravitation between two bodies, having irregular shape is taken to be the, distance between their, (a) centre of mass, (b) centre of the body, (c) Edge of the body, , Ans. (a) centre of mass, (ii), , If the distance between the two bodies is tripled,, how will the Gravitational force between them, change?, m1m2, Ans. F = G, d2, If the distance is tripled,, F' = G, , m1m2, (3d)2, , =G, , m1m2, 9d, , =, , 1, 9, , F, , The gravitational force between them becomes, 1, th the initial force., 9, , C, B, s, Sun, , A, , Fig 1.3: The orbit of a planet moving around the Sun., (a) In a given figure, area ESF is equal to ASB, then, what can you say about EF? (Use your brain power;, Textbook Page no. 4), Ans., (i), Distance EF is lesser than AB. Hence, speed will, be lesser at EF and greater at AB, (ii) Therefore, in an elliptical orbit, speed is not, constant., (b), , If x, y and z are the speeds of revolution of the, planet from A to B, C to D and E to F, then which, of the following relation is true?, (a) x > y > z (b) x < y < z (c) x < y > z (d) x > y < z, Ans. (a) x > y > z, (c) Correct and rewrite Kepler's third law., The period of revolution of a planet around the, sun is directly proportional to the cube of the, distance of the planet from the sun., Ans. The square of the period of revolution of a planet, around the sun is directly proportional to the, cube of the mean distance of the planet from the, Sun.

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Master Key Science and Technology Part-1 (Std. X), , 24, , (3), , If two objects are at a distance of d from each, others, m1, , m1, , (B) PE = ‒, , Then F =, Gm1m2, , (9), , 9d, , (A) KE = 0, (B) PE = 0, , Ans. If an artifical satellite is shifted from LEO to HEO,, the value of ‘g’ decreases. This is because as the, satellite is shifted from LEO to HEO, its distance, 1, from earth's surface increases. Because g α 2 , as, r, distance increases the value of g will decreases., , m2, , 2, , GMm, R, , At infinity, , (8) An artificial satellite is shifted from LEO to, HEO, how will the value of ‘g’ vary?, , 3d, , (4), , On Earth’s surface, 1, (A) KE = mv2, 2, , m2, , d, Gm1m2, Then F =, d2, If the distance between two bodies is 3d, , Ans., , Ans., , A, , B, , Ans. From Delhi to Moscow, the value of g will increase., As Delhi lies close to equator and Moscow is close, to pole, the value of g from equator to pole goes, on increasing., , F2, , F1, , How will the value of ‘g’ change if a person, travels from Delhi to Moscow?, , (10), C, AF1 + AF2 =, , m1, , =, , (5) , Which of the following best represent force, between Sun and Earth., S, S, S, E, E, E, (B), , (C), , Ans. B. Force is equally exerted by 2 bodies., (6) An elephant and a matchbox fall from a height, of 200m. If they are in a state of free fall, which, of them will reach the ground first and why?, Ans. Both will reach the ground at the same time., Since they are in a state of free fall, there is no, resistive force acting on them and the only force, acting is the force of gravity. The acceleration due, to gravity is same for all the bodies irrespective of, the mass of the body., (7) Complete the following for an object of mass 'm', On Earth’s surface, 1, (A) KE = mv2, 2, (B) PE = .............., , F1, , F2, 2d, , Ans. BF1 + BF2 ; CF1 + CF2, , (A), , A, , At infinity, , (A) KE = .............., (B) PE = 0, , m2, , m1, , B, F1, , F2, , m2, , d/3, , In which of the two cases, A or B, is the force, exerted stronger?, m1m2, Ans. F = G, d2, m1m2 1, Case A : F1 = G, = F, 4d2, 4, m1m2, Case B : F2 = G, , , (11), , d2 = 9 F, 9, The force exerted is maximum in case B., , If a traveler in a spacecraft orbiting the Earth, releases an object from his hand, it remains, stationary and appears to be in a state of, weightlessness. Does this mean there is no, force of gravity acting on the object?, , Ans., (i), No, there is a force of gravity acting on the body., Though the spacecraft is at a height from the, surface of the earth, the value of g there is not, zero, but only 11% less than its value on the, surface of the earth., (ii) , Thus, the weightlessness is not caused by the, height of the satellite but by being in a state of free, fall. Though the spacecraft is not falling on the

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Gravitation, , 25, , earth because of its velocity along the orbit, the, only force acting on it is the gravitational force of, the earth and therefore it is in a state of free fall., (iii) As the velocity of free fall does not depend on, the properties of an object, it is the same for the, spacecraft, the travellers and the objects in the, craft., (iv) Thus, if a traveller releases an object from his, hand, it will remain stationary with respect to, him and will appear to be weightless., (12), , (i) Will the velocity of a stone thrown vertically, upwards remain constant or will it change with, time? How will it change?, , (ii), , Why doesn't the stone move up all the time?, Why does it fall down after reaching a certain, height?, , (iii), , What does its maximum height depend on?, , Ans., (i) The velocity of the stone thrown vertically, upward will not remain constant. It will decrease, with time till it becomes zero., (ii) When the stone is thrown vertically upwards,, the gravitational force tries to pull it down, and reduces its velocity. Due to this constant, downward pull, the velocity becomes zero after, a while. The pull continues to be exerted and the, stone starts moving vertically downward towards, the centre of the earth under its influence., (iii) The maximum height the stone can achieve, depends on the initial velocity with which the, stone is thrown vertically upwards., Q.3.4. Complete the flow chart:, , Earth, , , , Low tide, , Ans., , Low tide, High, tide, , Earth, , Low tide, , High, tide, , Moon's, Gravitational, force, High, tide, , Q.4.1. Paragraph based Questions:, (1) Read the paragraph and answer the following, questions:, (1), ,  ou must be knowing about the high and low, Y, tides that occur regularly in the sea. The level, of sea water at any given location along sea, shore increases and decreases twice a day at, regular intervals. High and low tides occur at, different times at different places. The level, of water in the sea changes because of the, gravitational force exerted by the moon. Water, directly under the moon gets pulled towards, the moon and the level of water there goes up, causing high tide at that place. At two places, on the earth at 90° from the place of high tide,, the level of water is minimum and low tides, occur there., , (i), ,  ow many times does the sea level at the coast, H, change?, , Ans. , Sea level at the coast increases and decreases two, times a day. Two high tides and two low tides., (ii), , How does sea level get changed?, , Ans. T,  he sea-level at the coastal region gets changed, two times a day due to gravitational attraction of, the Moon., (iii), , Where is high tide and low tide caused?, , Ans. Due to gravitational attraction of the Moon, the, water mass swells towards the direction of the, Moon causing high-tide. The water mass shrinks, at 90° from this place which causes low tide., (2), ,  he centre of mass of an object is the point inside, T, or outside the object at which the total mass of, the object can be assumed to be concentrated., The centre of mass of a spherical object having, uniform density is at its geometrical centre. The, centre of mass of any object having uniform, density is at its centroid., , (i), ,  here can the total mass of an object be, W, assumed to be concentrated?, , Moon, , Ans. , At the centre of the mass of object., (ii), Moon, ,  here is the centre of mass located for an object, W, of uniform density?, , Ans. At its centroid., (iii), ,  here is the centre of mass located for a, W, spherical object?, , Ans. , At its geometrical centre.

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Master Key Science and Technology Part-1 (Std. X), , 26, , (iv), , What was the basis of Kepler’s laws?, , Ans. K,  epler’s laws were based on the motion of the, planets., (3), , (4), , Space travellers as well as objects in the, spacecraft appear to be floating. Why does this, happen? Though the spacecraft is at a height, from the surface of the earth, the value of g, there is not zero. In the space station the value, of g is only 11% less than its value on the surface, of the earth. Thus, the height of a spacecraft is, not the reason for their weightlessness., , Gravitational waves:, Waves are created on the surface of water when, we drop a stone into it. Similarly you must have, seen the waves generated on a string when both, its ends are held in hand and it is shaken. Light, is also a type of wave called the electromagnetic, wave. Gamma rays, X-rays, ultraviolet rays,, infrared rays, microwave and radio waves are, all different types of electromagnetic waves., Astronomical objects emit these waves and, we receive them using our instruments. All, our knowledge about the universe has been, obtained through these waves., Gravitational waves are a very different type, of waves. They have been called the waves, on the fabric of space-time. Insteine predicted, their existence in 1916. These waves are very, weak and it is very difficult to detect them., Scientists have constructed extremely sensitive, instruments to detect the gravitational waves, emitted by astronomical sources. Among these,, LIGO (Laser Interferometric Gravitational, Wave Observatory) is the prominent one., Exactly after hundred years of their prediction,, scientists detected these waves coming from, an astronomical source. Indian scientists have, contributed significantly in this discovery. This, discovery has opened a new path to obtain, information about the Universe., , (i), ,  hat are the different types of electro magnetic, W, waves?, Ans. , Gamma rays, X-rays, Ultraviolet rays, infrared, rays, microwave and radio waves., (ii), ,  hat are the waves on the fabric of space - time, W, called?, Ans. , They are called gravitational waves., (iii) Is it easy to detect Gravitational waves?, Ans. , No, these waves are very weak and it is very, difficult to detect them., (iv), ,  hat is the device, used to detect Gravitational, W, waves?, Ans. , LIGO (Laser Interferometric Gravitational waves, Observatory) is used to detect Gravitational, waves., , Weightlessness in space:, , Their weightlessness is caused by their being, in the state of free fall. Though the spacecraft is, not falling on the earth because of its velocity, along the orbit, the only force acting on it is the, gravitational force of the earth and therefore, it is in a free fall state. As the velocity of free, fall does not depend on the properties of an, object, the velocity of free fall is the same for, the spacecraft, the travellers and the objects in, the craft. Thus, if a traveller releases an object, from her hand, it will remain stationary with, respect to her and will appear to be weightless., (i), , Is the value of g zero in the space station?, , Ans. , No, the value of g is only 11% less than its value, on the surface of the Earth., (ii), , Why is weightlessness caused in a spacecraft?, , Ans. , The weightlessness is caused by them being in a, state of free fall., (iii), ,  hy doesn't the spacecraft fall towards the, W, Earth?, , Ans. , The spacecraft does not fall towards the Earth, because of its velocity along the orbit., (iv), , I f a traveller releases an object from her hand in, the spacecraft, what will happen?, , Ans. , The object will remain stationary with respect to, her, because, the velocity of free fall is the same for, the spacecraft, traveller and objects in the craft., Q.4.2. Activity based Questions:, (1) , Read the following paragraph carefully and, answer the following:, (1) Tie a stone to one end of a string. Take the other, end in your hand and rotate the string so that the, stone moves along a circle., As long as we are holding the string, we are, pulling the stone towards us i.e. towards the, centre of the circle and are applying a force, towards it. The force stops acting on it if we

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Gravitation, , 27, , release the string. In, this case, the stone will, fly off along a straight, line which is the, tangent to the circle at, the position of the, stone when the string, (a), is released, because, that is the direction of, its velocity at that, instant of time. You, may recall a similar, activity in which a 5, rupee coin kept on a, rotating circular disc, (b), flies off the disc along, the tangent to the disc. Fig 1.4: A stone tied to, Thus, a force acts on a string, moving along a, any object moving circular path and its velocity, in tangential direction, along a circle and it is, directed towards the centre of the circle. This is, called the Centripetal force., (i), ,  he impressed force on the stone is in which, T, direction?, , in a free fall. In free fall, the initial velocity of, the object is zero and goes on increasing due to, acceleration due to gravity of the earth. During, free fall, the frictional force due to air opposes the, motion of the object and a buoyant force also acts, on the object. Thus, true free fall is possible only, in vacuum. For a freely falling object, the velocity, on reaching the earth and the time taken for it, can be calculated by using Newton’s equations, of motion. For free fall, the initial velocity u = 0, and the acceleration a = g. Thus, we can write the, equations as, 1, v = gt; s = gt2; v2 = 2gs, 2, For calculating the motion of an object thrown, upwards, acceleration is negative, i.e. in a, direction opposite to the velocity and is taken, to be – g. The magnitude of g is the same but, the velocity of the object decreases due to -ve, acceleration., The moon and the artificial satellites are moving, only under the influence of the gravitational field, of the earth. Thus they are in free fall., , , (Try this; Textbook Page no. 11), , Ans. , On the stone, the force is directed towards the, centre of the circle., , (i), ,  hich force acts on the stone when held in the, W, hand?, , (ii), , Ans. Gravitational force in downward direction and, reaction force of hand in upward direction., , What happens if the string is released?, , Ans. T,  he stone will fly off in a straight line along the, tangent to the circle., (iii), , (ii), , What is centripetal force?, ,  hich force acts on the stone in free fall after, W, you release it?, , Ans. The force exerted towards the centre of the circle, on an object moving along a circular path is called, centripetal force. It is a centre seeking force., , Ans. Only gravitational force acts on it when it is, released., , (2) Take a small stone. Hold it in your hand., , Ans. When an object falls towards the Earth under the, influence of the Earth’s gravity alone it is called, free fall., , We know that the force of gravity due to the earth, acts on each and every object. When we were, holding the stone in our hand, the stone was, experiencing this force, but it was balanced by a, force that we were applying on it in the opposite, direction. As a result, the stone remained at rest., Once we release the stone from our hands, the, only force that acts on it is the gravitational force, of the earth and the stone falls down under its, influence. Whenever an object moves under the, influence of the force of gravity alone, it is said, to be falling freely. Thus the released stone is, , (iii), , (iv), , What is free fall?, ,  hat is the initial velocity and what is the effect, W, of gravitational acceleration on the object in, free fall?, , Ans. In free fall, the initial velocity of an object is zero, and as the time progresses, velocity increases due, to uniform gravitational acceleration., (v), , Write kinematic equations used in free fall?, 1 2, Ans. The kinematic equations are, v = gt, s =, gt and, 2, v2 = 2gs, , vvv

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Master Key Science and Technology Part-1 (Std. X), , 28, , ASSIGNMENT - 1, Time : 1 Hr., , (1), , Marks : 20, , Q.1. (A) Answer the following questions:, Fill in the blanks:, , , , (i), , , , (ii) Mass is also a measure of, , (2), , , (3), , Value of g at the centre of Earth is, , ., of an object., , Complete the analogy:, At poles : 9.83 m/s2 : : At equator :, , ., , Q.1. (B) Choose and write the correct option: , , (2), , (1), , The gravitational force of attraction between two objects is given by, m1m2, m1m2, m1m2, d2, (a) F α, (b), (c), (d) F α, F, α, F, α, 2, 2, d, d3, mm, d, , (2), ,  he gravitational force on the surface of the Moon is, T, Earth., , 1, , (a) five, , , 2, , (b) one fifth, , ., , times than that on the surface of the, , (c) one sixth, , (d) six, , Q.2. Answer the following: (Any 2), , (4), , (1), , Mahendra and Virat are sitting at a distance of 1 metre from each other. Their masses are 75 kg and, 80 kg respectively. What is the gravitational force between them?, , (2), , Define: (i), , (3), , Distinguish between: Weight and Mass, , , (1), , Centre of mass, , (ii), , Free fall, , Q.3. Answer the following: (Any 2), , (6), , (2), , If a person weighs 750 N on Earth, how much would be his weight on the Moon given that Moon’s, 1, 1, mass is, of that of the Earth and its radius is, of that of Earth., 81, 3.7, Define: The Universal law of gravitation and derive mathematically., , (3), , Observe the figure and answer the following questions:, , , , m1, , F2, , F1, , m2, , d, , , , , (i), , Write gravitational force between them., , , , (ii) What will happen to gravitational force if mass of one of the objects is doubled?, , , , (iii) What is the value of universal constant in SI system?, , , , Q.4. Answer the following: (Any 1), , (5), , (1), , Write the three laws given by Kepler. How did they help Newton to arrive at the inverse square law of, gravity?, , (2), , Define: Escape velocity and derive mathematically., , vvv

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Fig 2.1 Modern Periodic Table, , Atomic number of the element, Symbol of the element, Name of the element, Atomic mass of the element, , 2, Periodic Classification of Elements, , (29)

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Master Key Science and Technology Part-1 (Std. X), , 30, , CONCEPT MAP, Elements are arranged with, increasing Atomic mass, , Classification of elements, , Group of three, elements having, similar chemical, properties., Dobereiner’s, Triad, , upto Ca, out of 56, elements, , Limitations, led to, , Newlands', Octave, , 63, elements, , Modified, further, , Elements are arranged with, increasing Atomic number, Elements are arranged 118 elements, with increasing, Atomic mass, , Mendeleev’s, Periodic Table, , Research, led to, , Main, features, , Main, features, , Main, features, , Atomic mass of, middle element is, the arithmetic mean, of atomic mass of, the other two., , Every eighth, element had, properties similar, to that of the first, element., , Elements are, arranged in, vertical groups and, horizontal periods., , Modern, Periodic Table, , Main, features, Elements are, arranged on the, basis of electronic, configuration into 4, blocks., , Points to Remember:, zz Today 118 elements are known to the scientific world., zz In the year 1817, a German scientist Dobereiner suggested that properties of elements are related to their, , atomic masses., , zz Group of three elements having similar chemical properties are known as Triads., zz In the year 1866, English scientist John Newlands arranged the elements known at that time in an increasing, , order of their atomic masses., , zz Newlands’ Octaves or Law of Octaves: When the elements are arranged in an increasing order of their, , atomic masses, every eighth element had properties similar to those of the first., , zz The Russian scientist Dmitri Mendeleev developed the periodic table of elements during the period 1869 to, , 1872 A.D., , zz He arranged 63 elements known at that time in an increasing order of their atomic masses., zz Mendeleev’s Periodic Law: Properties of elements are periodic function of their atomic masses., zz In 1913 A.D., an English scientist Henry Moseley revealed through experiments done using x-ray tube that, , atomic number (Z) is a more fundamental property of an element than its atomic mass., , zz Modern Periodic Law: Properties of elements are a periodic function of their atomic numbers., zz Modern Periodic table contains seven horizontal rows called the periods 1 to 7 and the eighteen vertical, , columns are the groups 1 to 18., , zz Entire Modern Periodic table is divided into four blocks, viz, s-block, p-block, d-block and f-block., zz Electronic configuration of an element: The distribution of electrons in the various orbits around the nucleus, , of an atom of an element is called its electronic configuration., , zz It is the electronic configuration of an element which decides the group and the period in which the element, , is to be placed.

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Periodic Classification of Elements, , 31, , zz Valency: Combining capacity of an element., zz Valence electrons: Electrons present in the outermost shell of an atom., zz Atomic radius is the distance between the nucleus of the atom and its outermost shell. It is expressed in the, , unit picometer (pm). (1 pm = 10–12 m), , zz Atomic radius goes on decreasing while going from left to right within a period and goes on increasing while, , going down (from top to bottom) in a group., , zz Metal atoms have a tendency to form a cation by losing their valence electrons. This property is called, , electropositivity of an element., , zz An atom of a non-metal has a tendency to form an anion by accepting electrons from outside into its valence, , shell. This property is called electronegativity., , MASTER KEY QUESTION SET - 2, Q.1. (A) 1. Fill in the blanks and rewrite the complete, statement:, (1), ,  he arrangement of elements in a group of three, T, is known as, ., , (2), ,  he law used by Newlands’ to arrange elements, T, is known as, ., , (3), , Mendeleev arranged, at that time., , (4), ,  he vertical columns in the periodic table are, T, called, , while the horizontal rows are, called, ., , (5), ,  he element eka-boron in Mendeleev’s periodic, T, table is known as, in the Modern, Periodic table., , elements known, , (6), ,  here was no place for, T, original periodic table., , in Mendeleev’s, , (7), ,  endeleev created the, M, group without, disturbing the original periodic table., , (8), , In the Modern Periodic table, the elements are, arranged in accordance with their, ., , (9), ,  wo rows shown separately at the bottom of the, T, Modern Periodic table are called, and, ., , (10), , The, , contains the group 1 and 2 elements., , (11), , The d-block elements are called, , (12), ,  lements showing properties of both metals and, E, non-metals are called, ., , (13), , Newlands, , , (14), , In the family of alkali metals, the number of, valence electrons is, ., , (15), ,  here are seven electrons in the outermost shell, T, of the elements, such elements belong to the, family of, ., , ., , could arrange elements upto, out of total 56 elements known., , (16), ,  s we go down a group, the number of shells, A, goes on, ., , (17), , I n the Modern Periodic table, the elements placed, at the bottom of the periodic table are called as, elements., , (18), , Atomic radius is expressed in the unit, , (19), , Inert gases belong to, , (20), ,  lements with atomic numbers 90 to 103 in the, E, Modern Periodic table are called, , (21), ,  he valency of an element is determined by the, T, number of, electrons present in the, outermost shell of an atom., , (22), , From left to right, the atomic size of an atom, ., , (23), , An element is placed in 2nd period, so it has, shells., , (24), , Valency across a period, , (25), ,  n atom is said to be a non-metal if it, A, electrons., , (26), , group., , gradually., , is the only element in duplet state., , (27), , Elements of group 1A are called, , (28), , Group II A elements are called as, , (29), , Group VII A elements are called as, , (30), , First period contains, , (31), , Moseley, , , (32), , The number of shells, , (33), ,  he formula of chloride of a metal is MCl2, the, T, metal M belongs to, group., , (34), , , group contains all gases at room, temperature., , arranged, ., , ., ., ., , elements., elements, , according, , to, , down the group., , Ans. , (1) Triads (2) Newlands’ Law of Octaves (3) 63, (4) groups, periods (5) Scandium (6) noble gases, (7) zero (8) atomic numbers (9) Lanthanide series,, Actinide series (10) s-block (11) transition elements

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Master Key Science and Technology Part-1 (Std. X), , 32, , (12) metalloids (13) Calcium (Ca) (14) 1 (one), (15) halogens (16) increasing (17) inner transition, (18) picometer (19) zero (20) actinides (21) valence, (22) decreases (23) 2 (24) varies (25) gains or shares, (26) Helium (27) alkali metals (28) alkaline earth, metals (29) halogens (30) 2 (31) atomic number, (32) increases (33) 2 (II A) (34) 18 or zero, Q.1. (A) 2. Find the odd word out., (1), , Lithium, Beryllium, Boron, Chlorine, , Ans. Chlorine: It belongs to the third period while, others are second period elements., (2), , Helium, Radon, Argon, Boron, , Ans. Boron: It is a metal while others are inert or noble, gases., (3), , Chlorine, Bromine, Iodine, Oxygen, , Ans. Oxygen: It is a gas belonging to group 16 while, others are halogens belonging to group 17, (4), , Sodium, Lithium, Copper, Beryllium, , Ans. , Copper: Others are normal elements, while, copper is a transition element., (5), , Dalton, Dobereiner, Moseley, Newlands, , Ans. Dalton: He put forth the Dalton’s theory for atomic, structure, whereas all others classified elements., (6), , Boron, Silicon, Potassium, Antimony, , Ans. Potassium: It is a metal while others are metalloids., (7), , Aluminium, Argon, Xenon, Sodium, , Ans. , Sodium: It belongs to s-block, whereas others, belong to p-block., (8), , Boron, Silicon, Neon, Polonium, , Ans. Neon: It is an inert gas, whereas all others are, metalloids., (9), , Lithium, Magnesium, Sodium, Potassium, , Ans. Magnesium: It belongs to group II A (group 2),, whereas others belong to group I A (group 1)., (10), , Chlorine, Bromine, Arsenic, Fluorine, , Ans. Arsenic: It is a metalloid and others are halogens., Q.1. (A) 3. Complete the analogy., (1), , Dobereiner: Triad :: Newlands :, , Ans. Octaves - Doberiner arranged three elements in a, triad whereas Newlands arranged the elements, in increasing order of their atomic masses. He, found that first element and the eighth element, had similar properties and called it law of octaves., (2), ,  endeleev’s Periodic Table : Atomic mass ::, M, Modern Periodic table :, , Ans. Atomic number - Atomic mass is the basis for, , Mendeleev’s periodic table whereas the atomic, number is the basis for Modern Periodic Table., (3), , Hydrogen : First period :: Lithium :, , Ans. Second period - The first period contains Hydrogen, (1) and Helium (2) Lithium’s atomic number is 3,, therefore, it is put in the second period., (4), , Fluorine : 2, 7 :: Chlorine :, , Ans. 2, 8, 7 - Chlorine’s atomic number is 17 therefore, its electronic configuration is (2, 8, 7), (5), ,  roup 1: Alkali metals ::, G, earth metals, , : Alkaline, , Ans. Group 2 - The outermost shell of these elements, has 2 electrons, therefore, they are put in group 2., (6), Transition elements : d-block :: Inner transition, elements :, Ans. f-block - The d-block elements have 2 incomplete, outermost shells whereas the f-block elements, have 3 outermost shells which are incomplete., (7), Tellurium :, :: Radium : Metal, Ans. Metalloid - Tellurium has properties of both metals, and non metals, therefore, it is called a metalloid., (8), Transition elements :, :: Inner transition, elements : Three incomplete outermost shells, Ans. Two incomplete outermost shells - The d-block, elements have 2 incomplete outer most shells, whereas the f-block elements have 3 outermost, shells which are incomplete., (9), Lanthanides : Ce to Lu :: Actinides:, Ans. Th to Lr - Ce to Lu are called Lathanides and Th, to Lr are called Actinides., (10) Ca : Alkaline earth metal :: Cs :, Ans. Alkali metal - Cs has one electron in its outermost, shell and Ca has 2 electrons in its outermost shell., (11) Fe : Electropositive :: Cl :, Ans. Electronegative - Fe has a tendency to form, a cation by losing its valence electrons and it, is called electropositive property whereas Cl, has a tendency to gain electron and it is called, electronegative property., (12) (Li, Na, K) :, :: (F, Cl, Br) : Group 17, Ans. Group 1 - Li, Na, K have one electron in their, outermost shell whereas F, Cl, Br have 7 electrons, in their outermost shell., (13) Valency of Na (2,8,1) : 1 (One) :: Valency of P(2,8,5), :, Ans. 3 (Three) - Na loses 1 electron to get the, configuration of He (2b) and P requires 3 electrons, to get a configuration 2, 8, 8. (Ar). Therefore, the, valency of P is 3.

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Periodic Classification of Elements, , 33, , Q.1. (A) 4. Match the columns:, *(1), Column A, , Column B, , Column C, , (i), , Triad, , (a), , Lightest and negatively charged particle in all the atoms., , (1), , Mendeleev, , (ii), , Octave, , (b), , Concentrated mass and positive charge, , (2), , Thomson, , (iii) Atomic number (c), , Average of the first and the third atomic mass, , (3), , Newlands, , (iv) Period, , (d), , Properties of the eighth element similar to the first, , (4), , Rutherford, , (v), , Nucleus, , (e), , Positive charge on the nucleus, , (5), , Dobereiner, , (vi) Electron, , (f), , Sequential change in molecular formulae, , (6), , Moseley, , Ans. (i – c – 5), (ii – d – 3), (iii – e – 6), (iv – f – 1), (v – b – 4), (vi – a – 2), , (1), , Sodium, , (a), , Non-metal, , Q.1. (A) 5. State whether the following statement, is ‘True‘ or ‘False’. If false; write the correct, statement for the same., , (2), , Sulphur, , (b), , Lanthanide, , (1), , (3), , Manganese, , (c), , Metal, ,  ithium, Potassium and Sodium are elements, L, forming Dobereiner’s triad., , (4), , Cerium, , (d), , Transition metal, , (2), , ccording to Mendeleev’s periodic law, the, A, properties of elements are a periodic function of, their atomic numbers., , (3), , Periods are the horizontal rows of elements., , (4), , Group 17 elements are known as Noble gases., , (5), , Silicon is a metalloid., , (6), ,  s we move from left to right in a period, A, of Modern Periodic table, atomic size of the, elements gradually increases., , (7), ,  roup 1 elements in Modern Periodic table are, G, referred as “alkali metals”., , (8), , Argon is not an inert gas., , (9), , he d-block elements are called transition, T, elements., , (10), ,  pto Calcium, the law of octaves was found to, U, be applicable., , (11), ,  etals are electronegative and non-metals are, M, electropositive., , (12), ,  ka-aluminium was discovered and named as, E, Gallium (Ga)., , (13), ,  tomic number is the number of protons or, A, electrons present in the nucleus., , (14), , Eka boron is known as Germanium., , (15), , Tellurium, Polonium are metalloids., , (2), , Column A, , Column B, , Ans. (1 – c), (2 – a), (3 – d), (4 – b), (3), , Column A, , Column B, , (1), , Alkali metals, , (a), , Valency 4, , (2), , Alkaline earth, metals, , (b), , Valency 0, , (3), , Argon, , (c), , Divalent, , (4), , Carbon, , (d), , Monovalent, , Ans. (1 – d), (2 – c), (3 – b), (4 – a), (4), , Column A, (1), , Column B, , Atomic size in a, period from left, to right, , (a), , Atomic mass, , (2), , Atomic size in a, group from top, to bottom, , (b), , Increases, , (3), , Modern, Periodic Law, , (c), , Atomic number, , (4), , Mendeleev’s, Periodic Law, , (d), , Decreases, , Ans. (1 – d), (2 – b), (3 – c), (4 – a), (5), , Column A, , Column B, , (16), , Isotopes have same atomic masses., , (1), , Alkali metals, , (a), , Group 2, , (17), , Isotopes have similar chemical properties., , (2), , Alkaline earth metals, , (b), , Group 1, , (18), , La stands for Lutetium., , (3), , Halogens, , (c), , Group 18, , (19), , f-block elements are metalloids., , (4), , Noble gases, , (d), , Group 17, , (20), , I n an atom of an element, extra-nuclear electrons, take part in the chemical reaction., , Ans. (1 – b), (2 – a), (3 – d), (4 – c)

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Master Key Science and Technology Part-1 (Std. X), , 34, , Ans. , (1) False. Lithium, Sodium and Potassium are elements, forming Dobereiner’s triad. (2) False. According to, Mendeleev’s periodic law, the properties of elements, are a periodic function of their atomic masses., (3) True (4) False. Group 17 elements are known as, Halogens. (5) True (6) False. As we move from left, to right in a period of the Modern Periodic table,, atomic size of the elements gradually decreases., (7) True (8) False. Argon is an inert gas, belongs to, group 18. (9) True (10) True (11) False. Metals are, electropositive and non-metals are electronegative., (12) True (13) True (14) Eka-boron is known as, Scandium. (15) True (16) False. Isotopes are atoms, of the same element having same atomic number, but different atomic masses. (17) True (18) False. La, stands for Lanthanum. (19) False. f-block elements are, metals. (20) True, Q.1. (A) 6. Answer the following., *(1) Write the name and symbol of the elements from, the description., (i), , The atom having the smallest size., , Ans. , Helium (He)., (ii) The atom having the smallest atomic mass., Ans. , Hydrogen (H)., (iii) The most electronegative atom., Ans. , Fluorine (F)., (iv) The noble gas with the smallest atomic radius., Ans. , Helium (He)., (v) The most reactive nonmetal., Ans. , Fluorine (F)., *(2), , Write the names from the description., , (i), The period with electrons in the shell, K, L and M., Ans. , Period 3., (ii) The group with valency zero., Ans. , Group 18, (iii) The family of non-metals having valency one., Ans. , Halogens., (iv) The family of metals having valency one., Ans. , Alkali metals., (v) The family of metals having valency two., Ans. , Alkaline Earth metals., (vi) The metalloids in the second and third periods., Ans. , Boron and Silicon., (vii) Non-metals in the third period., Ans. , Phosphorus, Sulphur, Chlorine and Argon., (viii) Two elements having valency 4., Ans. , Carbon, Silicon., , Q.1. (A) 7. Name the following:, *(1), ,  hree elements having a single electron in their, T, outermost shell., , Ans. (i) Lithium (Li) (2,1) (ii) Sodium (Na) (2, 8 1), (iii) Potassium (K) (2,8,8,1)., *(2), , Three elements with filled outermost shell., , Ans. (i) Helium (2) (ii) Neon (2,8) (iii) Krypton (2,8,18,8)., *(3), , hree elements having 7 electrons in their, T, outermost shell., , Ans. ( i) Fluorine (2,7) (ii) Chlorine (2,8,7) (iii) Bromine, (2,8,18,7)., (4), Three elements which are metalloids., Ans. (i) Boron (B) (ii) Silicon (Si) (iii) Germanium (Ge)., (5), , hree alkaline earth metals with electronic, T, configuration., Ans. , (i) Beryllium (Be) (2,2) (ii) Magnesium (Mg), (2,8,2) (iii) Calcium (Ca) (2,8,8,2)., (6), Two pairs of Dobereiner’s triad., Ans. , (i) Lithium (Li), Sodium (Na), Potassium (K), (ii) Chlorine (Cl), Bromine (Br), Iodine (I)., (7), ,  he scientist who classified elements on the basis, T, of atomic mass., , Ans. Dmitri Mendeleev., (8), ,  he scientist who classified elements on the basis, T, of atomic number., , Ans. Henry Moseley., (9), , he 3 elements which were predicted by, T, Mendeleev for which he left blank spaces in the, periodic table., , Ans. Eka-Boron, Eka-Silicon, Eka-Aluminium., (10), ,  eries of 14 elements placed below the periodic, S, table having atomic no. from 58 to 71 [Ce to Lu], , Ans. Lanthanides., (11), ,  eries of 14 elements placed below the periodic, S, table having atomic no. from 90 to 103 [Th to Lr], , Ans. Actinides., Q.1. (B) Choose and write the correct options:, *(1), ,  he number of electrons in the outermost shell of, T, alkali metals is, (a) 1, , *(2), , (b) 2, , (c) 3, , (d) 7, ,  lkaline earth metals have valency 2. This means, A, that their position in the Modern Periodic table, is in, (a) Group 2, (c) Period 2, , (b) Group 16, (d) d-block

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Periodic Classification of Elements, , *(3), ,  olecular formula of the chloride of an element, M, X is XCl. This compound is a solid having high, melting point. Which of the following elements, be present in the same group as X., (a) Na, , *(4), , (b) Mg, , (c) Al, , (b) p-block, (d) f-block, , , halogens., , (b) Ca, Sr, Ba, (d) Cl, Br, I, , resembles alkali metals as well as, , (a) Lithium, (c) Hydrogen, (7), , First period consists of, (a) 1, , (8), , (b) 8, , elements., (c) 2, , (d) 4, , Electronic configuration of Mg is, (a) (2, 8, 4), , (b) (2, 8, 1), , (c) (2, 8, 3), , (d) (2, 8, 2), , (9), , (10), , (b) Sodium, (d) Silicon, , is in liquid state at room temperature., (a) Fluorine, , (b) Chlorine, , (c) Bromine, , (d) Iodine, , Atomic radius of Lithium is, (a) 41, , (b) 151, , (c) 152, , pm., (d) 157, , Ans. , (1) (a) 1 (2) (a) Group 2 (3) (a) Na (4)(b) p-block, (5) (c) Mg, Ca, Sr (6) (c) Hydrogen (7) (c) 2, (8) (d) (2, 8, 2) (9) (c) Bromine (10) (c) 152, Q.2.1. Solve the following:, (1), ,  , Y and Z are the elements of a Dobereiner’s, X, Triad. If the atomic mass of X is 7 and that of Z, is 39, what should be the atomic mass of Y?, Ans. Atomic mass of X = 7, Atomic mass of Z = 39, ∴, , , According to Dobereiner’s Triad, atomic mass of Y, =, , X+Z, 2, , =, , 7 + 39, 2, , =, , 46, 2, , = 23, , ∴ Atomic mass of Y = 23, (2), ,  wo elements X and Y have atomic number, T, 12 and 16 respectively. Write the electronic, configuration for these elements. To which, period of the Modern Periodic table do these, two elements belong? [HOTS], , Element, , Atomic, number, , Electronic, configuration, , X, , Z = 12, , 2, 8, 2, , Y, , Z = 16, , 2, 8, 6, , Both these elements belong to third period., *(3), , , triad does not follow Dobereiner’s, law of triad., (a) Li, Na, K, (c) Mg, Ca, Sr, , (6), , Ans., , (d) Si, , In which block of the Modern Periodic table are, the non-metals present?, (a) s-block, (c) d-block, , (5), , 35, , I dentify Dobereiner’s triads from the following, groups of elements having similar chemical, properties? [Can you tell?; Textbook Page 16], (i) Mg (24.3), Ca (40.1), Sr (87.6), (ii) S (32.1), Se (79.0), Te (127.6), (iii) Be (9.0), Mg (24.3), Ca (40.1), , Ans., (i), (a) Mg (24.3), (b) Ca (40.1), (c) Sr (87.6), a+c, 111.90, ∴ mean of, =, = 55.95, 2, 2, which is not approximately equal to the atomic, mass of middle element (Ca). Therefore, it does, not form Dobereiner’s triads., (ii), , (a) S (32.1), (b) Se (79.0), (c) Te (127.6), a+c, 159.70, ∴ mean of, =, = 79.85, 2, 2, which is approximately equal to atomic mass of, Se. Hence, it forms Dobereiner’s triads., (iii), , (a) Be (9.0), (b) Mg (24.3), (c) Ca (40.1), a+c, 49.10, ∴ mean of, =, = 24.55, 2, 2, which is approximately equal to atomic mass of, Mg. Hence, it forms Dobereiner’s triads., Q.2.2. State the laws / Define:, (1), Newlands’ Law of Octaves:, Ans. , Newlands’ Law of Octaves states that “When, the elements are arranged in an increasing order, of their atomic masses, the properties of every, eighth element are similar to those of the first.”, (2), Mendeleev’s Periodic law:, Ans. Mendeleev’s Periodic law states that “properties, of elements are periodic function of their atomic, masses.”, (3), Modern Periodic Law:, Ans. Modern Periodic law states that “properties of, elements are periodic function of their atomic, numbers.”, (4), Dobereiner’s Triad, Ans. When elements with similar chemical properties, are arranged in increasing order of their atomic, mass in group of three called as triads, the atomic, mass of the middle element is approximately, equal to the mean of the atomic masses of the, other two elements.

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Master Key Science and Technology Part-1 (Std. X), , 36, , (5), Periods, Ans. The Horizontal rows of elements in the periodic, table are called periods., (6), , Groups, , Ans. The Vertical columns of elements in the periodic, table are called groups., (7), , Normal Elements, , Ans. , The elements in which only last shell is, incompletely filled are called as normal elements., (8), Lanthanide Series, Ans. The series of fourteen elements from Cerium [Ce, (58)] to Lutetium [Lu (71)] is called Lanthanides, Series., (9), Actinide Series, Ans. The series of Fourteen elements from Thorium, [Th (90)] to Lawrencium [Lr (103)] is called, Actinide Series., (10) Transition elements, Ans. , The elements in which last two shells are, incompletely filled are called Transition elements., They belong to d-block of Modern Periodic Table., (11) Inner Transition elements, Ans. The f-block elements are placed at the bottom of, the periodic table and they have last three shells, incompletely filled. These elements are called, Inner Transition Elements., (12) Atomic radius, Ans. For an isolated atom, atomic radius is the distance, between the centre of the nucleus of an atom to, its outermost shell., (13) Valency, Ans. The number of electrons accepted, donated or, shared by an atom so as to complete the octet in, its outermost shell is called valency., , OR, , Valency is the combining capacity of an element., (14) Inert elements, Ans. , Elements present in group 18 on the extreme, right of the Modern Periodic table are called Inert, elements or Noble gases. They have all shells, including the outermost shell completely filled., (15) Isotopes, Ans. Different atoms of the same element having, same atomic number but different atomic mass, number are called Isotopes., (16) Metalloids, Ans. , Elements which show the properties of both, metals and non-metals are called as metalloids., e.g. Silicon (Si), Arsenic (As), Tellurium (Te)., , Q.2.3. Answer the following questions in one or two, sentences:, (1), What is meant by periodicity?, Ans. , The repetition of same properties after certain, interval is called periodicity., (2), ,  ow will the tendency to gain electrons change, H, as we go from left to right across a period? Why?, Ans. , Tendency to gain electrons increases from left, to right in a period because atomic size goes on, decreasing and effective nuclear charge increases., (3), ,  hy was Dobereiner’s classification of elements, W, not useful?, Ans. , Dobereiner’s Triad was not useful for classification, of elements because he could identify only some, triads from the known elements as other triads, did not obey Dobereiner’s rule., (4), , I n the periodic table where are the metalloids, placed?, Ans. , The metalloids are elements which lie along the, border of the zig zag line which separates metals, placed on the left side from non-metals placed on, the right side., (5), ,  lements of which group are called as alkali, E, metals?, Ans. , Elements of group 1 are called alkali metals., (6), ,  hich is the incomplete period in the Modern, W, Periodic table?, Ans. , Since the Modern Periodic table is complete there, is no incomplete period., (7), ,  hich law was modified into Modern Periodic, W, law?, Ans. , Mendeleev’s Periodic law was modified into, Modern Periodic law., (8), What are periodic properties?, Ans. , The properties which show gradual variation in a, group and in a period and they repeat themselves, after a certain interval of atomic number are, called periodic properties., (9), What are normal elements?, Ans. , Elements of group 1, 2 and 13 to 17 are called, normal elements. These elements have only the, outermost shell incompletely filled., (10), ,  n what basis is Potassium (Z = 19) placed in, O, 4th period and group 1?, Ans. , The electronic configuration of potassium is, (2, 8, 8, 1). It has four shells so it belongs to the, period 4 . The period 4 has elements with the 4th, shell being filled. Potassium has 1 electron in its, valence shell. Hence, it is placed in group 1. Group, 1 has elements with 1 electron in the valence shell.

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Periodic Classification of Elements, , 37, , (11), ,  ow will you differentiate between metals and, H, non-metals by the number of valence electrons?, Ans. E,  lements which have 1, 2 or 3 valence electrons, are metals and those which have 4, 5, 6 or 7, valence electrons are non-metals., *(12) W,  hat are the types of matter? (Can you recall?;, Textbook Page 16), Ans. T,  here are three types of matters viz elements,, compounds and mixtures., *(13) W,  hat are the types of elements? (Can you recall?;, Textbook Page 16), Ans. T,  he three types of elements are metals, nonmetals and metalloids., *(14) W,  hat are the smallest particles of matter called?, (Can you recall?; Textbook Page 16), Ans. T,  he smallest particles of matter are called atoms, and molecules., *(15) W,  hat is the difference between the molecules, of elements and compounds? (Can you recall?;, Textbook Page 16), Ans. M,  olecules of elements contain atoms of the same, , element where as molecules of compounds contain, atoms of two or more different elements., *(16) W,  hat are the values of ‘n’ for the shells K, L and, M? (Can you recall?; Textbook Page 24), Ans. K,  -1, L-2, M-3, *(17) W,  hat is the maximum number of electrons, that can be accommodated in a shell? Write the, formula. (Can you recall?; Textbook Page 24), Ans., , Shell, K, L, M, N, , 2n2, 2 × 12 = 2, 2 × 22 = 8, 2 × 32 = 18, 2 × 42 = 32, , Electron Capacity, 2, 8, 18, 32, , *(18) D,  educe the maximum electron capacity of the, shells K, L and M? (Can you recall?; Textbook, Page 24), Ans. K - 2, (2n2 = 2 × 12 = 2), L-8, , (2n2 = 2 × 22 = 2 × 4 = 8), , M - 18, , (2n2 = 2 × 32 = 2 × 9 = 18), , *(19) T,  here are some vacant places in Mendeleev’s Periodic table. In some of these places, the atomic masses are, seen to be predicted. Enlist three of these predicted atomic masses along with their group and period. (Think, about it; Textbook Page 19), Ans., , Predicted atomic mass, 44, 68, 72, , Group, III, III, IV, , *(20) D,  ue to uncertainty in the names of some of the, elements, a question mark is indicated before, the symbol in the Mendeleev’s periodic table., What are such symbols? (Think about it; Textbook, Page 19), Mendeleev arranged the elements in the, Ans. , increasing order of atomic masses. In some places, the atomic mass of elements would not be in right, order to put them in the correct group. Tellurium, is heavier than Iodine, but he put Iodine after, Tellurium because Iodine has similar properties, like F, Cl, Br. He placed the elements according, to their properties and put a question mark to, indicate that the atomic mass needs to be checked., for e.g. ?Yt = 88 in Mendeleev’s periodic table was, corrected as Yt = 88.906 in Modern Periodic table., Q.2.4. Write short notes on:, *(1), , Mendeleev’s Periodic law:, , Ans., (i), ,  he Russian scientist Dmitri Mendeleev developed, T, the periodic table of elements during the period, 1869 to 1872 A.D., , Period, 4, 5, 5, , Predicted Name, Eka boron, Eka aluminium, Eka silicon, , Present Names, Scandium, Gallium, Germanium, , (ii), ,  endeleev’s periodic table is the most important, M, step in the classification of elements., , (iii), ,  endeleev considered the fundamental property, M, of elements, namely the atomic mass as standard, and arranged 63 elements known at that time in, an increasing order of their atomic masses., , (iv), ,  hen, he transferred this into the periodic table, T, of elements in accordance with the physical and, chemical properties of these elements., , (v), ,  e organized the periodic table on the basis of the, H, chemical and physical properties of the elements., , (vi), ,  hese were the molecular formulae of hydrides, T, and oxides of the elements, melting points,, boiling points and densities of the elements and, their hydrides and oxides., , (vii) M,  endeleev found that the elements with similar, physical and chemical properties repeat after a, definite interval., (viii) O,  n the basis of this finding, Mendeleev, stated the, following periodic law “Properties of elements, are periodic function of their atomic masses”.

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Master Key Science and Technology Part-1 (Std. X), , 38, , *(2), , Structure of the Modern Periodic table., , (iv), , (i), , he Modern Periodic table contains seven, T, horizontal rows called the periods 1 to 7., ,  ased on chemical properties, they were to be, B, placed in same group, while based on atomic, masses they were to be placed in different groups., , (v), , Hence, there was no definite place for isotopes., , (ii), , imilarly, eighteen vertical columns are the, S, groups 1 to 18., , (vi), , (iii), , he arrangement of the periods and groups, T, results into formation of boxes., , his anomaly arose in Mendeleev’s periodic, T, table because the elements were arranged in an, increasing order of their atomic masses., , (iv), ,  tomic numbers are serially indicated in the, A, upper part of these boxes., , (v), , Each box corresponds to the place for one element., , (vi), ,  part from these seven rows, two separate rows, A, are placed at the bottom of the periodic table., , Ans., , (vii) I n 1913 A.D., Henry Moseley revealed that atomic, number is a more fundamental property of an, element than its atomic mass., (viii) H,  ence, he modified Mendeleev’s Periodic Table, and arranged the elements in an increasing order, of their atomic number in the Modern Periodic, Table., , (vii) T,  hese are called Lanthanide series and Actinide, series respectively., , (ix), ,  y doing so, the anomaly of Mendeleev’s Periodic, B, Table regarding isotopes was removed., , (viii) T,  here are 118 boxes in the periodic table including, the two series., , (x), , herefore, in the Modern Periodic Table the, T, isotopes could be placed in the same group, and the radioactive isotopes were placed in, two separate series below at the bottom of the, periodic table., Halogens or group 17 elements., , (ix), , I t means that there are 118 places for elements in, the Modern Periodic table., , (x), ,  he Modern Periodic table is now completely filled,, T, means all the 118 elements are now discovered., , (4), , (xi), ,  he entire periodic table is divided into four, T, blocks, viz, s-block, p-block, d-block and f-block., , (i), ,  he second last column in the periodic table is, T, group 17 (VII – A) which contains halogens., , (ii), ,  he members of this group from top to bottom, T, are as given in the following table:, , (xii) The s-block contains the groups 1 and 2 elements., (xiii) The groups 13 to 18 constitute the p-block elements., (xiv) T, he groups 3 to 12 constitute the d-block, elements. The d-block elements are called, transition elements., (xv) T,  he lanthanide and actinide series at the bottom, form the f-block elements., (xvi) A,  zig-zag line can be drawn in the p-block of the, periodic table., (xvii) T,  he metalloid elements lie along the border of, the zig-zag line., , Ans., , Group Symbol Atomic, Electronic Valency, 17, Number configuration, elements, (Z), Fluorine F, , 9, , 2, 7, , 1, , Chlorine Cl, , 17, , 2, 8, 7, , 1, , Bromine Br, , 35, , 2, 8, 18, 7, , 1, , Iodine, , 53, , 2, 8, 18, 18, 7, , 1, , 85, , 2, 8, 18, 32, 18, 1, 7, , I, , (xviii) All the metals lie on the left side of the zig-zag, line, while all the non-metals lie on the right side., , Astatine At, , *(3) Position of isotopes in the Mendeleev’s and the, Modern Periodic table., Ans., (i), Different atoms of the same element having same, atomic number but different atomic mass number, are called isotopes., , (iii), ,  he valence shell of a halogen contains seven, T, electrons., , (iv), , Thus, it needs one electron to complete its octet., , (v), , (ii), , I sotopes were discovered long time after, Mendeleev put forth the periodic table., ,  he halogen complete their octet and attain the, T, stable inert gas configuration by gaining one, electron., , (vi), , (iii), ,  s isotopes have the same chemical properties but, A, different atomic masses, a challenge was posed in, placing them in Mendeleev’s periodic table., ,  ence, the valency of halogens is one, i.e. they, H, are monovalent., , (vii) A,  ll of them have general formula X2. Fluorine (F2), and Chlorine (Cl2) are gases. Bromine (Br2) is a, liquid, while Iodine (I2) is a solid.

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Periodic Classification of Elements, , (5), , 39, , Transition Elements., , (7), Metallic and Non-metallic properties, Ans., (i), Metals show tendency to lose electrons. Therefore,, they are said to be electropositive., , Ans., (i), ,  lements present in groups 3 to 12 in the middle, E, of the periodic table are called transition elements., , (ii), , Their last two outermost shells are incomplete., , (iii), , They belong to d-block of Modern Periodic table., , (iv), ,  ll these elements are solid metals at room, A, temperature (except Mercury and Gallium)., , (6), , Inner-Transition Elements., , Ans., (i),  lements placed in two separate rows at the, E, bottom of the periodic table are called inner, transition elements., (ii), ,  hey include two series of elements: Lanthanide, T, series and Actinide series. They belong to f-block, elements of Modern Periodic table., , (iii), , 1 4 elements with atomic number 58 to 71, (Cerium to Lutetium) are called lanthanides., These elements are placed along with lanthanum, (La = 57) in group 3 and period 6 because of very, close resemblance in properties between them., , (iv), , 1 4 elements with atomic number 90 to 103, (Thorium to Lawrencium) are called actinides., These elements are placed along with actinium, (Ac = 89) in group 3 and period 7 because of very, close resemblance in their properties., , (v), , They have last three shells incompletely filled., , (vi), , All these elements are metals., , (ii), ,  on-metals show the tendency to accept electrons, N, or share electrons with another atoms. Therefore,, they are said to be electronegative., , (iii), ,  cross the period from left to right, metallic, A, character decreases and non-metallic character, increases., , (iv), ,  his is because the atomic size decreases due to, T, greater nuclear pull because of which valence, electrons cannot be easily removed., , (v), ,  own the group-metallic character increases, D, and non-metallic character decreases from top to, bottom in a group., , (vi), ,  s atomic number increases from top to bottom, A, in a group, newer shells get added which in, turn increases its atomic size. Thus nuclear pull, decreases due to addition of newer shells and, thus the valence electrons present in outermost, shell can be easily removed from the atom., , (vii) T,  he three types of elements can be easily shown, in the Modern Periodic table with the help of zigzag line., (viii) T,  he metalloid elements lie along the border of, the zig-zag line. All the metals lie on the left side, of zig-zag line while all the non-metals lie on the, right side., , Q.2.5. Distinguish between:, (1) Mendeleev’s Periodic table and Modern Periodic table., Ans., , Mendeleev’s Periodic Table, , Modern Periodic Table, , (i), , This table is based on the atomic mass of the (i), elements., , This table is based on the atomic number of the, elements., , (ii), , Elements are arranged in increasing order of (ii), their atomic mass., , Elements are arranged in increasing order of their, atomic number., , (iii) Mendeelev’s Periodic table consists of only, 63 elements., , (iii) Modern Periodic table consists of 118 elements., , (iv) Mendeelev’s Periodic table is based on, Mendeleev’s Periodic Law, ‘Properties of, element are a periodic function of their, atomic masses.’, , (iv) Modern Periodic table is based on Modern Periodic, Law,‘Properties of elements are periodic function, of their atomic number.’, , (v), , (v), , There was no space for inert gases and, radioactive isotopes., , The table contains all elements including inert, gases and radioactive isotopes.

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Master Key Science and Technology Part-1 (Std. X), , 40, , (vi) Some elements having similar properties, were placed in different groups. Some, elements in different groups were found to, have similar properties., , (vi) All elements placed in the same group have similar, properties., , (vii) Elements could not be divided into four, blocks., , (vii) Elements are divided into four blocks – i.e. s, p, d, and f blocks., , (2) Transition elements and Inner transition elements, Ans., , , , Transition elements, , (i), , Elements placed in groups 3 to 12 in the, middle of the Modern Periodic table are, called transition elements., , (i), , (ii), , These elements have last two incompletely, filled outermost shells., , (ii) These elements have last three incompletely filled, outermost shells., , (iii) They belong to the d-block in the Modern, Periodic table., (3), , Inner transition elements, Elements placed in two separate rows at the, bottom of the Modern Periodic table are called, inner transition elements., , (iii) They belong to the f-block in the Modern Periodic, table., , Inert gases and Normal Elements, , Ans., , Inert gases, , Normal Elements, , (i), , In the atoms of inert gases all the shells are, completely filled including the outermost, shell., , (i), , (ii), , They are stable and chemically inert (i.e., chemically non-reactive), , (ii) They are unstable and are chemically reactive., , (iii) They are included in the p-block of the, Modern Periodic table., , In the atoms of normal elements, all shells are, completely filled except the outermost shell., , (iii) They are included in the s-block as well as p-block, of the Modern Periodic table., , (iv) They are placed in zero group (Group 18) of (iv) These elements are placed in groups 1, 2 and 13 to, the Modern Periodic table., 17 of the Modern Periodic table., (4), , Groups and Periods of Modern Periodic Table, , Ans., , Groups, (i), , Periods, , Vertical columns of elements in the Modern, Periodic table are called Groups., , (i), , Horizontal rows of elements in the Modern Periodic, table are called periods., , (ii) There are 18 groups in the Modern Periodic, table., , (ii), , There are 7 periods in the Modern Periodic table., , (iii) The group number indicates the valence, electrons in the outermost shell., , (iii) The period number indicates the number of shells in, the atoms of an element., , (iv) The elements in the same group have similar (iv) The elements in a period exhibit gradual change in, chemical properties., properties from left to right., (5), , Metallic character and Non-metallic character, , Ans., , Metallic character, , Non-metallic character, , (i), , It is the tendency of an atom to lose electrons. (i), , It is the tendency of an atom to accept or share electrons., , (ii), , Metallic character decreases from left to right, in a period., , Non-metallic character increases from left to right in, a period., , (ii)

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Periodic Classification of Elements, , (6), , (iii) In a group, metallic character increases from, top to bottom., , (iii) In a group, non-metallic character decreases from, top to bottom., , (iv) Elements having metallic character are said, to be electropositive., , (iv) Elements having non-metallic character are said to, be electronegative., , s-block elements and p-block elements, , Ans., , s-block elements, , p-block elements, , (i), , Elements in group 1 and 2 are called s-block, elements including hydrogen., , (ii), , They contain 1 or 2 electrons in the outermost (ii) They contain 3 to 8 electrons in the outermost shell., shell., , (iii) They are all metals (except hydrogen)., (7), , 41, , (i), , Elements in group 13 to 17 and zero group are, called p-block elements., , (iii) They are metals, non-metals and metalloids., , Alkali metals and Alkaline earth metals, , Ans., , Alkali metals, , Alkaline earth metals, , (i), , In the Modern Periodic Table, elements in, Group 1 are alkali metals., , (i), , (ii), , Atoms of these elements have one electron, in the valence shell., , (ii) Atoms of these elements have two electrons in their, valence shell., , (iii) They are monovalent., , In the Modern Periodic Table, elements in Group 2, are alkaline earth metals., , (iii) They are divalent., , (iv) Their oxides and hydroxides dissolve readily (iv) Their oxides and hydroxides dissolve slightly in, in water., water., Q.2.6. Give Scientific Reasons:, *(1) Atomic radius goes on decreasing while going, from left to right in a period., Ans., (i), Atomic radius is the distance between the centre, of the atom and its outmost shell., (ii), , Atomic radius decreases across a period., , (iii), ,  his is because valence electrons are being added, T, to the same outermost shell or to the same energy, level because of which the electrons added, experience greater pull from the nucleus., , (iv), , his results in the electrons moving closer to, T, the nucleus. Hence, the atomic radius goes on, decreasing while going from left to right in a period., , *(2) Metallic character goes on decreasing while, going from left to right in a period., Ans., (i), Metallic character is the tendency of an atom to, lose electrons., (ii), , (iii), , I n a period, electrons are added to the same shell, and hence, these electrons experience greater, pull from the nucleus. As a result outermost shell, comes closer to nucleus, atomic size decreases.,  hus it becomes difficult to remove an electron, T, from the outermost shell of an atom. Hence,, , metallic character decreases from left to right in, a period., *(3) Atomic radius goes on increasing down a group., Ans., (i), Atomic radius is the distance between the centre, of the atom to its outermost shell., (ii), ,  ew shells are added to the atoms of the elements, N, as we go down from top to bottom in a group., , (iii), ,  he outermost electrons go further and further, T, away from the nucleus, extending the radius and, ultimately increase in the size of the atom. Hence,, atomic radius goes on increasing down a group., , *(4) Elements belonging to the same group have the, same valency., Ans., (i), Valency is the number of electrons donated,, , accepted or shared by the atoms of an element so, as to complete the octet in the outermost orbit., (ii), ,  alency is dependent upon the number of valence, V, electrons present in the outermost shell of an atom, called valence electrons., , (iii), ,  lements in the same group have same number, E, of valence electrons, irrespective of the number, of orbits. Hence, elements in the same group have, the same valency.

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Master Key Science and Technology Part-1 (Std. X), , 42, , *(5) The third period contains only eight elements, even though the electron capacity of the third, shell is 18., Ans., (i), Period 3 of the periodic table consists of elements, whose atom has three shells occupied by electrons., (ii), , (iii), ,  s per the electron holding capacity of shells,, A, 2 elements are present in the first period, and, 8 elements in the second period.,  he third period also contain only eight elements, T, due to the law of electron octet and also electrons, will fill orbitals in increasing order of energy, i.e., the lowest energy orbital first., , (6) Atomic number is a more fundamental property, of an element than its atomic mass., Ans., (i), The atomic number of an element indicates the, number of protons in the nucleus or the number, of electrons in the atom., (ii), ,  ll the atoms of an element have the same atomic, A, number., , (iii), ,  he number of electrons present in the outermost, T, shell of an atom is responsible for the formation, of compounds either by sharing, accepting or, donating electrons., , (iv), ,  he chemical properties of an element are decided, T, by its atomic numbers. Hence, atomic number is a, more fundamental property of an element than its, atomic mass., , (7), Alkali metals are placed in Group 1., Ans., (i), The group number indicates the number of, , valence electrons present in the outermost shell of, an atom of an element., (ii), , ll alkali metals have one electron in their, A, outermost shell. Hence, they are placed in Group 1., , (8), Inert gases exist in the form of free atoms., Ans., (i), Formation of molecules is based on the electronic, configuration of the combining elements., (ii), , In the case of inert gases, all the shells, including, the outermost shell are completely filled., , (iii), ,  hey have electronic configuration with complete, T, duplet or complete octet., , (iv), ,  ue to their stable electronic configuration,, D, atoms of these elements do not lose, gain or share, electrons with other atoms. Hence, inert gases, exist in the form of free atoms., , *Q.3.1. Answer the following:, (1) An element has its electronic configuration as, 2, 8, 2. Now answer the following questions:, (i), What is the atomic number of this element?, Ans. Electronic configuration of this element is (2, 8, 2), ∴ The atomic number of this element is 12., (ii), , What is the group of this element?, , Ans. A,  tomic number of element is 12 and its electronic, configuration is (2, 8, 2). As it has 2 valence, electrons, this element belongs to group 2 and its, name is Magnesium (Mg)., (iii), , To which period does this element belong?, , Ans. , Atomic number of this element is 12 and its, electronic configuration is (2, 8, 2). It has 3, shells. The number of shells indicates the period, number. Hence it belongs to period 3., (iv), ,  ith which of the following elements would, W, this element resemble? (Atomic numbers are, given in the brackets) N (7), Be (4), Ar (18), Cl (17), , Ans. , The atomic number of the given element is 12, and this element is Magnesium which belongs to, group 2 and Period 3., Electronic configuration of Mg (12) = (2, 8, 2), Electronic configuration of N (7) = (2, 5), Electronic configuration of Be (4) = (2, 2), Electronic configuration of Ar (18) = (2, 8, 8), Electronic configuration of Cl (17) = (2, 8, 7), , Number of valence electrons in this element, (Magnesium) is 2. Similarly the number of, valence electron in Be is 2. As they have same, number of valence electrons they belong to same, group i.e. group 2 and will resemble in some of, their properties., (2) Write down the electronic configuration of, the following elements from the given atomic, numbers. Answer the following questions with, explanation., (i), 3Li, 14Si, 2He, 11Na, 15P Which of these elements, belong to period 3?, Ans., Name of the, element, , Atomic, Number, , Lithium (Li), 3, Silicon (Si), 14, Helium (He), 2, Sodium (Na), 11, Phosphorous (P) 15, , Electronic, Configuration, K, L, M, 2, 1, 2, 8, 4, 2, 2, 8, 1, 2, 8, 5

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Periodic Classification of Elements, , 43, ,  s the elements Silicon (Si), Sodium (Na) and, A, Phosphorous (P) have electrons in the three shells;, K, L and M, therefore these elements belong to, period 3 as the number of shells indicates the, period number., (ii), , 1H, 7N, 20Ca, 16S, 4Be, 18Ar. Which of these, elements belong to the second group?, , Ans., Name, of the, element, , Symbol, , Atomic, Number, , Electronic, Configuration, K, , L, , M, , Hydrogen H, , 1, , 1, , Nitrogen, , N, , 7, , 2, , 5, , Calcium, , Ca, , 20, , 2, , 8, , 8, , Sulphur, , S, , 16, , 2, , 8, , 6, , 4, , 2, , 2, , 18, , 2, , 8, , Beryllium Be, Argon, , Ar, , N, , 2, , 8, , From the above electronic configurations,, elements Calcium (Ca) and Beryllium (Be) have, same number of valence electrons, i.e. 2, therefore, Calcium and Beryllium belong to group 2. We can, infer from this that the group follows the trend of, same valence electrons., (iii), , 7N, 6C, 8O, 5B, 13Al. Which is the most, electronegative element among these?, , Ans., , (iv), Ans., , 4Be, 6C, 8O, 5B, 13Al. Which is the most, electropositive element among these?, , Name of the, element, , Electronic, Atomic Configuration, Symbol, Number, K, L, M, , Beryllium, , Be, , 4, , 2, , 2, , Carbon, , C, , 6, , 2, , 4, , Oxygen, , O, , 8, , 2, , 6, , Boron, , B, , 5, , 2, , 3, , Aluminium, , Al, , 13, , 2, , 8, , 3, , Tendency of atoms of an element to lose electrons, and form positive ion, cation, is known as, electropositivity. These elements lose valence, electrons to form cations having a stable noble, gas configuration. It is the metallic character, of that element. Electropositivity decreases as, we move from left to right in a period. Among, the elements given, Beryllium, Carbon, Oxygen, and Boron belong to the second period. As, Beryllium is the left most element, it will show, high electropositive characteristic. Therefore,, Beryllium is the most electropositive element., (v), Ans., , 11Na, 15P, 17Cl, 14Si, 12Mg. Which of these has the, largest atom?, , Name of the, element, , Electronic, Atomic Configuration, Symbol, Number, K, L, M, , Electronic, Atomic Configuration, Symbol, Number, K, L, M, , Sodium, , Na, , 11, , 2, , 8, , 1, , Nitrogen, , N, , 7, , 2, , 5, , Phosphorous, , P, , 15, , 2, , 8, , 5, , Carbon, , C, , 6, , 2, , 4, , Chlorine, , Cl, , 17, , 2, , 8, , 7, , Oxygen, , O, , 8, , 2, , 6, , Silicon, , Si, , 14, , 2, , 8, , 4, , Boron, , B, , 5, , 2, , 3, , Magnesium, , Mg, , 12, , 2, , 8, , 2, , Aluminium, , Al, , 13, , 2, , 8, , Name of the, element, , 3, ,  mong the elements given, Nitrogen, Carbon,, A, Oxygen and Boron belong to the second period., As we know in a period from left to right, the, electronegativity increases. Electronegativity, means the tendency of an element to form anion, by accepting the electrons. Therefore, Oxygen, is the most electronegative element, since it is, present at the extreme right amongst the given, elements of period 2., , Atomic size or radius is the distance between the, nucleus of an atom and its outermost shell. Atomic, size gradually decreases from left to right across, a period. All the above elements belong to period, 3, having electrons in the three shells. Atomic, size decreases across a period, because valence, electrons get added to the same outermost shell., Electrons added experience greater pull from the, nucleus. Due to the increased nuclear charge,, the electrons are pulled towards the nucleus to, a greater extent and thereby, the size of the atom, decreases.

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Master Key Science and Technology Part-1 (Std. X), , 44, , , In case of sodium, third shell is having only, 1 electron. There is less force of attraction for, the valence electron towards the nucleus and, therefore, size of the atom is larger as compared, to remaining elements. Hence, Na is the largest, atom., (vi), Ans., , 19K, 3Li, 11Na, 4Be. Which of these atoms has, smallest atomic radius?, , Name, of the, element, , decreases, and it becomes difficult to remove an, electron from the atom. As in Sodium (2,8,1), only, one electron is present in the outermost orbit. So,, it easily loses this one electron because of weaker, nuclear attraction, hence, it possesses highest, metallic character., (viii) 6C, 3Li, 9F, 7N, 8O. Which of the above elements, has the highest non-metallic character?, Ans., , Atomic, Symbol, Number, , Potassium K, , Electronic, Configuration, K, , L, , 19, , 2, , 8, , Lithium, , Li, , 3, , 2, , 1, , Sodium, , Na, , 11, , 2, , 8, , Beryllium Be, , 4, , 2, , 2, , M, 8, , N, 1, , 1, , , From the above elements, potassium, lithium, and sodium belong to group 1. As we know,, atomic radius increases down a group or from, top to bottom. This is because while going down, a group, a new shell is added. Therefore, the, distance between the outermost electron and, the nucleus goes on increasing. Hence, Element, Lithium in group 1 and Beryllium in group 2, have the smallest atomic radius as they have, only two shells. In period: compared to Lithium,, Beryllium has the smallest radius., (vii) 13Al, 14Si, 11Na, 12Mg, 16S. Which of the above, elements has the highest metallic character?, Ans., Name of the, element, , Name of the, element, , Electronic, Atomic Configuration, Symbol, Number, K, L, M, , Carbon, , C, , 6, , 2, , 4, , Lithium, , Li, , 3, , 2, , 1, , Fluorine, , F, , 9, , 2, , 7, , Nitrogen, , N, , 7, , 2, , 5, , Oxygen, , O, , 8, , 2, , 6, , All the elements given above belong to second, period. While going from left to right in a period,, non-metallic character increases. The tendency of, an element to form anion or the electronegativity, is the non-metallic character of an element. As, Fluorine is present in extreme right of period 2,, it will also possess highest non-metallic character, and also Fluorine has 7 electrons in its outermost, shell, that means 1 electron is less to complete its, octet., (3) The following table shows the position of six, elements A, B, C, D, E and F in the periodic, table. (Activity Based), Groups /, Periods, , 1, A, , Aluminium, , Al, , 13, , 2, , 8, , 3, , 2, , Silicon, , Si, , 14, , 2, , 8, , 4, , 3, , Sodium, , Na, , 11, , 2, , 8, , 1, , Magnessium, , Mg, , 12, , 2, , 8, , 2, , Sulphur, , S, , 16, , 2, , 8, , 6, , All the elements given belong to period 3. Metallic, character of an element can be defined as how, readily an atom can lose electron. From left to, right across a period, metallic character decreases., This is because, electrons are added to the same, shell, and hence, these electrons experience, greater pull from the nucleus, thus atomic size, , Electronic, Atomic Configuration, Symbol, Number, K, L, , 2, , 3, to, 12, , 13, , 14, , 15, , 16, , 17, , B, D, , E, , 18, C, F, ,  sing the above table, answer the following, U, questions:, (i), , Which element is a metal with valency 2?, , Ans. D, (ii), , Which element is a non-metal with valency 3?, , Ans. B, (iii), ,  ut of D and E, which one has a bigger atomic, O, radius and why?, , Ans. , D, because the atomic size decreases along a, period from left to right.

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Periodic Classification of Elements, , (iv), , 45, ,  rite a common name for the family of elements, W, C and F?, , Ans. Noble gases, *(1) Chlorine has two isotopes, viz, Cl - 35 and, Cl - 37. Their atomic masses are 35 and 37. Their, chemical properties are same. Where should, these be placed in Mendeleev’s periodic table?, In different places or in the same place? (Use, your brain power; Textbook Page 19), Ans. , Mendeleev arranged the elements in increasing, order of atomic mass. Since the atomic masses of, the isotopes of chlorine are 35 and 37, they should, be kept in different positions in the Mendeleev, Periodic table., (2) Write the molecular formulae of oxides of, the following elements by referring to the, Mendeleev’s periodic table. (Use your brain, power; Textbook Page 20), (i), , Element, Na, , Na2O, , (ii), , Ca, , CaO, , (iii), , Rb, , Rb2O, , (iv), , Ba, , BaO, , (v), , Sn, , SnO2, , (vi), , Si, , SiO2, , (vii), , C, , CO2, , (viii) P, , P2O5, , (ix), , ClO, ClO2, , Cl, , Valency = 3, *(5), , Q.3.2. Explain the following, , Ans., , Electronic configuration = (2, 8, 3), , Oxides, ,  hat is the relationship between the electronic, W, configuration of an element and its valency?, (Think about it; Textbook Page 24), Ans. , From electronic configuration, we come to know, the number of valence electrons, i.e. the number, of electrons in the outermost shell. The valency of, an element is determined from valence electrons., That is, how many electrons it has to lose or gain, or share to complete its outermost orbit., *(6), ,  he atomic number of beryllium is 4 while, T, that of oxygen is 8. Write down the electronic, configuration of the two and deduce their, valency from the same. (Think about it; Textbook, Page 24), Ans. , Atomic number of Beryllium = 4, Electronic configuration, , = (2, 2), , Atomic number of Oxygen, , = 8, , Electronic configuration, , = (2, 6), , Valency of Beryllium is 2,, Valency of Oxygen is = 8 ‒ 6 = 2, Q.3.3. Answer the questions based on Modern, Periodic table., *(1), , (i), , *(3) Write the molecular formula of the compounds, of the following elements with hydrogen., (C, S, Br, As, F, O, N, Cl) (Use your brain power;, Textbook Page 20), Ans. Elements Compound Elements Compound, formed, formed, with H, with H, C, , CH4, , S, , H2S, , Br, , HBr, , As, , AsH3, , F, , HF, , O, , H2O, , N, , NH3, , Cl, , HCl, , (4) A metal M forms an oxide having the formula, M2O3. It belongs to 3rd period in the Modern, Periodic table. Write the atomic number and, valency of the metal., Ans. Atomic number = 13, , Use your brain power; Textbook Page 25, Element, , O, , B, , C, , N, , Be, , Li, , Atomic, radius (pm), , 66, , 88, , 77, , 74, , 111 152, ,  y referring to the Modern Periodic table find, B, out the period to which the above elements, belong?, , Ans. , Elements O, B, C, N, Be, Li belong to period 2., (ii), ,  rrange the above elements in a decreasing, A, order of their atomic radii., Ans. , The above elements in the decreasing order of, their atomic radii are as follows:, Elements, , (iii), , Li, , Be, , B, , C, , N, , O, , Atomic radii 152 111, , 88, , 77, , 74, , 66, ,  oes this arrangement match with the pattern of, D, the second period of the Modern Periodic table., Ans. , Yes. As we move from left to right within a period, the atomic number increases one by one, meaning, the positive charge on the nucleus increases by, one unit at a time, but the electrons are added, to the same orbit, thereby, increasing the pull, towards the nucleus which decreases the size of, the atom.

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Master Key Science and Technology Part-1 (Std. X), , 46, , (iv), ,  hich of the above elements have the biggest, W, and the smallest atom?, Ans. , Li has the biggest atom and O has the smallest, atom., (v), ,  hat is the periodic trend observed in the, W, variation of atomic radius while going from left, to right within a period?, Ans. , Atomic radius decreases from left to right within, a period., *(2), , Use your brain power; Textbook Page 25, Element, Atomic radius, , K, 231, , N, 186, , Rb, 244, , Cs, 262, , Li, 152, , (i), ,  y referring to the Modern Periodic table find, B, out the group to which above elements belong?, Ans. , Above elements belong to group 1., (ii), ,  rrange the above elements vertically, A, downwards in an increasing order of atomic, radii., Ans. , The above elements vertically downwards in an, increasing order of atomic radii are as follows:, Ans., , Li, Na, K, Rb, Cs, , 152, 186, 231, 244, 262, , *(4), (i), , Use your brain power; Textbook Page 27., What is the cause of non - metallic character of, element?, Ans. , When any element shows tendency to accept or, share electrons and forms a negative charge, they, are said to be electronegative. Non-metals are, electronegative., (ii), ,  hat is the expected trend in the variation of, W, non-metallic character of element from left to, right in a period?, Ans. , Non-metallic character increases from left to, right in a period., (iii), ,  hat would be the expected trend in the, W, variation of non-metallic character of elements, down a group?, Ans. , Non-metallic character decreases as we move, down a group., *(5), , Can you tell; Textbook Page 22, , (i), ,  o through the Modern Periodic table (fig 2.1), G, and write the names one below the other of the, elements of group 1., , Ans. , H - Hydrogen, Li - Lithium, Na - Sodium, K - Potassium, , (iii), ,  oes this arrangement match with the pattern, D, of the group 1 of the Modern Periodic table?, Ans. , Yes. As we go down a group number of shells, increases, therefore the atomic radii also increases., (iv), ,  hich of the above elements have the biggest, W, and the smallest atom?, Ans. , 'Cs' has the biggest atom and 'Li' has the smallest, atom., (v), ,  hat is the periodic trend observed in the, W, variation of atomic radii down a group?, Ans. , Atomic radii increases down a group, as number, of shells are added one at a time., , Rb - Rubidium, Cs - Caesium, Fr - Francium, are the elements of group 1., (ii), Ans., , Use your brain power; Textbook Page 26, Classify the elements of the third period into, metals and non - metals., Ans. Elements of third period are:, Mg, Metals, , Al, , Si, Metalloid, , P, , S, , Cl, , Ar, , Non - Metals, , (ii), ,  n which side of the period are the metals? Left, O, or Right?, Ans. Left., (iii), ,  n which side of the period did you find the, O, Non - metals?, Ans. Right., , Element Atomic No., , Electronic, configuration, K, , *(3), (i), , Na, ,  rite the electronic configuration of the first, W, four elements in this group., , (iii), , L, , M, , H, , 1, , 1, , Li, , 3, , 2, , 1, , Na, , 11, , 2, , 8, , 1, , K, , 19, , 2, , 8, , 8, , N, , 1, ,  hat similarity do you find in their electronic, W, configuration?, , Ans: The outermost shell contains one electron in all, these elements., (iv), ,  ow many valence electrons are there in each, H, of these elements?, , Ans: Number of valence electrons in each of the these, elements is 1.

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Periodic Classification of Elements, , 47, , *(6), , Can you tell; Textbook Page 23, , (i), ,  n going through the Modern Periodic table, O, (fig 2.1) it is seen that the elements Li, Be, B, C,, N, O, F and Ne belong to the period-2. Write, down the electronic configuration., , Ans., , Element, , Atomic no., , Electronic, configuration, K, , L, , Li, , 3, , 2, , 1, , Be, , 4, , 2, , 2, , B, , 5, , 2, , 3, , C, , 6, , 2, , 4, , N, , 7, , 2, , 5, , O, , 8, , 2, , 6, , F, , 9, , 2, , 7, , Ne, , 10, , 2, , 8, , M, , N, , numbers, So, the position of isotopes of elements, was decided by arranging the elements in the, ascending order of their atomic numbers., (ii), , As isotopes have the same atomic number they, do not need to be given different positions (slots)., , (3) Can there be an element with atomic mass 53 or, 54 in between the two elements, Chromium, 52 Cr and Manganese 55 Mn?, 24, 25, Ans. , It is not possible. Since their atomic numbers are, continuous, there cannot be an element between, Chromium and Manganese., (4) What do you think? Should hydrogen be placed, in the group 17 of halogens or group 1 of alkali, metals in the Modern Periodic table?, Ans. , (i), , Hydrogen fits in its current position in the, periodic table (group 1) much better than in, group 17 (though is not a perfect position). For, example Hydrogen forms oxide (H2O) just like, oxides of alkali metals (Na2O, K2O). It has the, same common charge (+ 1) as all alkali metals., , (ii), , It resembles halogens in forming diatomic, molecules H2, Cl2, F2, Br2., , *Q.3.4. Write answers with explanation:, (Use your brain power; Textbook Page 21), (1) How is the problem regarding the position of, cobalt (59Co) and nickel (59Ni) in Mendeleev’s, periodic table resolved in Modern Periodic table?, Ans., (i), According to Mendeleev’s periodic law, elements, are arranged in the order of their increasing, atomic masses., , (iii), ,  does not behave like halogens, being mainly, It, in the +1 oxidation state. It does not happen to, be like halogens, the most stable compounds of, halogens have either -1 or a very high oxidation, state like -7 or +5.), , (iv), , Hydrogen has one electron in the outermost shell, and it is placed in top of group/based on atomic, number., , (ii), , (1) Consider isotopes of oxygen 16O and 18O. Would, you be able to place them in Mendeleev’s, Periodic Table?, , (ii), , I s the number of valence electrons same for all, these elements?, Ans. No, they are not the same., (iii) Is the number of shells the same in these ?, Ans. , Yes, the number of shells are the same. Each one, of this has 2 shells., , (iii), , (iv), , When Co and Ni were put in the group on the, basis of their chemical properties it was found, that Co having higher mass of 58.9 comes first, and Nickel with slightly lower atomic mass of, 58.7 comes later., Modern Periodic table resolved this problem., According to the Modern Periodic law, elements, are arranged in their increasing order of atomic, number., , The atomic number of Cobalt is 27 comes first and, Nickel with atomic number 28 comes next even if, their atomic masses are in the reverse order., (2) How did the position of 35 Cl and 37 Cl get fixed, 17, 17, in the Modern Periodic table?, Ans., (i), Modern Periodic table is based on atomic, , , Q.3.5. Activity Based Questions:, , Ans. No, we cannot place 16O and 18O in Mendeleev’s, Periodic Table., (2) Find resemblance between hydrogen and alkali, metals by writing compounds of both with, chlorine, sulphur and oxygen., Ans. Alkali metals are : Li, Na, K, Rb, Cs, Fr., H2O, Na2O, H2S, Na2S, HCl, NaCl, , Hydrogen and alkali metals form similar, formulae with oxygen, sulphur and chlorine., (3) How are isotopes of different elements placed, in the Modern Periodic Table?, Ans. Isotopes have same atomic number, but different

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Master Key Science and Technology Part-1 (Std. X), , 48, , atomic mass. In Modern Periodic table the, elements are arranged in the increasing order, of their atomic numbers. So, all isotopes of an, element can be given the same position., , with oxygen and hydrogen. He selected, hydrogen and oxygen as they were very, reactive and formed compounds with most, elements. The formulae of the hydrides and, oxides formed were treated as one of the basic, properties of an element for its classification., He sorted out the elements with similar, properties and pinned the cards together on, a wall. He observed that most of the elements, got a place in the periodic table and were, arranged in the order of their increasing atomic, masses. It was also observed that there occurs, a periodic recurrence of elements with similar, physical and chemical properties. On this basis, Mendeleev formulated a periodic law which, states that “The properties of the elements are, the periodic function of their atomic masses”., , (4) What should be the position of Hydrogen in the, Modern Periodic Table? Why?, Ans. , In the Modern Periodic Table, elements are, arranged in the increasing order of their atomic, number, so Hydrogen must be the first element, in the Modern Periodic Table, because it has the, least atomic number., (5) How do you calculate valency of an element, from its electronic configuration?, Ans. , (i), Electronic configuration tells about the valence, electrons. If the number of valence electrons are, 1, 2 or 3, then such elements will donate 1, 2 or 3, electrons. Their valency is also 1, 2 or 3 respectively., (ii) If an element has 4 valence electrons, it shares all, 4 electrons, so the valency is 4. If the number of, valence electrons are 5, 6 or 7 electrons, then such, elements accept 3, 2 or 1 electrons respectively,, thus their valency is 3, 2 or 1., (iii) When the valence electrons are 2 (for only, one shell) or 8 then it is a stable electronic, configuration. Thus, the valency is zero., (6) What is the valency of elements with atomic, number 8, 14, 17 and 20?, Ans., , Atomic, number, , Electronic, configuration, , Valency, , 8, , (2, 6), , 2, , 14, , (2, 8, 4), , 4, , 17, , (2, 8, 7), , 1, , 20, , (2, 8, 8, 2), , 2, , Q.3.6. Paragraph Based Questions:, (1) Read the following paragraph and answer the, questions given below:, Dmitri Ivanovich Mendeleev was the first, to classify elements on the basis of their, fundamental property, the atomic mass, and, also on the similarity of chemical properties., When Mendeleev started his work, 63 elements, were known. He examined the relationship, between the atomic masses of the elements and, their physical and chemical properties. Among, chemical properties Mendeleev concentrated, on the compounds formed by the elements, , (i), , What was the basis of Mendeleev's classification?, , Ans. Mendeleev's classification was based on the, fundamental property of atomic mass and also, on the similarity of chemical properties., (ii), ,  hat type of relationship of elements was, W, examined by Mendeleev?, , Ans. Mendeleev examined the relationship between, the atomic masses of the elements and their, physical and chemical properties., (iii), ,  ow many elements were known when, H, Mendeleev started his work?, , Ans. , 63 elements were known when Mendeleev, started his work., (iv), ,  hat is meant by periodicity according to, W, Mendeleev?, Ans. , (i), Mendeleev observed that elements with similar, physical and chemical properties reoccur when, elements were arranged in the increasing order, of their atomic masses., (ii), , So according to Mendeleev, the recurrence of, elements with similar physical and chemical, properties when they are arranged in an, increasing order of their atomic masses, is called, periodicity., , (iii), , What is Mendeleev's Periodic Law?, , Ans. Mendeleev’s Periodic Law is "The properties of, elements are the periodic function of their atomic, masses”.

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Periodic Classification of Elements, , 49, , (2) Read the following paragraph and answer the, questions given below:, In the Modern Periodic table, the elements, are arranged in the increasing order of their, atomic numbers. This arrangement is based on, the Modern Periodic law, which states that the, properties of elements are a periodic function of, their atomic numbers. In the Modern Periodic, table, each column is called a group and each, row is called a period. Elements within the, same group show similarity and gradation in, properties. This is due to the same number of, electrons in the outermost shell., (i), , (ii), ,  endeleev kept vacant spaces in the periodic, M, table for elements not discovered till then., Three of these unknown elements were given, the names eka-boron, eka-aluminium and ekasilicon from the known neighbours and their, atomic masses were indicated as 44, 68 and 72,, respectively. Not only this but their properties, were also predicted. Later on these elements, were discovered and named as scandium (Sc),, gallium (Ga) and germanium (Ge) respectively., The properties of these elements matched well, with those predicted by Mendeleev., , (iii), ,  here was no place reserved for noble gases in, T, Mendeleev’s original periodic table. However,, when noble gases such as helium, neon and, argon were discovered towards the end of the, nineteenth century, Mendeleev created the ‘zero’, group without disturbing the original periodic, table in which the noble gases were fitted very, well., , State the law on which modern periodic table, is based., , Ans. , Modern periodic table is based on modern, periodic law which states that 'Properties of, elements are a periodic function of their atomic, numbers'., (ii), , so as to give them proper place in the periodic, table in accordance with their properties. For, example, the previously determined atomic mass, of beryllium, 14.09, was changed to the correct, value 9.4, and beryllium was placed before boron., ,  ow many groups are there in the modern, H, periodic table?, , Ans. There are 18 groups in Modern periodic table., (iii), ,  hat is the number of valence electrons in an, W, element of group 1 and group 18 respectively?, , Ans. The number of valence electrons in an element of, group 1 is 1 and the number of valence electrons, in an element of group 18 is 8., , (2) What are the demerits of Mendeleev’s periodic, table?, , (iv), , (i), ,  o fixed position could be given to hydrogen, N, which resembles alkali metals and halogens in, the periodic table., , (ii), , I sotopes have different atomic masses but, same chemical properties. Based on chemical, properties, they were to be placed in same group,, while based on atomic masses, they were used to, be placed in different groups. Hence, there was, no definite place for isotopes., , (iii), , In certain places, elements with higher atomic, mass were placed before elements of lower, mass e.g. Co (At. Mass 58.93) was placed before, Ni (At. Mass 58.71), , (iv), ,  ome elements with different properties were, S, placed in the same group. e.g. (Mn) Manganese, was placed with halogens which totally differ in, properties., , What is the trend in the variation of valency, while going down a group?, , Ans. All elements in a group have the same number of, electrons in the outermost shell. Therefore, down, the group, valency remains the same. Valency, of group 1 elements is 1 and that of group 2, elements is 2. The valency of group 18 elements is, zero because they are noble gases with completed, outermost shell., (v), , Which pair of elements do you think will have, similar properties?, , Ans., , (a) Sodium and Argon, (b) Sodium and Potassium, (c) Potassium and Neon, Ans. The pair of elements having similar properties are,, (b) Sodium and Potassium as they belong to the, same group i.e. group 1 of Modern periodic table., Q.4.1. Answer in detail:, (1), State the merits of Mendeleev’s Periodic Table., Ans., (i), Atomic masses of some elements were revised, , (3) Merits of Modern Periodic, Mendeleev’s periodic table., , table, , over, , Ans. , In Modern Periodic table elements were arranged, on the basis of their atomic number.

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Master Key Science and Technology Part-1 (Std. X), , 50, , (i), ,  ll isotopes of the same element have different, A, masses but same atomic number. Therefore, they, occupy same position in the Modern Periodic, table., , (ii), ,  hen the elements are arranged according to, W, their atomic numbers, the anomaly regarding, certain pairs of elements in Mendeleev’s periodic, table disappears. e.g. atomic numbers of Cobalt, and Nickel are 27 and 28 respectively. Therefore,, Cobalt will come first and then Nickel, although, its atomic mass is greater., , (iii), ,  lements are classified according to their, E, electronic configuration into different blocks., , *(4), , Think about it; Textbook Page 24, , (i) What is the relationship between the electronic, configuration of an element and its valency?, , Ans. , Electronic configuration means the arrangement, of electrons in their respective shells including, valence shell. The number of electrons in the, valence shell determines its valency., Thus, electronic configuration in a way, determines the valency of an element., (ii) The atomic number of beryllium is 4 while, that of oxygen is 8. Write down the electronic, configuration of the two and deduce their, valency from the same., Ans. , ∴, , ∴, , , Atomic number of Beryllium is 4, Its electronic configuration is 2, 2, Atomic number of oxygen is 8, Its electronic configuration is 2, 6, Valency of Beryllium is 2, , , , and valency of Oxygen is also 2., , (iii) Write the Electronic configuration and valency of first 20 elements., Ans., , Hydrogen H, , Electronic, Atomic, Configuration Valency, No., K L M N, 1, 1, 1, , Sodium Na, , Helium He, , 2, , 2, , 0, , Magnesium Mg 12, , 2, , 8, , 2, , 2, , Lithium Li, , 3, , 2, , 1, , 1, , Aluminium Al, , 13, , 2, , 8, , 3, , 3, , Beryllium Be, , 4, , 2, , 2, , 2, , Silicon Si, , 14, , 2, , 8, , 4, , 4, , Boron B, , 5, , 2, , 3, , 3, , Phosphorous P, , 15, , 2, , 8, , 5, , 3, , Carbon C, , 6, , 2, , 4, , 4, , Sulphur S, , 16, , 2, , 8, , 6, , 2, , Nitrogen N, , 7, , 2, , 5, , 3, , Chlorine Cl, , 17, , 2, , 8, , 7, , 1, , Oxygen O, , 8, , 2, , 6, , 2, , Argon Ar, , 18, , 2, , 8, , 8, , 0, , Fluorine F, , 9, , 2, , 7, , 1, , Potassium K, , 19, , 2, , 8, , 8, , 1, , 1, , Neon Ne, , 10, , 2, , 8, , 0, , Calcium Ca, , 20, , 2, , 8, , 8, , 2, , 2, , Elements/, Symbol, , (iv) What is the periodic trend in the variation of, valency while going from left to right within, a period. Explain the answer with reference to, period 2 and period 3., Ans. Period 2, Elements Li Be B C N O F Ne, Valency, 1, 2, 3, 4, 3, 2, 1, 0, Period 3, Elements, Valency, , Na, 1, , Mg Al Si, 2, 3 4, , P, 3, , S, 2, , Cl Ar, 1, 0, , , For periods 2 and 3, on going from left to right,, the valency increases from 1 to 4, then starts, decreasing upto 0., , Elements/, Symbol, , Electronic, Atomic, Configuration Valency, No., K L M N, 11, 2 8 1, 1, , (v) What is the periodic trend in the variation of, valency while going down a group? Explain the, answer with reference to group 1, group 2 and, group 18., Ans. , (i), , All the elements in a group have the same number, of electrons in the outermost shell. Therefore,, down the group, valency remains the same., , (ii), , Valency of group 1 elements is 1 and that of group, 2 elements is 2., , (iii), , The valency of group 18 elements is zero because, they are noble gases with completed outermost, shell.

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Periodic Classification of Elements, , 51, , (5) Comparative study of all the four-blocks of Modern Periodic table., Ans. Comparative study of blocks, s-block, , p-block, , d-block, , f-block, , No. of, shells, incomplete/, valence, electrons, , Outermost shell, incomplete (1 or 2, valence electrons)., , Outermost shell, incomplete except zero, group elements that have, completely filled shells, (3 to 8 valence electrons)., , Last two shells are, incomplete., , Last three shells are, incomplete., , Position, , Group 1, Group 2 and, Hydrogen, , Group 13 to Group 17 and, zero group, , Group 3 to Group 12, , Lanthanides series and, actinides series placed, separately at the bottom, of periodic table., , Includes, , All metals except, , Metals, non-metals,, , Metals., , Metals., , hydrogen., , metalloids, zero, , Transition elements, , Inner-transition, , group elements., Types of, elements, , Normal elements, , Normal and inert, elements., , elements., , (6) How could the Modern Periodic Table remove, various anomalies of Mendeleev’s table?, Ans., (i), The Modern Periodic Table was classified on, the basis of atomic number. All isotopes of the, same elements have different masses but same, atomic number. Therefore, they occupy the same, position in the Modern Periodic table., (ii), ,  hen elements are arranged according to their, W, atomic numbers, the anomaly regarding certain, pairs of elements in Mendeleev’s Periodic Table, disappears e.g. atomic number of cobalt and, nickel are 27 and 28 respectively. Therefore,, cobalt will come first and then nickel, although, atomic mass of cobalt is greater., , (iii), ,  lements are classified according to their, E, electronic configuration into different blocks., , (iv), , Hydrogen has one electron in the outermost, shell. It is placed on top of the group 1 with alkali, metals which have one electron in the outermost, shell. This is because in the Modern Periodic Table, elements are arranged on the basis of electronic, configuration., , Q.4.2., (1), , Answer the questions based on the given data., Elements, Atomic, radius (pm), , K, , Na, , Rb, , Cs, , Li, , 231, , 186, , 244, , 262, , 152, , (i), ,  o which group do the elements belong? What is, T, the family called?, Ans. The elements belong to group 1. The family is, called as alkali metals., (ii), , Arrange the above elements in an increasing, order of atomic radii. Does this arrangement, match with the pattern of the group in the above, answer?, Ans. , The above elements in an increasing order of, atomic radii are as follows., Elements, Atomic, radius (pm), , Li, , Na, , K, , Rb, , Cs, , 152, , 186, , 231, , 244, , 262, , This arrangement match with the pattern of the, group in the above answer (i.e. group 1), (iii), ,  hich of the above elements have the biggest, W, and the smallest atom?, Ans. The element with biggest atom is Caesium (Cs), and the element with smallest atom is Lithium, (Li)., (iv), , What is the periodic trend observed in the, variation of atomic radii down a group?, Ans. While going down the group the atomic radii, goes on increasing, this is because new shells, are added to the atoms of the elements as we, go down from top to bottom in a group. The, outermost electron go further and further away, from the nucleus, extending the radius and

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Master Key Science and Technology Part-1 (Std. X), , 52, , ultimately increasing the size of the atom inspite, of the increased nuclear charge., (2), , In the following table, seven elements P, Q,, R, S, T, U and V (here letters are not the usual, symbols of the elements) of the modern periodic, table with their atomic numbers are given., 3, , 4, , 5, , 6, , 7, , P, , 8, , 9, , T, , 11, , 12, , Q, , R, , 13, , 14, S, , 15, , 16, , 10, V, , 17, , 18, , U, , (i), Which of these is an inert gas? Name it., Ans. Element 'V' is an inert gas. It is Neon (Ne)., (ii) Which of these is a halogen? Name it., Ans. Element 'U' is a halogen. It is Chlorine (Cl)., (iii) Which of these are metals? Name them., Ans. Elements P, Q and R are metals. Element P is, Lithium metal. Element Q is Sodium metal., Element R is Magnesium metal., (iv), , If Q combines with U, what would be the, formula of the compound formed? If Q and U, are replaced by their respective metal what will, be the formula of the compound formed., Ans. , If Q combines with U the formula of the, compound formed is QU. If Q and U are replaced, by their respective metal and non - metal. Element, Q is sodium (Na). Element U is chlorine (Cl). The, formula of the compound formed is NaCl., (v), , Write the electronic configuration of R and T, and, the type of bond formed by their combination., Ans. Electronic configuration of : R is (2, 8, 2) and T, is (2, 6). They combine with each other forming, Ionic bond., (3), , eighteen vertical columns (1 to 18) called groups., The arrangement of the periods and groups, results into formation of boxes. Atomic numbers, are serially indicated in the upper part of these, boxes., (ii) How are blocks indicated?, Ans. On the basis of the electronic configuration, the, elements in the Modern Periodic table are divided, into four blocks, viz, s - block, p - block, d - block, and f - block. The s - block constitutes the groups, 1 and 2. Groups 13 to 18 constitute the p - block., Groups 3 to 12 constitute the d - block, while the, lanthanide and actinide series at the bottom form, the f - block. The d - block elements are called, transition elements. A zig-zag line is shown in the, p - block of the periodic table., (iii), ,  hich elements are present near the zig zag, W, line?, Ans. , The zig-zag line shows the three traditional, types of elements, i.e. metals, non-metals and, metalloids. The metalloid elements lie along the, border of the zig-zag line. All the metals lie on, the left side of the zig-zag line while all the non metals lie on the right side., (iv), , Draw the electronic configuration of the period, 2 elements of first group in the periodic table., Ans. The electronic configuration of the second period, element of the first group in the periodic table is, shown below:, , Write the answers of the questions with, reference to the structure of the periodic table., , Li, 1, 2, (2, 1), (4), , A part of periodic table is shown in the following, figure., 1, , (i), , Which points are considered for the arrangement, of the Modern Periodic table?, Ans. In the Modern Periodic table, the elements are, arranged in the order of their increasing atomic, number. In the Modern Periodic table there, are seven horizontal rows called periods and, , 1, 2, 3, 4, , 18, 2, P, , 13, , 14, , 15, Q, , 16, R, S, , 17, , (i), Write the symbol of the element 'Q'., Ans. Symbol of the element 'Q' is N (Nitrogen), , T

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Periodic Classification of Elements, , 53, , (ii), , Will elements 'R' and 'S' have same number of, valence electrons?, Ans. Yes elements 'R' and 'S' will have the same number, of valence electrons (six electrons) as they belong, to the same group i.e. group 16., (iii), ,  rrange elements 'P', 'Q' and 'R' in increasing, A, order of their metallic character., Ans. , The elements in the increasing order of their, metallic character are as follows: P > Q > R, (iv), , What is the number of electrons in L shell of, element T?, Ans. Element T is in the 3rd period of periodic table,, hence it has three shells K, L and M. As it also, belongs to 18th group i.e., Zero group all it's, shells are completely filled. Therefore number of, electrons in it's L shell will be 8., (v), , Name any two elements that will have properties, similiar to that of element 'P'., Ans. The two elements that will have properties, similar to that of element 'P' are Magnesium (Mg), and Calcium (Ca) as they also belong to group 2, of the modern periodic table., (5), , Study the below given periodic table in which, four elements are indicated by alphabets: A, B,, C and D, , 1, , 18, 2, , A, , 13 14 15 16 17, 3, , 4, , 5, , 6, , 7, , 8, , 9 10 11 12, , B C D, , (ii), , Among 'A' and 'C' which element has larger, atomic radius? Why?, Ans. Element A and C belong to the same period. So, according to the periodic trends in the modern, periodic table, the atomic radius goes on, decreasing while going from left to right within, a period. Therefore atomic size of element 'A' is, larger than atomic size of element 'B'., (iii), , Identify element 'D' and write its electronic, configuration. Also write the electronic, configuration of the elements above and below, 'D' in the same group., Ans. , Element 'D' is sulphur (S) and its electronic, configuration is (2, 8, 6). The electronic, configuration of element above 'D' in the, same group will be (2, 6) and the electronic, configuration of element below 'D' in the same, group will (2, 8, 18, 6), (6), , A scientist was studying reactions of metals and, non-metals. He knew group 1 and 2 elements, are metals while group 17 elements are nonmetals. So, he chooses different elements from, group 1, group 2 and group 17., , (i), What is the valency of magnesium?, Ans. As Magnesium belongs to group 2, its valency is 2., (ii), , Name the element in group 17 which forms a, diatomic molecule and exists in solid state at, room temperature., Ans. The element in group 17 which forms a diatomic, molecule and exists in solid state at room, temperature is Iodine., (iii), , (i), , Which element is a metalloid? Name this, element and also mention the metalloids in, modern periodic table., Ans. Element 'B' is a metalloid. This element is Silicon, (Si). The metalloids in periodic table are : Boron, (B), Silicon (Si), Germanium (Ge), Arsenic (As),, Antimony (Sb), Tellurium (Te) and Polonium, (Po)., ,  ame the element in group 17 which belongs to, N, the same period as sodium., Ans. The element in group 17 which belongs to the, same period as sodium (Na) is chlorine (Cl)., (iv), , Write the formula of compound formed in the, reaction between lithium and bromine., Ans. The formula of compound formed in the reaction, between lithium (Li) and bromine (Br) is LiBr, (Lithium bromide)., (v), , Write the formula of compound formed in the, reaction between calcium and fluorine., Ans. The formula of compound formed in the reaction, between Calcium (Ca) and fluorine (F) is CaF2, (Calcium Fluoride)., , vvv

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Master Key Science and Technology Part-1 (Std. X), , 54, , ASSIGNMENT - 2, Time : 1 Hr., , (1), , Marks : 20, , Q.1. (A) Answer the following questions:, , (3), , Fill in the blanks:, , , , (i) The element eka-boron in Mendeleev's periodic table is known as, periodic table., , , , (ii) In the family of alkali metals, the number of valence electrons is, , (2), , , in the Modern, ., , Name the following:, The most electronegative non-metal., , Q.1. (B) Choose and write the correct option: , (1), , The number of electrons in the outermost shell of alkali metals is, (a) 1, , (2), , (b) 2, , (c) 3, , (d) 7, , Alkaline earth metals have valency 2. This means that their position in the Modern Periodic table is in, (a) Group 2, , , , (2), , (b) Group 16 (c) Period 2, , (d) d-block, , Q.2. Answer the following: (Any 2), , (4), , (1), , (i), , (2), , Distinguish between Transition and Inner transition elements., , (3), , Give scientific reason: Inert gases exist in the form of free atoms., , , (1), , , State the Modern Periodic law., , (ii), , Define Dobereiner's Triad., , Q.3. Answer the following: (Any 2), , (6), , Identify Dobereiner's triads from the following:, (i) Mg (24.3), Ca (40.1), Sr (87.6), , (ii) S (32.1), Se (79.0), Te (127.6), , (iii) Be (9.0), Mg (24.3), Ca (40.1), , (2), , Draw the atomic structure of Calcium and Fluorine., , (3), , Write the molecular formula of the compound of the following elements with Hydrogen., , , , (i), , , , Q.4. Answer the following: (Any 1), , (1), , (ii), , C, , (iii), , F, , (iv), , N, , (v), , Cl, , (vi), , P, (5), , (a) 4Be, 6C, 8O, 5B. Which is the most electronegative element among these? Justify your answer., (b), , (2), , S, , 13, , Al, 14Si, 11Na, 12Mg, 16S. Which of these elements has the highest metallic character?, , What are the merits and demerits of Mendeleev's Periodic table?, , vvv

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3, , Chemical Reactions and Equations, CONCEPT MAP, CHANGE, PHYSICAL, Identified by, permanent, changes in, , Chemical, properties, , Evolution of, gas, Change in, temperature, , CHEMICAL, , CHEMICAL, EQUATION, Consists of, , Various, types, , Chemical, Process, Like, , Represented by, , REACTION, , Takes, place by, , Colour, State of, matter, , CHEMICAL, Shown by, , Reactants, (L. H. S.), Combination, A+B, , Endothermic, (heat taken), , AB, , Arrow head, in (middle), , Decomposition, , Exothermic, (heat given out), , AB, , Oxidation (gain, of O2, loss of H2), , Energy, , Products, (R. H. S.), , A+B, , Displacement, A+BC, , Reduction (gain, of H2, loss of O2), , AC + B, , Double decomposition, , Precipitation, (formation of a precipitate), , A B–+C+D–, +, , A+D–+C+B–, , Neutralization, (formation of salt and water), , Points to Remember:, zz, , Compounds are formed by chemical combination of elements., , zz, , Many physical and chemical changes occur in our surroundings., , zz, , When chemical changes occur, a chemical reaction takes place., , zz, , Chemical change is irreversible whereas physical change is reversible., , zz, , Representation of a chemical reaction with the help of chemical formulae, is a chemical equation., , zz, , A chemical equation is to be balanced, because total mass of reactants should be equal to total mass of products., , zz, , A chemical reaction involves breaking and making of the bonds between the atoms to produce new substances., , zz, , Chemical reactions are classified into four types. in accordance with the nature and the number of the, reactants and products., (55)

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Master Key Science and Technology Part-1 (Std. X), , 56, , zz, , Four types of chemical reactions - (a) Combination (b) Decomposition (c) Displacement (d) Double, displacement., , zz, , When heat or light is evolved in a chemical reaction, this reaction is known as Exothermic reaction. e.g., burning of Mg wire, dissolving NaOH in water., , zz, , When heat, light or electricity is absorbed, then this type of reaction is known as Endothermic reaction. e.g., electrolysis of water, melting of ice., , zz, , Many types of substances give reactions called oxidation and reduction., , zz, , When a reactant combines with oxygen or loses hydrogen to form the product, this reaction is called oxidation, reaction., , zz, , In a reduction reaction, oxygen is removed from a substance or hydrogen combines with a substance., , zz, , When oxidation and reduction occur simultaneously in the same reaction, then this type of reaction is known, as redox reaction. e.g. 2H2S + SO2 → 3S + 2H2O, Here, SO2 is reduced and H2S is oxidized., , zz, , Oxidation of vegetable oil causes rancidity., , zz, , Antioxidants are used to prevent oxidation of fats and oil., , zz, , The slow process of decay or destruction of metal due to the action of air, water, salt or acid, alkali is known, as corrosion., , zz, , Corrosion of metals can be prevented by coating them with suitable paints or plastic coat, galvanizing,, electroplating with non-corrosive metal., , zz, , When an acid reacts with an alkali, we get salt and water, we call it neutralization., , MASTER KEY QUESTION SET - 3, Q.1. (A) 1. Fill in the blanks and rewrite the complete, statements:, (1), (2), , (11), ,  he reaction in which heat is given out along, T, with products is known as, reaction., , (12), , AgNO3 + NaCl → AgCl +, , ., , (13), ,  he reaction CaCO3 → CaO + CO2 is a, T, reaction., ,  redox reaction takes place during, A, respiration., , (14), ,  he reaction in which oxygen is added to the, T, substance is called, reaction., ,  ancidity in the food stuff cooked in oil or ghee is, R, prevented by using, ., , (15), ,  he chemical reaction during which H2(g) is lost is, T, termed as, ., , (3), , Corrosion can be prevented by using, , ., , (4), ,  he chemical reaction in which heat is absorbed, T, is called, reaction., , (16), ,  hen acids and alkalis react together,, W, and, is formed., , (5), , The substance undergoing oxidation acts as a, , agent., , (17), , Fe + CuSO4 → FeSO4 +, , (18), , ., , The symbol 'aq' in a chemical equation represents, ., , (6), , ormation of Nitric oxide from nitrogen and, F, oxygen is a, reaction., , (19), , (7), ,  hen Oxidation and Reduction takes place, W, simultaneously in a given chemical reaction, it is, known as, ., ,  ll chemical reactions obey law of conservation, A, of mass, hence, chemical equations should be, ., , (20), , (8), ,  he substances taking part in a chemical reaction, T, are called, , whereas the substances, formed as a result of a chemical reaction are, called, ., , ouble displacement reaction in which an, D, insoluble salt is formed is also called, reaction., , (9), , A chemical reaction is represented by writing a, ., , (10), ,  he reaction of a vegetable oil with hydrogen gas, T, takes place in the presence of, as a, catalyst to form vanaspati ghee., , Ans. , (1) decomposition (2) oxidation (3) anti-rust solution, (4) endothermic (5) reducing (6) combination (7) redox, reaction (8) reactants, products (9) chemical equation, (10) Nickel (Ni) (11) Exothermic (12) NaNO3, (13) cellular (14) anti-oxidants (15) oxidation (16) salt,, water (17) Cu(s) (18) aqueous solution (19) balanced, (20) precipitation

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Chemical Reactions and Equations, , 57, , Q.1. (A) 2. Match the columns:, (I), , (5), , Rusting of iron is an oxidation reaction., , (6), ,  ize of the particles of reactants does not affect, S, the rate of chemical reaction., , (7), ,  ombustion is the rapid reaction between carbon, C, dioxide and fuel., , (8), ,  precipitate is an insoluble solid formed from, A, solution during a chemical reaction., , (9), ,  chemical equation shows a chemical reaction, A, using symbols and chemical formulae instead of, words., , (10), Types of, Reaction, ,  ecomposition of compost is an endothermic, D, reaction., , (11), , (i) Displacement, , nlike physical changes, chemical changes, U, cannot be easily reversed., , (12), ,  burning of match stick is an example of, A, chemical change., , (2) 2 AgCl(s), , (b) FeSO4(aq) + (ii) Combination, Cu(s), , (13), , (3) CuSO4(aq), + Fe(s), , (c) BaSO4 ↓ + (iii) Decomposition, ZnCl2(aq), ,  he reaction Zn(s) + CuSO4(aq) → ZnSO4(aq), T, + Cu(s) is an example of double displacement, reaction., , (14), , Chemical change is a temporary change., , (4) H2O(l) +, CO2(g), , (d)2Ag(s) +, Cl2(g), , (15), , (g) indicates the physical state of a substance as, solid., , (16), ,  onversion of quick lime into slaked lime is an, C, example of displacement reaction., , (17), , Calcium oxide is also called lime or quicklime., , (18), , In a chemical equation, the symbol ↓ is used to, denote precipitation formation., , Reactants, (1) Fe + S, , Types of, Reaction, , Products, , (a) NaCl + H2O (i) Oxidation, , (2) CuSO4 + (b) 2CuO, Zn, , (ii) Neutralization, , (3) 2Cu + O2 (c) ZnSO4 + Cu (iii)Displacement, (4) HCl +, NaOH, , (d)FeS, , (iv)Combination, , Ans. , (1-d-iv), (2-c-iii), (3-b-i), (4-a-ii), *(II), , Reactants, , Products, , (1) BaCl2(aq) + (a) H2CO3(aq), ZnSO4(aq), , (iv) Double, displacement, , Ans. , (1-c-iv), (2-d-iii), (3-b-i), (4-a-ii), (III), , Column A, , Column B, , (1) Heating of, (a) Turns lime water, Potassium Chlorate, milky, , Ans. , (1-d), (2-f), (3-e), (4-a), (5-c), (6-b), , Ans. (1) True (2) False. Catalyst speed up the rate of reaction., (3) False. Reaction that releases energy is called, Exothermic. (4) True (5) True (6) False. Size of the, particles of reactants affect the rate of chemical reaction., (7) False. Combustion is the rapid reaction between, oxygen and fuel. (8) True (9) True (10) False., Decomposition of compost is an exothermic reaction., (11) True (12) True (13) False. It is a displacement, (single) reaction (14) False. It is a permanent change., (15) False. (g) indicates the physical state of a substance, as gas. (16) False. It is an example of combination, reaction. (17) True (18) True., , Q.1. (A) 3. State whether the following statements, are True or False. Write the correct statements:, , (1), , (2) Depositing a layer, of zinc on iron, , (b) Physical change, , (3) Souring of milk, , (c) Rust, , (4) Carbon dioxide, , (d) MnO2 is used as, catalyst, , (5) Iron oxide, , (e) Chemical change, , (6) Dissolving common (f) Galvanisation, salt in water, , (1), , Digestion of food is a chemical change., , (2), ,  catalyst slows down the rate of reaction to, A, make a better product., , (3), ,  eaction that releases energy is called, R, Endothermic., , (4), , Ammonium chloride is a sublimable salt., , Q.1. (A) 4. Name the following:,  change that takes place due to change in the, A, parameters such as temperature, pressure., , Ans. Physical change, (2), ,  process in which some substances undergo, A, bond breaking and are transformed into new, substances by formation of new bonds., , Ans. Chemical reaction

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Master Key Science and Technology Part-1 (Std. X), , 58, , (3), , epresentation of a chemical reaction in a, R, condensed form using chemical formulae., , Ans. Chemical equation, (4), , If the number of atoms of each element is not, same on the two sides of an equation., , Ans. Unbalanced equation, , (c) Formation of precipitate, (d) Lowering of temperature, (5), , What is rust?, (a) Sodium oxide, (c) Copper oxide, , Ans. Biogas, ,  ecause of the formation of which of the following,, B, lime water turns milky when carbon dioxide is, passed through it?, , (6), , (a) Calcium Carbonate, , (5), ,  uel formed when organic waste is decomposed, F, by micro-organism., , (6), , (b) Iron oxide, (d) Silver oxide, ,  he substance in whose presence the rate of a, T, chemical reaction changes without causing any, chemical change to it., , (b) Calcium bicarbonate, (c) Calcium hydroxide, , Ans. Catalyst, (7), ,  he life on earth is protected from ultraviolet, T, radiation of the sun., , (d) Sodium Carbonate, (7), , Ans. Ozone layer, (8), , (a) Calcium Chloride, , A state prior to the formation of the O2 molecule., , (b) Hydrogen Chloride, , Ans. Nascent oxygen., Q.1. , (A) 5. Complete the following reactions and, name the products:, (1), , CuSO4(aq) + Fe(s) → FeSO4(aq) +, Ferrous, Sulphate, , Cu(s), Copper, , (2), , CuSO4(aq) + Pb(s) → PbSO4(aq) +, Lead, Sulphate, , Cu(s), Copper, , A chemical reaction involves, , (c) Sodium hydroxide, (d) Sodium Chloride, (8), , (b) Milk is set into curd, (c) Ripening of fruit, (d) Respiration process, (9), ., , (a) Only breaking of bonds., (b) Only formation of bonds., (c) Both breaking and formation of bonds., , (10), , A balanced chemical equation always obeys, , ., (a) Law of conservation of Mass, (b) Law of thermal equilibrium, , (11), , (c) Law of conservation of energy,  ily food kept out for few days gives a bad taste, O, and a bad smell because of, ., (a) Corrosion, (c) Heating, (4), , The sign ↓ indicates., (a) Release of gas, (b) Dissolution of gas, , (b) Displacement, (d) Rancidity, , (a) Sulphur dioxide, , (b) Hydrogen, , (c) Oxygen, , (d) Zinc dioxide, , Heating of sugar is called a, , (12), , reaction., , Combination reaction, Displacement reaction, Double displacement reaction, Decomposition reaction, ,  ntioxidants are used to prevent, A, food containing fats and oils., (a), (b), (c), (d), , (d) All of the above, (3), ,  hen sulphuric acid is poured over zinc, which, W, of the following gas is formed?, , (a), (b), (c), (d), , (d) None of these., (2), , is a physical change., (a) Ice changes into water, , Q.1. (B) Choose and write the correct option:, (1), ,  hich of the following is formed when Sodium, W, hydroxide reacts with hydrochloric acid?, , of, , Reduction, Oxidation, Oxidation and reduction, Decomposition, , In the reaction given CuO(s) + H2(g) → Cu(s) + H2O(l)., Copper oxide is, and hydrogen is, ., (a) Oxidized, reduced (b) Reduced, oxidized, (c) Unaffected, reduced (d) Unaffected, oxidized

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Chemical Reactions and Equations, , (13), ,  hich among the following is double displacement, W, reaction?, (a), (b), (c), (d), , Pb + CuCl2 → PbCl2 + Cu, Na2SO4 + BaCl2 → BaSO4↓ + 2NaCl, C + O2 → CO2, CH4 + 2O2 → CO2 + 2H2O, , Ans. , (1) (c) Both breaking and formation of bonds., (2) (a) Law of conservation of Mass (3) (d) Rancidity, (4) (c) formation of precipitate (5) (b) Iron oxide, (6) (a) Calcium carbonate (7) (d) Sodium chloride, (8) (a) Ice changes into water (9) (b) Hydrogen, (10) (d) Decomposition Reaction (11) (b) oxidation, (12) (b) reduced, oxidized (13) (b) Na2SO4 + BaCl2 →, BaSO4 ↓+ 2NaCl, Q.2.1.Answer the following:, (1), , What is a balanced chemical equation?, , Ans. , The chemical equation in which the number of, atoms of different elements are equal in reactant, side (L.H.S.) and product side (R.H.S.) is called a, balanced chemical equation., (2), , “We need to balance an unbalanced skeletal, chemical equation”. Justify the statement., Ans. , , 59, , (5), What is rust? Write the chemical formula., Ans. , (i), Due to moisture and various components of, atmosphere, oxidation of iron takes place., (ii) A certain type of reddish coloured solid layer, collects on the iron surface. This layer is called rust., (iii) The chemical formula is Fe2O3. xH2O., (6), ,  hy respiration is considered an exothermic, W, reaction?, Ans. , In the process of respiration glucose is broken, into Carbon dioxide and energy is released. That, is why, it is considered as an exothermic reaction., C6H12O6(aq) + 6O2(g) → 6CO2(g) + 6H2O(l) + Energy (ATP), (7), ,  ame the factors on which the speed of a, N, chemical reaction depends., Ans. , The speed of a chemical reaction depends on factors, like size of the particles of reactants, concentration, of the reactants, nature of the reactants, temperature, of the reaction and presence or absence of catalyst., (8), What is meant by reactants?, Ans. , The chemicals that undergo chemical change in a, chemical reaction are called reactants., (9), ,  hat is meant by the term products of a, W, chemical reactions?, Ans. , The substances that are formed after a chemical, reaction takes place, are called products of a, chemical reaction., , (i), , Skeletal chemical equations are either self, balanced or unbalanced., , (ii), , We need to balance the unbalanced chemical, equation because of law of conservation of mass, which states that “In any reaction, the total mass, of each of the elements in the reactants is the same, as that of the total mass of each of the respective, elements in the products”., , (iii), , Therefore, chemical equation must be balanced, in each and every chemical reaction., , (11), , (3), , iving an example list two information, G, which make a chemical equation more useful., (informative)., , (12), , Ans. , (i) Physical state of reactants must be mentioned,, , e.g. 2H2(g) + O2(g) → 2H2O(l)., (ii) Condition in which a reaction takes place is, written on the arrow head,, burning, , e.g. 2H2(g) + O2(g), 2H2O(l)., (4), ,  hat type of reaction takes place when, W, vegetable matter is converted to compost?, , Ans. , During composting complex substances are, broken down into simpler substances with the, evolution of heat. Hence, it is an exothermic, decomposition reaction., , (10), ,  hen carbon dioxide is passed through lime, W, water it turns milky. Why?, Ans. , Lime water (Calcium hydroxide) combines with, carbon dioxide to form a suspension of calcium, carbonate which turns lime water milky., Ca(OH)2(aq) + CO2(g) → CaCO3(s) + H2O(l), , W, hat, type, of, reaction, Na2SO4 + BaCl2 → BaSO4 + 2NaCl., Ans. , It is a double displacement reaction., , is, , this:, ,  hat do you understand by precipitation, W, reaction? Explain with examples., , Ans., (i),  hen two ionic compounds react and two ionic, W, compounds (product) are formed of which one, product is insoluble and settles as a solid, it is, called a precipitate., (ii) Such reactions in which precipitate is formed are, called precipitation reactions., (iii) , Example: When aqueous solution of Sodium, Sulphate is mixed with the aqueous solution of, Barium Chloride, it forms Barium Sulphate in the, form of white precipitate., Na2SO4(aq) + BaCl2(aq) → BaSO4↓ + 2NaCl(aq), (White)

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Master Key Science and Technology Part-1 (Std. X), , 60, , (13) Which soloution is used in voters ink?, Ans. Solution of Silver nitrate (AgNO3) is used in the, voters ink., (14), ,  hat happens when a piece of zinc metal is, W, added to copper sulphate solution?, Ans. , (i), When a piece of zinc metal is added to copper, sulphate solution, more reactive zinc metal, displaces the less reactive copper from copper, sulphate solution and a solution of zinc sulphate, is obtained while copper gets deposited on zinc, making it reddish brown in colour., (ii) Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s), *(15) W,  hich is the oxidant used for purification of, drinking water? (Use your brain power! Textbook, page number 42), Ans., (i) Chlorine is a strong oxidant commonly used in, water treatment for oxidation and disinfection., (ii) Even Potassium permanganate can be used as, oxidant for purification of water., , *(16) W,  hy is potassium permanganate used for, cleaning of water tanks? (Use your brain power!, Textbook page number 42), Ans. , (i), Potassium permanganate oxidises dissolved, iron, manganese and hydrogen sulphide into, solid particles., (ii) KMnO4 is used to remove organic matter from, the water tank by oxidation which also removes, unwanted taste and odour., (iii) It also kills harmful bacteria and fungi., *(17) S,  ome examples of redox reaction are given., Identify the reductants and oxidants from them., (Use your brain power! Textbook page number 43), 2H2S + SO2 → 3S↓ + 2H2O, Ans., (i) , In a redox reaction, the reductant transfers, electrons to the oxidant., (ii) Then in the reaction, the reductant or reducing, agent loses electrons and is oxidised and the, oxidant or oxidizing agent gains electrons and is, reduced., (iii) , ∴ Hence, in the above equation, reductant is SO2, and oxidant is H2S, MnO2 + 4HCl → MnCl2 + 2H2O + Cl2↑, (i), Reductant: MnO2 - as it loses oxygen., (ii) Oxidant: HCl - as it gains oxygen., *(18) If oxidation means losing electrons, what is, meant by reduction? (Use your brain power!, Textbook page number 43), Ans. , If oxidation means losing electrons, then, reduction means gaining electrons., , *(19) W,  rite the reaction of formation of Fe2+ by the, reduction Fe3+ by making use of the symbol (e-)?, (Use your brain power! Textbook page number 43), Ans. Fe3+(aq) + eFe2+ (aq)., reduce, (20) Write two conditions necessary for rusting of, iron., Ans. Rusting of iron takes place in (i) the presence of, air and (ii) presence of moisture., (21), ,  ame the term used for the solution of a, N, substance in water., Ans. Aqueous (aq) is the term used for the solution of, a substance in water., (22), ,  hat are the changes that take place when fats, W, and oils are oxidized?, Ans. When fats and oils are oxidized, they become, rancid and their taste changes., *(23) Identify the reactants and products of equation:, 60°C, Vegetable Oil(l) + H2(g), Vanaspati ghee(s), Ni Catalyst, , (Use your brain power! Textbook page number 35), Ans. , Reactants: Vegetable oil and hydrogen gas, Product: Vanaspati ghee, *(24) Use your brain power! (Textbook page number 36), (i) What is the number of reactants in the following, reactions., NH3(g) + HCl(g) → NH4Cl(s), , - (1), , 2Mg + O2 → 2MgO, , - (2), , CaO + H2O → Ca(OH)2 + Heat - (3), Ans. , In all the above equations, there are two reactants., (ii) What is the number of molecules of reactants, taking part in the above reactions?, Ans., NH3, +, HCl, →, NH4Cl, 1 molecule , 2Mg, , +, , 2 molecules, CaO, , 1 molecule, O2, , 2MgO, , →, , Ca(OH)2 + Heat, , 1 molecule, , +, , 1 molecule , , →, , H2O, 1 molecule, , (iii) How many products are formed in each of the, above reaction?, Ans. In all the above reactions, only one product is, formed., Q.2.2.Define the following/write notes:, (1), , Physical Change, , Ans. It is a change in which only the physical, properties of substances are changed and no new, substance is formed. This change is a temporary, and reversible change.

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Chemical Reactions and Equations, , (2), , Chemical Change, , Ans. I t is a change in which substances are converted, into new substance or product possessing, properties altogether different from the original., This is a permanent and irreversible change., (3), , Chemical Reaction, , Ans. , A chemical reaction is a process in which some, substances undergo bond breaking and are, transformed into new substances by formation of, new bonds., (4), , Chemical Equation, , Ans. T, he representation of a chemical reaction in, a condensed form using chemical formulae is, called as the chemical equation., *(5), , Combination Reaction, , Ans., (i), When two or more reactants combine in a, reaction to form a single product, it is called a, combination reaction., (ii) e.g.: When a Magnesium strip is burnt in air, it, combines with oxygen, to form white ash of, magnesium oxide., 2Mg(s), +, O2(g)→ 2MgO(s) (white ash), (iii), Magnesium, Oxygen Magnesium, oxide, (6), , Decomposition Reaction, , Ans., (i) , The chemical reaction in which two or more, products are formed from a single reactant is, called Decomposition Reaction., (ii) e.g.: When Calcium carbonate is strongly heated, it decomposes to form Calcium oxide and Carbon, dioxide gas., ∆ CaO, +, CO2↑, (iii) e.g. CaCO3(s), (s), Calcium, Calcium, Carbon, Carbonate, Oxide, dioxide, *(7), , Displacement Reaction, , Ans. , (i), The reaction in which the place of the ion of a, less reactive element in a compound is taken by, another more reactive element by formation of its, own ions, is called displacement reaction., Zn, + CuSO4(aq) → ZnSO4(aq) + Cu(s), (ii), (s), Zinc Copper, Zinc, Copper, sulphate, sulphate, (iii), , In the above equation when zinc dust is added, to blue coloured copper sulphate solution, a, colourless solution of zinc sulphate is formed, and copper is deposited on zinc., , 61, , (iv), , The Zn2+ ions formed from Zn atom take the place, of Cu2+ ions in copper sulphate, and Cu atoms, formed from Cu2+ ion come out. It means that Zn, displaces Cu from CuSO4., , (8), , Double displacement Reaction, , Ans., (i),  he reaction in which the ions in the reactants are, T, exchanged to form a precipitate are called double, displacement reactions., (ii) BaSO4(aq) + K2CrO4(aq) → BaCrO4(s) + K2SO4(aq), Barium Potassium, Barium Potassium, sulphate chromate chromate sulphate, (iii), ,  otassium chromate solution reacts with Barium, P, sulphate solution to form Potassium sulphate, solution and a yellow precipitate of Barium, chromate (BaCrO4)., , *(9), , Endothermic Reaction, , Ans., (i),  reaction in which heat is either absorbed from the, A, surrounding or has to be supplied continuously, from outside is called as endothermic reaction., (ii) e.g. melting of ice, dissolution of potassium, nitrate in water. During the process of dissolution, of potassium nitrate (KNO3) in water, heat from, the surroundings is absorbed and therefore, the, temperature of the resulting solution is less., (iii) The process in which heat is absorbed from, outside is called endothermic process., (iv) KNO3(s) + H2O(l) + Heat → KNO3(aq), Potassium Water, Potassium nitrate, nitrate, solution, (10), , Exothermic Reaction, , Ans., (i), A reaction in which heat is given away when, reactants are transformed into the products is, called as exothermic reaction., (ii) e.g. formation of ice from water, dissolution of, sodium hydroxide in water., (iii) When solid NaOH is dissolved in water, heat, is given out and therefore, the temperature, increases., (iv) NaOH(s) + H2O(l) → NaOH(aq) + Heat, Sodium, Water, Sodium hydroxide, hydroxide, solution, *(11) Oxidation Reaction, Ans. , The chemical reaction in which a reactant, combines with oxygen or loses hydrogen to form, a product is called oxidation reaction., e.g. (1) 2Mg(s), Magnesium, , +, , O2(g), → 2MgO(s), Oxygen Magnesium oxide

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Master Key Science and Technology Part-1 (Std. X), , 62, , (2) 2Cu(s), Copper, , +, , O2(g) → 2CuO(s), Oxygen Copper oxide, , +, H2(g), (3) MgH2 → Mg(s), Hydrogen, Magnesium Magnesium, hydride, (12), , Reduction Reaction, , Ans. , The chemical reaction in which reactants gain, hydrogen or loses oxygen to form the product is, called reduction reactions., e.g. (1) HgO(s) → Hg(l) + O2(g), Mercuric Mercury Oxygen, oxide, (2) H2(g) + Cl2(g) → 2HCl(g), Hydrogen Chlorine Hydrogen, Chloride, (13) Redox Reaction, , *(17) Balanced Equation, Ans., (i), The number of atoms of the elements in the, reactants is same as the number of atoms of those, elements in the products. Such an equation is, called a 'balanced equation'., (ii) In any reaction, the total mass of each of the, respective elements in the reactants is same as the, total mass of each of the respective elements in, the products., (iii) This is in accordance with the law of conservation, of mass., (iv), , Ans. , A chemical reaction in which oxidation and, reduction takes place simultaneously is called, redox reaction., Oxidation, e.g. CuO(s) + H2(g) → Cu(s) + H2O(l), Copper Hydrogen Copper Water, Oxide, Reduction, (14), , Catalyst, , Ans., (i), The substance in whose presence the rate of a, chemical reaction changes, without causing any, chemical change to it is called a catalyst., (ii), , e.g. SO2 + 2H2S → 3S + 2H2O, On reactant side, , On product side, , No. of atoms of, Sulphur (S) = 3, , No. of atoms of, Sulphur (S) = 3, , No. of atoms of, Oxygen (O) = 2, , No. of atoms of, Oxygen (O) = 2, , No. of atoms of, hydrogen (H) = 4, , No. of atoms of, hydrogen (H) = 4, , ∴ The given equation is balanced., Q.2.3.Complete the following chemical equations,, balance and state the type of reactions:, (1), HSO + HO →, 2 2, , 7, , 2, , Ans. H S O + H O → 2 H SO, 2 2 7, 2, 2, 4, It is a Combination reaction., ∆, KClO3, , +, , 2KClO3, , +, , (2), , (ii), , e.g. Potassium chlorate (KClO3) decomposes, rapidly in presence of manganese dioxide (MnO2), to liberate O2 gas, where manganese dioxide acts, as a catalyst., ∆, 2KClO3, 2KCl + 3O2, , (3), , Sunlight, , (15), , Neutralisation Reaction, , Ans. 2 AgCl, , Sunlight, , MnO2, , Ans., (i), The reaction in which an acid and a base, react together to form salt and water is called, Neutralisation reaction., (ii) e.g. HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l), Sodium, Hydrochloric Sodium, Water, chloride, acid, hydroxide, (Salt), (acid), (base), (Water), (16), , Rancidity, , Ans. , Rancidity is the oxidation of oils and fats, resulting in change in taste and giving out foul, smell. Rancidity in the food stuff is prevented by, antioxidants., , Ans., , ∆, , 2 KCl, , O2, 3 O2, , It is a Decomposition reaction., Ag + Cl2, 2 Ag, , + Cl2, , It is a Decomposition reaction., (4), Ans., (5), Ans., (6), Ans., , Al +, , HCl →, , +, , 2 Al + 6 HCl → 2 AlCl3, It is a Displacement reaction., , H2, , + 3 H2, , NaOH + H2SO4 →, , +, , H2O, , + 2 H2O, 2 NaOH + H2SO4 → Na2SO4, It is a Double displacement reaction., N2 +, , →, , N2 + 3 H2, , →, , It is a Combination reaction., , NH3, 2 NH3

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Chemical Reactions and Equations, , 63, , Q.2.4.Distinguish between:, (1), , Physical Change and Chemical Change., , Ans., , Physical Change, (i), , Chemical Change, , In this change, the composition of substance (i), does not change and no new substance is, formed., , In this change, the composition of substance changes, and a new substance is formed., , (ii) There is a change only in physical, properties., , (ii), , There is a change in both physical and chemical, properties., , (iii) This change is temporary., , (iii) This change is permanent., , (iv) The original substance can be recovered by, simple means., , (iv) The original substance cannot be recovered by, simple means., , (v) e.g.: Steam is condensed., , (v), , e.g.: Ripening of fruit, Milk is converted to curd., , (2) Displacement and Double displacement Reaction, Ans., , Displacement Reaction, (i), , Double displacement Reaction, , The reaction in which the place of the ion (i), of a less reactive element in a compound, is taken by another more reactive element, by formation of its own ions, is called, displacement reaction., , The reaction in which the ions in the reactants are, exchanged to form a precipitate are called double, displacement reactions., , (ii) These reactions are usually slow and take, longer time for their completion., , (ii) These reactions are usually fast and take place, instantaneously., , (iii) During these reactions, change of colour, takes place., Fe(s) + CuSO4(aq) → FeSO4(aq) + Cu(s), , (iii) During these reactions, precipitates are formed., AgNO3(aq) + NaCl(aq)→ AgCl(s)↓ + NaNO3(aq), , Iron, , Copper, sulphate, , Iron, sulphate, , Silver, nitrate, , copper, , Sodium, chloride, , Silver, chloride, , Sodium, nitrate, , (3) Combination reaction and Decomposition reaction, Ans., , Combination reaction, (i), , Decomposition reaction, , When two or more reactants combine in a, (i), reaction to form a single product, it is called a, combination reaction., , The chemical reaction in which two or more, products are formed from a single reactant is called, decomposition reaction., , (ii) It is a synthesis reaction., , (ii) It is a degradation reaction., , (iii) New substances are synthesized in industry, using this reaction., , (iii) In nature, complex substances are broken down, into simpler compounds., , (iv) CaO(s) + H2O(l) → Ca(OH)2(aq), , (iv), , Calcium, oxide, , Water, , Calcium, hydroxide, , CaCO3(s) ∆ CaO(s) +, Calcium, Calcium, carbonate, oxide, , CO2(g)↑, Carbon, dioxide, , (4) Exothermic reaction and Endothermic reaction, Ans., , Exothermic reaction, (i), , Endothermic reaction, , The reaction where heat energy is evolved or (i), liberated is called as exothermic reaction., , (ii) Temperature of the reaction mixture rises., , The reaction where heat energy is absorbed is called, as endothermic reaction., , (ii) Temperature of the reaction mixture lowers.

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Master Key Science and Technology Part-1 (Std. X), , 64, , (5), , (iii) Generally, the reactions are fast., , (iii) Generally, the reactions are slow., , (iv) e.g.: When NaOH dissolves in water heat is, evolved., , (iv) e.g.: When Calcium carbonate is strongly heated,, calcium oxide is formed., , Oxidation reaction and Reduction reaction, , Ans., , Oxidation reaction, (i), , Reduction reaction, , It is a chemical process in which a substance, gains oxygen or loses hydrogen., , (i), , It is a chemical process in which a substance gains, hydrogen or loses oxygen., , (ii) During oxidation reaction, an atom or an ion (ii) During reduction reaction, an atom or ion gains, loses electron., electron., (iii) Oxidizing agents are used to carry out this, reaction., , (iii) Reducing agents are used to carry out this, reaction., , (iv) Example: C + O2 → CO2↑, , (iv) Example: C + 2H2 → CH4, , Q.2.4.Give scientific reasons:, *(1) When the gas formed on heating limestone is, passed through freshly prepared lime water, the, lime water turns milky., Ans. , (i) , On heating, limestone undergoes thermal, decomposition to form calcium oxide (quick, lime) and carbon dioxide., ∆ CaO + CO ↑, CaCO, 3(s), , (s), , quick lime, , 2, , (ii) , Calcium oxide when dissolved in water forms, Calcium hydroxide (lime water)., CaO(s) + H2O(l) → Ca(OH)2(aq), Calcium, hydroxide, (iii) , When carbon dioxide is passed through lime, water it turns milky due to formation of white, precipitate of Calcium carbonate., Ca(OH)2(aq) + CO2(g) → CaCO3(s) + H2O(l), Calcium, carbonate, *(2) It takes time for pieces of Shahabad tiles to, disappear in HCl, but its powder disappears, rapidly., Ans. , (i) , When HCl is added to Shahabad, effervescence of CO2 is formed., , stone, , (ii) The CO2 effervescence is formed slowly with the, pieces of Shahabad tiles while at the faster speed, with the powder., (iii) This is because the rate of a reaction depends, upon the size of the particles of the reactants, taking part in the reaction., (iv) Smaller the size of the reactant particles more, , is the surface area available for the reaction and, higher is the rate of reaction., (v) Hence, it takes time for pieces of Shahabad tiles, to disappear in HCl, but its powder disappears, rapidly., *(3) , While preparing dilute sulphuric acid from, concentrated sulphuric acid in the laboratory,, the concentrated sulphuric acid is added slowly, to water with constant stirring., Ans. , (i) , Concentrated sulphuric acid (H2SO4) is a, hazardous, highly corrosive chemical. It has a, higher density than water and is hygroscopic,, which means that it absorbs water from the air., (ii) , In the process of dilution of concentrated, sulphuric acid with water, very large amount of, heat is liberated., (iii) As a result, water gets evaporated instantaneously, along with splashing of acid which may cause an, accident., (iv) , To avoid this, required amount of water is, taken in a glass container and small quantity of, concentrated sulphuric acid at a time is added, with stirring., (v) Therefore, only a small amount of heat is liberated, at a time., *(4) It is recommended to use air tight container for, storing oil for long time., Ans. , (i) Air, heat and light affect the quality of edible oil,, which deteriorates through oxidation (rancidity)., (ii) The oxidation process is greatly enhanced when, edible oils are stored in containers that are not air, tight and in areas where oil is exposed to air, heat, and light.

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Chemical Reactions and Equations, , (iii) , Hence, it is recommended to use air tight, containers for storing oil for long time, to prevent, it from getting oxidised., (5) Grills of doors and windows are always painted, before they are used., Ans. , (i) The grills of doors and windows are generally, made of the metal iron., (ii) When they are exposed to air, due to the effect, of moisture in air, iron undergoes slow oxidation, reaction and forms a layer of reddish brown rust, which gets deposited on the surface of grills., (iii) Thus, due to the reaction of slow process of decay, or destruction of metal iron, the grills of doors, and windows made of metal iron get corroded., (iv) Hence, in order to prevent the rust and corrosion,, grills of doors and windows are always painted, before they are used., (v) By doing this, air and moisture cannot come in, contact with the iron and hence rusting doesn't, take place., (6) Digestion of food is an example of decomposition, reaction., Ans. , During digestion, the starch in the various food, items gets decomposed into simple sugar while, proteins get decomposed into amino acids. Hence, digestion of food is an example of decomposition, reaction., (7) Respiration is considered to be an exothermic, reaction., Ans. , (i) , When heat is given out during a chemical, reaction, it is an exothermic reaction., (ii) Glucose formed in our body during the process, of digestion, combines with oxygen in our body, and provides energy., , 65, , *Q.3.1.Choose the correct option from the bracket and, explain the statement giving reason:, , (oxidation, displacement, electrolysis, reduction, zinc,, copper, double displacement, decomposition), zinc, (1), To prevent rusting, a layer of, metal, , is applied on iron sheets., Ans. , Explanation: Due to moisture and various, components of atmosphere, oxidation of iron, takes place. A certain type of reddish coloured, solid layer collects on the iron surface. This layer, is called rust. Its chemical formula is Fe2O3. x H2O., To avoid rusting, a layer of zinc metal is applied, on iron sheets which prevents iron sheets from, directly coming in contact with the atmosphere, and hence, prevents rusting. This process of, giving a thin coating of zinc on iron to protect it, from rusting is called galvanisation., (2), ,  he conversion of ferrous sulphate to ferric, T, sulphate is oxidation reaction., , Ans. E,  xplanation: The conversion of ferrous sulphate, (FeSO4) into ferric sulphate Fe2(SO4)3 is an, oxidation reaction., 2FeSO4 → Fe2(SO4)3, Net ionic reaction: Fe2+ → Fe3+, (Ferrous) (Ferric), (a) , In the above ionic reaction, when ferric ion is, formed from ferrous ion, the positive charge, is increased by one unit. While this happens,, the ferrous ion loses one electron and losing of, electrons is an oxidation reaction., (3), , hen electric current is passed through, W, acidulated water electrolysis of water takes place., Ans. Explanation:, (i), , Water decomposes into hydrogen and oxygen, gases on passing electric current through, acidulated water., , (8) We store silver chloride in dark coloured bottles., Ans. , (i) A chemical reaction that proceeds by absorption, of light is called a photo chemical reaction., , (ii), , This decomposition takes place by means of, electrical energy., , (iii), , Therefore, it is called as electrolysis., , (ii) We store silver chloride in dark coloured bottles, in order to protect it from the effect of sunlight., , (4), ,  ddition of an aqueous solution of ZnSO4 to, A, an aqueous solution of BaCl2 is an example of, double displacement reaction., , (iii) , Therefore, respiration is considered to be an, exothermic reaction., , (iii) Sunlight decomposes silver chloride and forms, silver and chlorine., Sunlight, , 2AgCl(s), Silver chloride, (white), , 2Ag(s) + Cl2↑, Silver, Chlorine, (Grey), , (iv) As dark coloured bottles obstruct light, silver, chloride will get prevented from decomposition., , Ans. Explanation:, (i), , When aqueous solution of ZnSO4 is added to, aqueous solution of BaCl2, a reaction takes place, forming an aqueous solution of ZnCl2 and a white, precipitate of BaSO4., , (ii), , In this reaction, ions in the reactants are

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Master Key Science and Technology Part-1 (Std. X), , 66, , exchanged to form a precipitate. Such reactions, are called double displacement reactions., Q.3.2.Answer the following:, *(1) What is the reaction called when oxidation and, reduction take place simultaneously? Explain, with one example., Ans., (i) , When oxidation and reduction take place, simultaneously in a given reaction it is termed as, Redox reaction., (ii), , During oxidation a reactant combines with, oxygen or loses hydrogen and during reduction, it gains hydrogen or loses oxygen., , (iii), , Example of Redox reaction:, oxidation (O2 added), , 2H2S + SO2 → 3S + 2H2O, reduction (O2 lost), , (iv), , In the above reaction, SO2 is reduced and H2S is, oxidised., , *(2) , How can the rate of the chemical reaction, namely decomposition of hydrogen peroxide, be increased?, Ans. , (i) , The chemical reaction in which two or more, products are formed from a single reactant, is called decomposition reaction. Hydrogen, peroxide decomposes into water and oxygen., 2H2O2(l) → 2H2O (l) + O2↑, Hydrogen, peroxide, , Water, , Oxygen, , (ii) , The decomposition of hydrogen peroxide into, water and oxygen takes place slowly at room, temperature. However, the rate of this chemical, reaction can be increased on adding manganese, dioxide (MnO2) powder in it, here manganese, dioxide acts as a catalyst., *(3) Explain the term reactant and product giving, examples., Ans. , (i), A chemical reaction is a process in which some, substances undergo bond breaking and are, transformed into new substances by formation of, new bonds., (ii) The substances taking part in chemical reaction, are called reactants, whereas the substances, formed as a result of a chemical reaction by, formation of new bonds are called products., , (iii) For example :, (a) Formation of carbon dioxide gas by combustion, of coal in air is a chemical reaction. In this, reaction Coal (carbon) and Oxygen (from air) are, the reactants while carbon dioxide is the product., C + O2 → CO2, Reactants product, , (b), , CuSO4 + Zn → ZnSO4 + Cu, Copper Zinc, Zinc, Copper, sulphate, sulphate, , In the above equation, copper sulphate and zinc, are the reactants and zinc sulphate and copper, are the products., *(4) Explain the types of reaction with reference to, oxygen and hydrogen. Illustrate with examples., Ans. Types of reaction with reference to oxygen and, hydrogen are oxidation reaction and reduction, reaction., , Oxidation reaction: The chemical reaction in, which a reactant combines with oxygen or loses, hydrogen to form the product is called Oxidation, reaction., e.g. 2Mg + O2 → 2MgO, , Here O2 combines with magnesium (Mg) to form, magnesium oxide (MgO), MgH2 → Mg + H2, Here MgH2 loses hydrogen., , , , , Reduction reaction: The chemical reaction in, which a reactant combines with hydrogen or, loses oxygen to form product is called reduction, reaction., e.g. C + 2H2 → CH4, Here carbon combines with hydrogen and forms, compound CH4 (Methane)., HgO ∆ Hg + O, , , , 2, , Here Mercuric oxide loses oxygen to form, mercury., , , , *(5) Explain the similarity and difference in two, events, namely adding NaOH to water and, adding CaO to water., Ans. Similarity:, (i), , (ii), , Both NaOH and CaO dissolve in water. NaOH, dissolves in water forming aqueous NaOH, while CaO reacts with water to form Calcium, Hydroxide [Ca(OH)2], Both are bitter in taste, and turn red litmus to, blue. i.e. both are basic in nature.

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Chemical Reactions and Equations, , NaOH(s) + H2O(l) → NaOH(aq) + Heat, CaO(s) + H2O(l) → Ca(OH)2(aq) + Heat, Both are exothermic reactions., (iii) After adding water, the properties of aqueous, solution of sodium hydroxide (NaOH) and, aqueous solution of calcium oxide (CaO) are, different., , Difference:, (i), , (ii), (iii), , Sodium Hydroxide (NaOH) is highly soluble, in water, whereas calcium oxide (CaO) is less, soluble in water., Aqueous solution of NaOH is monovalent,, whereas aqueous solution of Ca(OH)2 is divalent., As NaOH is highly soluble in water, makes strong, base whereas Ca(OH)2 is a weak base., , (6) , Explain two ways by which food industries, prevent rancidity., Ans. , (i) , Rancidity can be prevented by adding, antioxidants to food containing fat and oil., (ii) , It can be prevented by packaging fat and oil, containing food in airtight containers or by filling, the extra space by nitrogen gas., (7) Will it be possible for you to decompose water, by heat or light energy? If you pass current, from a 6 volt battery, is decomposition of water, possible?, Ans. , Pure water cannot be decomposed by heat,, light or by passing current from a 6 volt battery., However, if few drops of acid is added to water it, can be decomposed by passing current from a 6, volt battery., *(8) A. Identify from the following reactions the, reactants that undergo oxidation and reduction., Ans. , (a) , Fe + S → FeS⋅Fe undergoes oxidation and S, undergoes reduction., (b) , 2Ag2O → 4Ag + O2↑ - Silver oxide undergoes, reduction., (c) , 2Mg + O2 → 2MgO - Mg undergoes oxidation., (d) , NiO + H2 → Ni + H2O Redox reaction., In the above equation, reactant NiO (Nickel oxide), loses oxygen, thus undergoes reduction whereas, H2 (Hydrogen gas) gains oxygen forming water, and undergoes oxidation., oxidation, NiO + H2 → Ni + H2O, Reduction, , 67, , (8) B. Classify the following reactions into different, types., (a) AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq), Ans. , (Double displacement reaction) Precipitation, reaction, (b) CaO(s) + H2O(l) → Ca(OH)2(aq) + Heat, Ans. Combination reaction [Exothermic reaction.], ∆ 2KCl + 3O, (c) 2KClO, 3(s), , (aq), , 2(g), , Ans. Decomposition reaction/Thermal decomposition, reaction., (d) CuO + H2 heat Cu + H2O, Ans. Displacement reaction [Redox reaction.], (e) CuSO4(aq) + Pb(s) → PbSO4(aq) + Cu(s), Ans. Displacement reaction, (f) 2H2O2 UV, 2H2O + O2(l), light, Ans. Photolytic decomposition is a chemical reaction, in which molecules are broken down into smaller, units through the absorption of light., (g) BaS(aq) + ZnSO4(aq) → BaSO4(s) + ZnS(aq), Ans. Double displacement reaction., *(9) Balance the following equation stepwise., (a) H2S2O7(l) + H2O(l) → H2SO4(l), Ans. Step 1: Write the equation and then count the, number of atoms of each element on the R.H.S., and L.H.S. of the arrow., H2S2O7 + H2O → H2SO4 ......................................(1), Elements, , No. of atoms, in reactants, (L.H.S.), , No. of atoms, in products, (R.H.S.), , H, , 4, , 2, , S, , 2, , 1, , O, , 8, , 4, ,  tep 2: Check whether no. of atoms on both sides, S, is equal. It is seen from the above table number of, atoms of all the elements on the two sides are not, the same., Step 3: To balance the number of atoms on both, sides, we write coefficient 2 for H2SO4., Now the equation becomes., H2S2O7 (l) + H2O (l) → 2H2SO4 (l) .......................(2), , Now see that the number of atoms of each, element are equal on both sides.

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Master Key Science and Technology Part-1 (Std. X), , 68, , No. of atoms No. of atoms, in reactants, in products, H, 4, 4, S, 2, 2, O, 8, 8, Hence, the equation (2) is a balanced equation., Elements, , , , (b) SO2(g) + H2S(g) → S(s) + H2O(l), , Ans. Step 1: Write the equation given., SO2 + H2S→ S + H2O ......................................(1), Count the number of atoms of each element in, products and reactants., Elements, , No. of atoms, in reactants, , No. of atoms, in products, , S, O, H, , 2, 2, 2, , 1, 1, 2, , Step 2: It is seen from the above table, number of, atoms on both sides are not the same. It means, that equation (1) needs to be balanced., Step 3: We will start with oxygen. On product, side we need 2 atoms of oxygen, hence coefficient, 2 for H2O. Now the equation will be, ∴ SO2 + H2S → S + 2H2O ............................(2), Again count the number of atoms of elements on, both the sides., Elements, , No. of atoms, in reactants, , No. of atoms, in products, , S, , 2, , 1, , O, , 2, , 2, , H, , 2, , 4, , , Still the equation is unbalanced. To balance, sulphur and hydrogen atoms on reactant side,, write coefficient 2 for H2S, ∴ SO2 + 2H2S → S + 2H2O ............................(3), Again count the number of atoms on both the, sides., Elements, , No. of atoms, in reactants, , No. of atoms, in products, , S, O, H, , 3, 2, 4, , 1, 2, 4, , Now if we count number of atoms of hydrogen, and oxygen, they are equal. But number of, atoms of sulphur is not balanced. Now there are, 3 atoms of sulphur on reactant side, whereas, , on product sides it is only one atom. Hence, we, write coefficient 3 for sulphur on R.H.S. Now the, equation becomes., SO2 + 2H2S → 3S + 2H2O ............................(4), Again count number, of atoms in equation (4), Elements, , No. of atoms, in reactants, , No. of atoms, in products, , S, O, H, , 3, 2, 4, , 3, 2, 4, , It means equation (4) is a balanced equation., (c) Ag(s) + HCl(aq) → AgCl ↓+ H2↑, Ans. , Step 1: Write the chemical equation and count the, numbers of atoms in reactants and products., Ag + HCl→ AgCl + H2 ......................................(1), Elements, , No. of atoms, in reactants, , No. of atoms, in products, , Ag, 1, 1, H, 1, 2, Cl, 1, 1, , Step 2: From the above table, we see the number, of atoms of hydrogen on reactant side is not, balanced. So, to balance hydrogen atom, we write, coefficient 2 for HCl., Ag + 2HCl→ AgCl + H2 ......................................(2), Again count the number of atoms of elements on, both sides., Elements, , No. of atoms, in reactants, , No. of atoms, in products, , Ag, H, Cl, , 1, 2, 2, , 1, 2, 1, , , Step 3: Now to balance chlorine, we write, coefficient 2 for AgCl, Ag + 2HCl→ 2AgCl + H2 ......................................(3), Again count the number of atoms of elements on, both sides., Elements, , No. of atoms, in reactants, , No. of atoms, in products, , Ag, , 1, , 2, , H, , 2, , 2, , Cl, , 2, , 2, , , Step 4: Atoms of elements, hydrogen and, chlorine are now balanced. Number of atoms of, Ag on product side is 2, and on reactant side is, 1 to balance number of atoms of Ag on reactant

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Chemical Reactions and Equations, , 69, , side, we write coefficient 2 for Ag., , Elements, , No. of atoms, in reactants, , No. of atoms, in products, , Now count number of atoms in equation (4) again, , Na, , 2, , 2, , Elements, , No. of atoms, in reactants, , No. of atoms, in products, , O, , 6, , 6, , H, , 4, , 4, , Ag, , 2, , 2, , S, , 1, , 1, , H, , 2, , 2, , Cl, , 2, , 2, , ∴2Ag + 2HCl→ 2AgCl + H2 ...........................(4), , It means equation (4) is a balanced equation., (d), , NaOH(aq) + H2SO4(aq) → Na2SO4 (aq) + H2O(l)., , Ans. Step 1: Write the equation given., NaOH + H2SO4→ Na2SO4 + H2O , , ...(1), , Count the number of atoms of each element in, products and reactants., Elements, , No. of atoms, in reactants, , No. of atoms, in products, , Na, , 1, , 2, , O, , 5, , 5, , H, , 3, , 2, , S, , 1, , 1, , The number of oxygen and sulphur atoms are, equal on both sides. Hence, we will balance Na, atoms on both sides., , Step 2: To balance Na atom we write coefficient 2, for NaOH., 2NaOH + H2SO4→ Na2SO4 + H2O, , ...(2), , , Step 3: Count the number of Na atoms, H atoms,, O atoms and S atoms on both sides., Elements, , No. of atoms, in reactants, , No. of atoms, in products, , Na, , 2, , 2, , O, , 6, , 5, , H, , 4, , 2, , S, , 1, , 1, , Step 4: To balance H, we write coefficient 2 for, H2O on the product side. Hence the equation, becomes., 2NaOH + H2SO4→ Na2SO4 + 2H2O, , ...(3), , Again count the number of atoms of elements on, both sides., ,  ow the number of atoms of elements are equal, N, on both sides. Hence, equation (3) is a balanced, equation., *(e) Write down the steps in balancing the equation., (Use your Brain Power! Textbook page no.35), N2(g) + H2(g) → NH3(g)., , Ans. Step 1: Write the equation given., N2 + H2 → NH3 ...(1), Now count the number of atoms of each elements, on reactants and product side., Elements, , No. of atoms, of reactants, , No. of atoms, of product, , N, , 2, , 1, , H, , 2, , 3, ,  tep 2: Number of atoms of elements on both side, S, is not the same. So we need to balance it. First we, will balance Nitrogen by writing coefficient 2 for, the product NH3., N2 + H2 → 2NH3 ...(2), , Step 3: Again count the number of atoms of, elements on both sides., Elements, , No. of atoms, in reactants, , No. of atoms, in products, , N, , 2, , 2, , H, , 2, , 6, , Step 4: Number of atoms of Nitrogen is now, balanced on both sides, but number of atoms of, hydrogen is not balanced. Therefore, we write, coefficient 3 for Hydrogen on reactant side. The, equation becomes:, N2 + 3H2 → 2NH3 ...(3), Now, count the number of atoms on both side., Elements, , No. of atoms, in reactants, , No. of atoms, in products, , N, , 2, , 2, , H, , 6, , 6, , It means the equation (3) is a balanced equation.

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Master Key Science and Technology Part-1 (Std. X), , 70, , *(f) Write down a balanced chemical equation for, the following reaction. (Use your Brain Power!, Textbook page no.35), , *(i) What is the difference in the process of, dissolution and a chemical reaction ? (Use your, Brain Power; Textbook page no.39), , , Calcium chloride + Sulphuric acid → Calcium, sulphate + Hydrogen chloride, Ans. CaCl2(s) + H2SO4 (aq) → CaSO4(s) + 2HCl(aq), Calcium, Sulphuric, Calcium, hydrogen, chloride, acid, sulphate, chloride, , Ans. , In dissolution, solute dissolves in the solvent, without reacting chemically with it. In chemical, reaction, reactants will react chemically with, each other and new products are formed. The, bonds in the reactants are broken and new bonds, are formed., , *(g) Write down the physical states of reactants and, products in following reactions. (Use your Brain, Power! Textbook page no.35), , (j) , Does a new substance form when a solute, dissolves in a solvent ? (Use your Brain Power;, Textbook page no.39), , (i) SO2 + 2H2S → 3S + 2H2O, , Ans. , No new substance is formed when a solute, dissolves in a solvent., , Ans. SO2(g) + 2H2S(g) → 3S(s) + 2H2O(l), (ii) 2Ag + 2HCl → 2AgCl + H2, , *(k) , Take into account the time required for, following processes. Classify them into two, groups and give titles to the groups. (Can you, tell; Textbook page no.39), , Ans. 2Ag(s) + 2HCl(l) → 2AgCl(s) + H2(g), *(h) , Is it possible to produce hydrogen by, decomposition of water by means of heat,, electricity or light ? (Can you recall; Textbook page, no.37), , (i), , (ii) Iron article undergoes rusting., , Ans. , Yes. It is possible to produce hydrogen by, decomposition of water. Water decomposes, into hydrogen and oxygen gases on passing, electric current through acidulated water. This, decomposition takes place by means of electrical, energy., 2H O(l) electric current 2H ↑ + O ↑, 2, , 2, , (iii) Erosion of rocks takes place to form soil., (iv) Alcohol is formed on mixing yeast in glucose, solution under proper conditions., (v) Effervescence is formed on adding baking, soda into a test tube containing dilute acid., (vi) A white precipitate is formed on adding, dilute sulphuric acid to barium chloride solution., , 2, , It is called electrolysis., Ans., , Cooking gas starts burning on ignition ., , Fast reaction, (i), , Cooking gas starts burning on ignition ., , Slow reaction, (i), , Iron article undergoes rusting., , (ii) Effervescence is formed on adding baking (ii), soda into a test tube containing dilute acid., , Erosion of rocks takes place to form soil., , (iii) , A white precipitate is formed on adding (iii) , Alcohol is formed on mixing yeast in glucose, dilute sulphuric acid to barium chloride, solution under proper condition., solution., *(l) , Complete the following reactions and give, names of the products. (Use your Brain Power;, Textbook page no.38), , *(10) Observe the following picture write down the, chemical reaction with explanation., , (i) , CuSO4(aq) + Fe (s) → ............... + ..............., Ans. CuSO4(aq) + Fe (s) → FeSO4(aq) + Cu(s), Iron, copper, sulphate, , O2, , Collected rust, Fe2O3.H2O, , Cathode, O2 + 2H2O + 4e–, , (ii) , CuSO4(aq) + Pb(s) → ............... + ..............., Ans. CuSO4(aq) + Pb(s) → PbSO4(aq) + Cu(s), Lead, copper, sulphate, , O2 + 4H+ + 4e–, , Water drop, Fe2+ (aq), , e–, , Anode, 4OH, , –, , Fe, , Fe2+ + 2e–, , 2H2O, Iron, , Ans. 

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Master Key Science and Technology Part-1 (Std. X), , 72, , Q.4.1. Describe an activity to observe what happens, when quick lime is added to water taken in a, beaker. State two important observations and, name the type of reaction taking place., , (a) , Photochemical decomposition reaction: Those, reaction in which a compound breaks down into, simple substances in the presence of light are, called Photochemical decomposition reactions., The colour of salt will change from white to grey., sunlight, (b) 2AgCl(s), 2Ag(s) + Cl2(g), (c), , Silver chloride is used in photography., , (2), , Material Required: Quicklime (Calcium oxide),, water, beaker, Procedure:, , Check the given materials and chemicals, required for a chemical reaction and answer the, following questions., , , , Material - Copper wire, iron nail, beaker or big, test tube., , Take 5g of calcium oxide in a beaker., , , , Chemicals - Ferrous sulphate solution and, copper sulphate solutions., , (a), , Which reaction will you study with the help, of above material and solutions? Draw the, diagram of the experiment arrangement., , (b), , How would identify that the reaction is carried, out?, , , Ans. , Aim: To observe what happens when quicklime, is added to water taken in a beaker., , (i), , (c) Mention one commercial use of this salt., Ans. , , (ii) Add water to it slowly., (iii), , Touch the beaker., , (iv), , Note down the observations., Observation: The beaker becomes hot Calcium, oxide reacts with water vigorously, to form calcium hydroxide with, the evolution of heat., CaO(s), + H2O(l) → Ca(OH)2(aq) + Heat, Quick lime Water Slaked lime, , (c), , Write a balanced chemical equation for the, reaction., Ans. , , Conclusion: The reaction between CaO and H2O, is a combination reaction. It is an, exothermic reaction because heat is, evolved., , Cork, Thread, Test tube, , Q.4.2. Activity based questions:, (1), , (a) Identify the type of chemical reaction that will, take place and define it. How will the colour of, the salt change?, (b) Write the chemical equation of the reaction that, takes place., , Copper wire, , Iron nail, , The following diagram displays a chemical, reaction. Observe carefully and answer the, following questions., , , , Test tube, stand, , CuSO4, solution, , FeSO4, solution, , (a), , Reaction of metal with solution of salts of other, metals (Displacement reaction), , (b), , The reaction is carried out in the test tube where, iron nail is kept in copper sulphate solution. This, is because iron being more reactive than copper,, iron displaces copper from copper sulphate, solution. The colour of copper sulphate changes, from blue to light green due to formation of, ferrous sulphate and copper gets deposited on, iron nail making it reddish brown in colour., Fe(s), , + CuSO4(aq) → FeSO4(aq) + Cu(s), , Iron Copper, sulphate, , Ferrous, sulphate, , Copper

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Chemical Reactions and Equations, , 73, , Q.4.3.Answer the following questions in detail:, (1) What is the importance of a chemical equation?, OR Give the significance of a chemical equation., , (iv) It does not react with water or any of the gases, present in the atmosphere. Common acids do not, have any effect on gold., , Ans. The significance of a chemical equation is as, follows:, , (v)  Thus, gold ornaments do not corrode., , (i) Chemical equation describes briefly the names of, the reactants, the products and the state of matter., (ii) It explains the specific conditions of temperature,, pressure, catalyst, etc. under which the reaction, takes place., (iii) It also states whether energy is evolved or needs, to be supplied., , (5) What are the factors which affect the rate of a, chemical reaction., Ans. The rate of reaction depends upon the following, factors., (i) Nature of the reactants:, (a), , (iv) Chemical equation also represents the relative, number of molecules and moles of reactants and, products., (v) The relative masses of the reactants and products, can also be calculated using a chemical equation., Thus, a chemical equation describes a chemical, reaction qualitatively as well as quantitatively., (2) What is rancidity? Mention only two ways by, which rancidity can be prevented., Ans. The process in which taste and smell of food, gets spoiled is called rancidity. It happens due to, oxidation., Prevention of rancidity., (i) Antioxidants are added to fatty acids to prevent, oxidation., (ii) , Food should be kept in airtight containers in, refrigerator. Chips are packed in presence of, nitrogen gas which prevents spoilage by oxidation., (3) Give four uses of decomposition reaction., Ans., (i), Decomposition reaction helps all the living, , matter to return back to nature, after death., (ii), , Management of garbage., , (iii), , Photosynthesis → decomposition of water., , (iv), , Photography → decomposition of silver bromide., , (v), ,  sed in chemical industry to obtain elements, U, from complex compounds., , (vi), ,  ecomposition of agricultural waste leads to, D, formation of compost., , (4) What is corrosion? Do gold ornaments corrode?, Justify., Ans., (i) The slow process of decay or destruction of metal, due to the effect of air, moisture and acids on it is, known as corrosion., (ii) Gold ornaments do not get corroded., (iii), , Gold is a noble metal and is least reactive., , 2NO(g) + O2(g) → 2NO2(g), , Fast, , 2CO(g) + O2(g) → 2CO2(g), , Slow, ,  hese reactions appear to be similar but the first, T, is fast while second is slow. This is due to the, difference in the amounts of energies which are, required for breaking up of different bonds and, different amounts of energies are released in the, formation of different bonds., (b), , 2Al + 6HCl → 2AlCl3 + 3H2 ↑, Zn + 2HCl → ZnCl2 + H2 ↑,  l and Zn react with dilute hydrochloric acid,, A, H2 gas is liberated and water soluble salts of, the metals are formed. However, the reaction of, aluminium metal takes place faster as compared, to zinc metal. The nature of metal is responsible, for this difference. Al is more reactive than, Zn. Therefore, the rate of reaction of Al with, hydrochloric acid is higher than that of zinc., Nature of reactivity of reactants influences the, rate of a chemical reaction., , (ii) Concentration of the reactants:, , Greater the concentration of the reactants, faster, is the reaction conversely, as the concentrations, of the reactants decrease the rate of reaction also, decreases. Dilute HCl reacts slowly with CaCO3, and thereby, CaCO3 disappears slowly and CO2, also liberates slowly on the other hand the reaction, with concentrated HCl takes place rapidly and, CaCO3 disappears fast. Concentrated acid reacts, faster than dilute acid, which means that rate of, reaction is proportional to the concentration of, reactants., (iii) Temperature:, The rate of reaction increases with increase of, temperature. In most of the cases, the rate of, reaction becomes nearly double for 10K rise of, temperature. In some cases, reactions do not, take place at room temperature but take place at, higher temperature.

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Master Key Science and Technology Part-1 (Std. X), , 74, , (iv) Presence of catalyst:, , A catalyst generally increases the speed of a, reaction without itself being consumed in the, reaction. In case of reverse reaction, a catalyst, helps to attain the equilibrium quickly without, disturbing the state of equilibrium., (v) Size of the particles of reactants:, , , For a reaction involving a solid reactant or, catalyst, smaller the particle size, greater is the, surface area, and faster is the reaction., (vi) Presence of light:, , Some reactions do not take place in the dark, but can take place in the presence of light like, photosynthesis or photochemical reactions., , vvv, , ASSIGNMENT - 3, Time : 1 Hr., , (1), (2), (3), , Marks : 20, , Q.1. (A) Fill in the blanks: , , (3), , The reaction CaCO3 → CaO + CO2 is a, , reaction., , The chemical reaction in which heat is absorbed is called, A redox reaction takes place during, , reaction., , respiration., , Q.1. (B) Choose and write the correct options:, (1), , (2), ,  ecause of the formation of which of the following, lime water turns milky when carbon dioxide is, B, passed in it?, (a) Calcium Carbonate (b) Calcium bicarbonate (c) Calcium hydroxide (d) Sodium Carbonate, , (2), , When sulphuric acid is poured over zinc, which of the following gas is formed?, (a) Sulphur dioxide, , , , (b) Hydrogen, , (c) Oxygen, , (d) Zinc dioxide, , Q.2. Answer the following: (Any 2), , (4), , (1), , What happens when a piece of zinc metal is added to copper sulphate solution?, , (2), , Write a note on Endothermic Reaction., , (3), , Distinguish between physical change and chemical change., , , , Q.3. Answer the following: (Any 2), , (6), , (1), , Explain the term reactant and product giving examples., , (2), , Balance the given Chemical Equations:, , , , (a), , , , (b), , (3), , HNO3 + Ca(OH)2 → Ca(NO3)2 + H2O, heat, FeSO4, Fe2O3 + SO2 + SO3, , Identify the type of reaction:, , , , (a), , Fe(s) + CuSO4(aq) → FeSO4(aq) + Cu (s), , , , (b), , 2H2(g) + O2(g) → 2H2O(l), , , , (c), , Na2SO4(aq) + BaCl2(aq) → BaSO4(s) + 2NaCl(aq), , , , Q.4. Answer the following: (Any 1), , (5), , (1), , Explain the types of reaction with reference to oxygen and hydrogen. Illustrate with examples., , (2), , (a), , , , (b), , What is rancidity? Mention two ways by which rancidity can be prevented., What is corrosion? Do gold ornaments corrode? Justify your answer., , vvv

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4, , Effects of Electric Current, , Points to Remember:, zz Electric current is the flow of electrons in a conductor or amount of electric charge flowing through a, , particular cross sectional area in unit time., , zz A continuous path consisting of conducting wires and other resistances between the terminals of a battery,, , along which an electric current flows, is called an electric circuit., , zz The production of heat energy in a conductor by electric current flowing through it is called the heating effect, , of the electric current., , zz Force that exists in the space around a magnet of a current carrying conductor is called magnetic field., zz Right hand thumb rule is a convenient rule for finding out direction of the magnetic field produced by a, , current flowing through an electrical conductor., , zz When a copper wire with a resistive coating is wound in a chain of loop (like a spring) is called solenoid., zz Fleming’s left hand rule: According to this rule, the left hand’s thumb, index finger and the middle finger are, , stretched so as to be perpendicular to each other. If the index finger is in the direction of the magnetic field,, and the middle finger points in the direction of current, then the direction of thumb is the direction of force, on the conductor., , zz A device which converts electrical energy into mechanical energy is called an electric motor., zz The process by which a changing magnetic field in a conductor induces a current in another conductor is, , called electromagnetic induction., , zz Faraday’s law of electromagnetic induction states that whenever the number of magnetic lines of force, , passing through the coil changes, current is induced in the coil. The current produced in the coil is called as, Induced current., , zz Fleming’s Right hand rule: According to this rule when thumb, index finger and middle finger are stretched, , in the direction perpendicular to each other then the position of thumb indicates direction of motion of, conductor, the index finger indicates direction of magnetic field and middle finger indicates the direction of, induced current., , zz The device which converts mechanical energy into electrical energy is called an electric generator., zz A current that flows only in one direction is called direct current (D.C.)., zz A current whose direction reverses periodically with time is called alternating current. (A.C.), zz A generator which converts mechanical energy into electrical energy in the form of alternating current is, , called AC generator., , zz A generator which converts mechanical energy into electrical energy in the form of direct current is called, , DC generator., , MASTER KEY QUESTION SET - 4, Q.1. (A) 1. Fill in the blanks:, (1), ,  he generation of heat in a resistor when an, T, electric current is passed through the circuit is, called, ., (75), , (2), ,  he electric bulb consists of a filament made up, T, of, ., , (3), ,  he potential difference between a live wire and, T, neutral wire is, ., , (4), , I n the electric bill, the electricity usage is, expressed in, .

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Master Key Science and Technology Part-1 (Std. X), , 76, , (5), , 1 kWh electrical energy used, is also called, ., , (6), ,  hen a copper wire with a resistive coating, W, is wound in a chain of loop (like a spring), it is, called, ., , Q.1. (A) 2. Find the odd word out., *(1) Fuse wire, insulator, rubber gloves, generator, Ans. Generator (All others are concerned with safety, measures.), , (7), ,  he device which converts electrical energy into, T, mechanical energy is called, ., , *(2) Voltmeter, Ammeter, Thermometer, Galvanometer, Ans. Thermometer (All others are based on magnetic, effect of electric current.), , (8), , I n a straight current carrying conductor, if, the current through the wire is increased, the, intensity of magnetic field, ., , (9), , The right hand thumb rule is also called as, ., , *(3) Loudspeaker, Microphone, Electric motor, Magnet, Ans. Magnet (All others are related to magnet effect of, an electric current.), , (10), ,  he magnetic lines of force inside the solenoid are, T, parallel to each other, which means the magnetic, field is, ., , (11), , The frequency of AC current in India is, , ., , (12), ,  he direction of force acting on a current carrying, T, conductor depends on, and, ., , (13), ,  he force acting on a current carrying conductor, T, is maximum when the direction of current is, to the direction of magnetic field., , (14), ,  galvanometer is a sensitive device which works, A, on the same principle as that of an, ., , (15), , The commercial unit of power is, , (16), , Generator works on the principle of, , (17), , The electric oven is an example of, , (18), , The SI unit of heat energy is, , (19), ,  he electric current that reverses its direction, T, periodically with time is called, ., , (20), ,  hen the potential difference between two, W, terminals becomes, , the electric current, stops flowing., , ., ., ., ., , (4), Armature coil, Brushes, Magnet, Direct current, Ans. Direct current (All others are the parts of electric, motor.), (5), ,  efrigerator, Electric fan, Mixer, Electric water, R, heater, Ans. Electric water heater (All others are applications, of magnetic effect of current.), (6), Fuse wire, Neutral wire, Earthing wire, Live wire, Ans. Fuse wire (All others are concerned with electric, supply wires.), (7), ,  lectric heater, Electric bulb, Electric fuse, Electric, E, press, Ans. Electric fuse (All others are wires with high, melting points.), (8), ,  lectric oven, Electric bulb, Electric press, Electric, E, fan, Ans. Electric fan (All others are based on heating effect, of electric current.), (9),  lectric bell, Electric fuse, Solenoid, Microphone, E, Ans. Electric fuse (All others work on the principle of, magnetic effect of electric current.), Q.1. (A) 3. Answer the following in one sentence:, , (21), , The, is used in the electric circuit to, protect it from loss due to excessive current., , (22), , When live wire and neutral wire come in contact,, takes place., , (1), What is heating effect of an electric current?, Ans. , The generation of heat in a metallic conductor by, the effect of electric current flowing through it is, called heating effect of electric current., , (23), , The unit of electric resistance is, , (2), , ., , Ans. , (1) heating effect (2) Tungsten (3) 220 V, (4) kWh (5) 1 unit (6) Solenoid (7) Electric motor (8), increases (9) Maxwell’s cork-screw rule (10) uniform, (11) 50Hz (12) direction of the current, direction of, magnetic field (13) perpendicular (14) electric motor, (15) kilowatt (kW) (16) electromagnetic induction, (17) heating effect of electric current (18) joule, (19) Alternating Current (20) zero (21) Electric fuse, (22) Short circuit (23) ohm (Ω), ,  hat are the appliances that work on the, W, principle of heating effect of electric current?, Ans. , Electric bulb, electric press, electric heater, electric, oven, electric fuse, electric stove etc. are some, appliances that work on the principle of heating, effect of electric current., (3), What is solenoid?, Ans. When a copper wire with a resistive coating, is wound in a chain of loop (like a spring), it is, called solenoid.

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Effects of Electric Current, , (4), What is electric motor?, Ans. The device that converts electrical energy into, mechanical energy is called an electric motor., (5), ,  hat do you understand by the term ‘Direct, W, current (D.C.)?, Ans. , The electric current that flows in one direction,, i.e. from higher potential to lower potential is, called as direct current (D.C.)., (6), What is alternating current (A.C.)?, Ans. , The electric current that reverses its direction, periodically with time is called an alternating, current (A.C.)., , 77, , (16) Write the relation between watt hr and joule., Ans. 1 watt hr = 3.6 × 103 J, (17) When will the induced current in the electrical, conductor (coil) be maximum?, Ans. It will be maximum when the direction of motion, of the electric conductor is perpendicular to the, magnetic field., Q.1. (A) 4. Match the columns:, (1), , Column A, , Column B, , (1), , Fleming, , (a), , (7), Name two types of electric current., Ans. , Direct current and Alternating current are two, types of electric current., , Relation, between electric, energy and, magnetism, , (2), , Faraday, , (b), , (8), Name the types of electric generator., Ans. , AC generator and DC generator are the two types, of generators., ,  aw of, L, electromagnetic, induction, , (3), , Oersted, , (c), , Magnetic field,, direction of, electric current, direction of, motion of a, conductor., , (9), ,  ame the three types of wires or cables used in, N, domestic electric circuit., Ans. , There are three types of wires or cables used in, domestic electric circuit namely live wire (phase, wire), neutral wire and earth wire., , Ans. (1 – c), (2 – b), (3 – a), , (10), , (2), ,  hich electrical appliance converts electrical, W, energy into mechanical energy?, Ans. , The electric motor converts electrical energy into, mechanical energy.,  hich appliance converts mechanical energy, W, into electrical energy?, Ans. , The electric generator converts mechanical, energy into electrical energy., , Column A, , Column B, , (1), , Earthing wire, , (a), , Electric motor, , (2), , Overloading, , (b), , Electric, generator, , (3), , Electromagnetic, force, , (c), , Protection, against electric, shock, , (4), , Electromagnetic, induction, , (d), ,  xcessive, E, electric current, , (11), , (12), ,  hat is the frequency of alternating current in, W, India?, Ans. In India, the frequency of alternating current, (A.C.) is 50 Hz., (13) What do you mean by induced current?, Ans. , The electric current that is set up in the coil due, to the motion of either coil or a magnet (magnetic, field) is called induced current., (14), ,  ame the two components or devices used for, N, the safety of electrical gadgets?, Ans. Electric fuse and earth wire are the two devices, used for the safety of electrical components or, gadgets., (15), ,  rite the relation between kilowatt hour and, W, joule., Ans. 1 kWh = 3.6 × 106 J, , Ans. (1 – c), (2 – d), (3 – a), (4 – b), (3), , Column A, , Column B, , (1), , Right, hand, thumb rule, , (a), , Magnetic effect, of electric current, , (2), , Hans Oersted, , (b), , Properties of a, bar magnet, , (3), , Solenoid, , (c), , Do not intersect, each other, , (4), , Magnetic lines, of force, , (d), ,  irection of, D, current and, magnetic field, , Ans. (1 – d), (2 – a), (3 – b), (4 – c)

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Master Key Science and Technology Part-1 (Std. X), , 78, , (4), , Column A, , Column B, , (1), , Electric, generator, , (a), , Transmitted over, long distances, , (2), , Electric, motor, , (b), , Unidirectional, flow of current, , (3), , Direct, current, , (c), ,  onverting, C, mechanical energy, into electrical, energy, , (4), , Alternating, current, , (d), ,  sed in mixers,, U, refrigerator etc., , Ans. (1 – c), (2 – d), (3 – b), (4 – a), Q.1. (A) 5. State whether the following statement, ‘True‘ or ‘False’. If false; write the correct, statement for the same., (1), , The SI unit of electric charge is coulomb (C)., , (2), , The SI unit of electric current is volt (V)., , (3), ,  esistivity of a conductor does not depend upon, R, its temperature., , (4), ,  he filament in the electric bulb is made of a wire, T, with high melting point., , (5), ,  he direct current always flows in one direction, T, only., , (6), , The electric motor is used to generate electricity., , (7), ,  uring lightning, all the electrical appliances, D, must be switched off., , (8), ,  any high power rating electrical appliances can, M, be connected to a circuit at a time., , (9), , Direct current has a frequency of 50 Hz in India., , (10), , Electric power (P) = VI, , (11), , According to Ohm’s law, V = I × t, , (12), ,  he production of heat in a wire when connected, T, to electric circuit is called heating effect of an, electric current., , (13), , 1 kWh = 3.6 × 10 J, , (14), , 1 kWh electric unit = 10 units., , (15), ,  he device which converts electrical energy to, T, mechanical energy is called an electric generator., , (16), ,  lternating current is perfectly used for domestic, A, electric circuits., , (17), ,  hen more electrical appliances work together, W, in a circuit, it results in a short circuit., , (18), ,  hen a live wire and a neutral wire come in, W, contact with each other, it may cause short circuit., , 9, , Ans. , (1) True (2) False. SI unit of electric current is ampere., (A) (3) False. Resistivity depends on temperature., (4) True (5) True (6) False. Electric generator is used to, generate electricity. (7) True (8) False. Many high power, rating electrical appliances should not be connected, simultaneously in the electric circuit. (9) False. Direct, current flows in one direction. (10) True (11) False., According to Ohm’s law, V = I × R (12) True (13) False., 1 kWh = 3.6 × 106 J (14) False. 1 kWh energy is 1 unit., (15) False. The device which converts electric energy, into mechanical energy is called electric motor., (16) True (17) False. When more appliances are, connected simultaneously, it results in overloading,, causing fire. (18) True., Q.1. (B) Choose and write the correct option:, (1), ,  he direction of magnetic field due to electric, T, current is decided by, (a), (b), (c), (d), , (2), , Right hand thumb rule, Fleming’s left hand rule, Fleming’s right hand rule, None of the above, ,  he device which converts mechanical energy, T, into electrical energy is called, (a) Electric bell, (c) Electric fuse, , (3), , 1 mA =, (a) 10–6A, , (4), , (b) Electric generator, (d) Electric motor, , (b) 106A, , (c) 10–3A, , (d) 103A, , 1 watt =, (a) 1 joule/1 second, (b) 1000 J/s, (c) 1 calorie/1 second (d) 1 joule.second, , (5), , Electric power (P) =, (a) V.t/Q, , (6), , (7), , (8), , (b) Q.t/V, , (c) V.Q.t, , (d) V.Q/t, , 1 kWh =, (a) 36 × 106 J, , (b) 3.6 × 106 J, , (c) 3.6 × 109 J, , (d) 36 × 109 J, ,  he deflection of the pointer of, T, either side of zero mark., , is on, , (a) Voltmeter, , (b) Ammeter, , (c) Galvanometer, , (d) Thermometer, , If the potential difference across a wire is 2 V and, the current through the wire is 1 A, the electric, power is, ., (a) 4 W, , (b), , 1, 2, , W, , (c) 2 W, , (d), , 1, 4, , W

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Effects of Electric Current, , (9), , At the centre of which of the following four, circular rings is the magnetic field strongest, for, equal magnitude of current?, b, , (1) 2 I, , (a) 3, , (11), , (12), , (2) 2 I, , a, , (3) 2 I, , (10), , 79, , (4) 2 I, , (b) 1, , (13), , d, , G, , c, , (c) 4, , (a) The galvanometer will not deflect., (b) The galvanometer will deflect in one, direction when the key is inserted and in the, opposite direction when the key is removed., (c) The galvanometer will deflect in one direction, when the key is inserted and remain still, when the key is removed., (d) The galvanometer will deflect in one, direction when the key is inserted and in the, same direction when the key is removed., , (d) 2, ,  hich of the following is used to find the, W, direction of the magnetic lines of force around a, conductor?, (a) Bar magnet, , (b) Magnetic needle, , (c) Disc magnet, , (d) Spherical magnet, , A galvanometer can be used to measure, ., (a) only DC, , (b) only AC, , (c) both DC and AC, , (d) potential difference, , (14), , (a), , Conductor, loop, , +, A, –, , (15), , R, A, , A, , Magnetic, lines, , Conductor, loop, N, , S, , Cardboard, R, , B, , Magnetic field in A is stronger., Magnetic field in B is stronger., Magnetic fields in A and B are same., Magnetic fields in A and B are weaker., , V, , V, , +, , (b), , –, A, +, , +, , V, , –, , +, +, , (d), , –, A, +, , –, , –, , R, –, , R, +, , Cardboard, , –, , R, , +, , (c), , S, , A, , +, A, –, –, , N, , (a), (b), (c), (d), , Out of the four circuits shown, for studying the, dependence of current on the potential difference, across a resistor, the correct circuit is, +, , Write the correct option by observing the figures:, Magnetic, lines, , In the arrangement shown in figure, there are, two coils wound on a non-conducting cylindrical, rod. Initially the key is not inserted. Then the key, is inserted and later removed. Then,, , –, , R, +, , V, , –, , The intensity of magnetic field is expressed in, ., (a) ampere, , (b) volt, , (c) coulomb, , (d) oersted, , Ans. (1) (a) Right hand thumb rule (2) (b) Electric generator, (3) (c) 10–3A (4) (a) 1 joule / 1 second (5) (d) V.Q/t, (6) (b) 3.6 × 106 J (7) (c) Galvanometer (8) (c) 2 W (9), (d) 2 (10) (b) Magnetic needle (11) (c) both DC and, AC (12) (b) Magnetic field in B is stronger (13) (b), The galvanometer will deflect in one direction when, the key is inserted and in the opposite direction when, the key is removed. (14) (c) (15) (d) oersted, Q.2.1. Solve the following:, Type: A, , , , H = I2Rt joules, V2t, H=, joules, R, , P = I2R watt, V2, P=, watt, R, , H = VIt joules, , P = VI watt

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Master Key Science and Technology Part-1 (Std. X), , 80, , *(1) Heat energy is being produced in a resistance in, a circuit at the rate of 100 W. The current of 3 A, is flowing in the circuit. What must be the value, of the resistance?, Ans. Given:, , I = 3A, , , , P = 100 W, , , , R = 12 Ω, , , , V = 220 V, , To find: Power (P) = ?, V2, R, , Formulae: P =, , To find: R = ?, , Solution: Case (i) (For length 6m), , Formulae: P = I2R, , P=, , Solution: P = I2R, 100 = (3)2 × R, , ∴, , 100 = 9R, , Case (ii) (For length 3m), , Ans. Given:, , P1 = 100 W, , , , P2 = 60 W, , , , V = 220 V, , To find: I = ?, Formulae: P = VI, Solution:, Total power (P) = P1 + P2 = 100 + 60 = 160W, P = VI, I=, , P, 160, 8, =, =, V, 220 11, , I = 0.72A, Ans. , The current flowing in the main conductor is, 0.72A., *(3) A 6 m long wire made from an alloy, nichrome, is, shaped into a coil and given for producing heat., It has a resistance of 24 ohm. Can we get more, heat if the wire is cut into half of its original, length and shaped into a coil? For getting, Energy, the two ends of the wire are connected, to a source with a potential difference of 220 V., Ans. Given:, , Case (i) (For length 6m), , , , R = 24 Ω, , , , Case (ii) (For length 3m), , V2, (220)2, =, R, 12, 220 × 220, 12100, =, =, 12, 3, , P=, , R ≈ 11Ω, *(2) Two tungsten bulbs of wattage 100 W and 60, W power work on 220 V potential difference., If they are connected in parallel, how much, current will flow in the main conductor?, , 220 × 220, 55 × 110, 6050, =, =, 24, 3, 3, , P = 2016.67 W, , 100, R =, 9, R = 11.11Ω, Ans. The value of the resistance is 11Ω., , P=, , V2, (220)2, =, R, 24, , ∴, , P = 4033.33 W, Ans. H,  ence more heat will be obtained after cutting the, wire into half as power consumption increases., *(4) An electric iron uses a power of 1100 W when, set to higher temperature. If set to lower, temperature, it uses 330 W power. Find out the, electric current and the respective resistance of, 220 V., Ans. Given:, , P1 = 1100 W, , , , P2 = 330 W, , , , V = 220 V, , To find: I1 = ?, I2 = ?, R1 = ?, R2 = ?,, Formulae: P = VI, P =, Solution:, , V2, R, , P1 = VI1, I1 =, , P1, V, , 1100, 220, , =, , I1 = 5 A, I2 =, , P2, V, , =, , 330, 3, =, 220, 2, , I2 = 1.5 A, P1 =, , V2, R1, , ∴ R1 =, , V2, P1, , =, , (220)2, 1100, , =, , 220 × 220, = 2 × 22, 1100

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Effects of Electric Current, , 81, , current and (ii) heat energy produced in 12 sec., , Ans: (i) 0.25 A (ii) 750 J., , R1 = 44 Ω, R2 =, , V2, (220)2, 220 × 220, 440, =, =, =, 330, 3, P2, 330, , R2 = 146.67 Ω, Ans. , If power is 1100 W, then the current is 5 A and, resistance is 44 Ω and when power is 330 W, then, the current is 1.5 A and resistance is 146.67 Ω, *(5) A cell is connected to a 9 ohm resistance,, because of which heat of 400 J is produced, per second due to current flowing through it., Obtain the potential difference applied across, the resistance., , (6), ,  ind the resistance of a coil of power 60 W when, F, potential difference of 240 V is applied across it., , Ans: 960 Ω., , (7), , I f a bulb is rated 220 V and 100 W . Find, resistance. , Ans: 484 Ω, , (8), ,  n electric current of 5 A flows through a wire, A, of resistance 41.8 Ω. Find the time to obtain heat, of 3000 cal. , Ans: 12 seconds, Type: B, Energy = Power × Time, H=P×t, , Ans. Given: R = 9 Ω, , , H = 400 J, , , , t=1s, , To find: V = ?, , *(1) Who will spend more electrical energy, 500 W, TV set in 30 mins or 600 W heater in 20 mins?, , Formula: H =, , Ans. (i) For TV set, , V 2t, R, V2t, Solution: H =, R, 400 =, , V2 × 1, 9, , 3600 = V2, , , V = √ 3600, , V = 60 V, Ans. , The potential difference applied across the, resistance is 60 V., , NUMERICALS FOR PRACTICE, (1), ,  he resistance of a wire is 25 Ω. Due to electric, T, current passing through it, 6.25 J/sec heat, is generated in the wire. Find the potential, difference. , Ans: 12.5 V, , (2), , I f 90 W bulb is connected to a circuit with, potential difference of 360 V, find the current, flowing through the bulb. , Ans: 0.25 A, , (3), ,  n electric bulb is connected to 250 V. The, A, current passing through it is 0.27 A. What is the, power of the bulb? , Ans: 67.5 W, , (4), , I f the bulb of 60 W is connected across a source, of 220 V, find the electric current drawn by it., , Ans: 0.2727 A., , (5), , Note: P,  ower should be in kW, Time should be in hours, ,  potential difference of 250 V is applied across a, A, resistance of 1000 Ω in an electric iron. Find (i) the, , Electrical energy consumed = P × t, 500, 30, ×, 1000, 60, , , , =, , , , = 0.5 × 0.5, , , , = 0.25 kWh, , (ii) For Heater, Electrical energy consumed = P × t, 600, 20, ×, 1000, 60, , , , =, , , , = 0.6 ×, , , , = 0.2 kWh, , 1, 3, , Ans. Therefore, TV will consume more electrical, energy., *(2) An electric tungsten bulb is connected into a, home circuit. The home electric supply runs, at 220 V potential difference. When switched, on, a current of 0.45 A flows through the bulb., What must be power (wattage) of the bulb? If, it is kept on for 10 hours, how many units of, electricity will be consumed?, Ans. Given: V = 220 V, , , I = 0.45 A, , , , t = 10 h, , To find: P = ?, Unit of energy = ?

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Master Key Science and Technology Part-1 (Std. X), , 82, , 75, × 60 = 45 kWh, 100, ∴ Total Cost = 45 × 3, , Formulae: P = VI, H = P × t, , ∴H=, , Solution: P = VI, , , = 220 × 0.45, , P = 99 W, , , ∴P=, , , 99, kW, 1000, , H=P×t, 99, =, × 10, 1000, , , , H = 0.99 kWh, , Total cost = ` 135, Ans. , Cost of energy to operate electric iron for 30 days, is ` 135., (5) A washing machine rated 300 W operates one, hour/day. If the cost of unit is ` 3.00, find the, cost of the energy to operate a washing machine, for the month of March., Ans. Given:, , Ans. T,  he power of the bulb will be 99 W and the units, of electricity consumed is 0.99., , , , *(3) An electric iron of 1100 W is operated for 2 hrs, daily. What will be the electrical consumption, expenses for the month of April? (The electric, company charges ` 5 per unit of energy)., , , , Ans. Given: P = 1100 W, 1100, 11, , =, kW =, kW, 1000, 10, , t = 2 × 30 = 60 h, Unit cost = ` 5, , Solution: H = P × t, , ∴, , 11, =, × 60, 10, H = 66 kWh, , Total cost = 66 × 5, Total cost = ` 330, , Ans. T,  he electrical consumption expenses for the, month of April will be ` 330., , , , , =, , 750, 75, kW =, 1000, 100, , Solution: H = P × t, , Solution: H = P × t, , 3, 93, × 31 =, = 9.3 kWh, 10, 10, , ∴, , H=, , ∴, , Cost of energy = 9.3 × 3, Cost of energy = ` 27.9, , Ans. , The cost of the energy to operate a washing, machine for the month of March is ` 27.90., (6) If a TV of rating 100W is operated for 6 hrs per, day, find the number of units consumed in a, leap year., Ans. Given: Power (P) = 100 W, , , , P=, , 100, 1, kW =, kW, 1000, 10, , Time (t) = 6 × 366 = 2196 hrs, , To find: Unit consumed = ?, Formula: H = P × t, Solution: H = P × t, H=, , , 1, × 2196, 10, , H = 219.6 kWh, , Ans. Number of units consumed is 219.6., , NUMERICALS FOR PRACTICE, , Unit cost = ` 3, Formula: H = P × t, , (t) = 1 × 31 = 31 hrs, , Formula: H = P × t, , (t) = 2 × 30 = 60 hours, , To find: Total Cost = ?, , 300, 3, kW =, kW, 1000, 10, , To find: Total Cost = ?, , (4) An electric iron rated 750 W operates 2 hours/, day. If the cost of unit is ` 3.00 per kWh. Find, the cost of energy to operate electric iron for 30, days., Ans. Given: (P) = 750 W, , =, , Unit cost = ` 3, , To find: Total cost = ?, Formula: H = P × t, , (P) = 300 W, , (1), , , The, potential difference of 260 V is applied at, the domestic circuit. An LED is connected to, the circuit. An electric current of 0.35 A flows, through the LED.

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Effects of Electric Current, , 83, , Then find : (i) Power of LED Bulb, (ii) Units consumed if the bulb is operating for, 10 hr., , Ans: (i) 91 W (ii) 0.91 unit (kWh), , (2), ,  300 W T.V is operated every day for 6 hr. Find, A, the electrical energy (no. of units) consumed., , Ans: 657 kWh., , (3), ,  n electric iron rated 750 W is operated for, A, 2 hr/day. If the cost of unit is ` 3.00 per kWh,, find the cost of energy used to operate electric, iron for 30 days. , Ans: ` 135., , (4), , I f T.V. of rating 100 W is operated for 6 hr per, day. Find the number of units consumed in a, leap year. , Ans: 219.6 units, , Q.2.2. Write the definition / laws:, (1), AC generator:, Ans. , The device which converts mechanical energy, into electrical energy in the form of alternating, current is called AC generator., (2), DC generator:, Ans. , A device which converts mechanical energy into, electrical energy in the form of direct current is, called DC generator., (3), Right Hand Thumb Rule:, Ans. Imagine that you have held the conductor in, your right hand in such a way that your thumb, points in the direction of the current. Then turn, your fingers around the conductor, the direction, of the fingers is the direction of the magnetic, lines of force., , (4), Fleming’s left hand rule:, Ans. , According to the rule, the left hand thumb, index, finger, and the middle finger are stretched so as to, be perpendicular to each other. If the index finger, is in the direction of the magnetic field, middle, finger points in the direction of the current, then, the direction of the thumb is the direction of the, force on the conductor., (5), Heating effect of electric current:, Ans. , The production of heat energy in a conductor by, the electric current flowing through it is called, the heating effect of an electric current., (6), Fleming’s Right Hand Rule:, Ans. , Stretch the thumb, the index finger and the, middle finger in such a way that they will be, perpendicular to each other. In this position, the, thumb indicates the direction of motion of the, conductor, the index finger shows the direction of, the magnetic field, and the middle finger shows, the direction of the induced current., *(7) Maxwell’s cork - screw rule:, Ans. , If a right handed cork screw is assumed to be, held along the conductor, and screw is rotated, such that it moves in the direction of the current,, then the direction of magnetic field is same as, that of the rotation of screw., (8), Faraday’s law of induction:, Ans. , If the number of magnetic lines of force passing, through the coil changes, current is induced in the, coil. This is known as Faraday’s law of induction., The current produced in the coil is called the, induced current., , Q.2.3. Distinguish between:, *(1), , AC generator and DC generator, , Ans., , AC generator, (i), , AC generator converts mechanical energy into (i), electrical energy in the form of alternating current., , (ii) In this, brass slip rings are used., (2), , DC generator, DC generator converts mechanical energy into, electrical energy in the form of direct current., , (ii) In this, brass split rings are used., , Electric motor and AC generator, , Ans., , Electric motor, , AC generator, , (i), , The device which converts electrical energy into, mechanical energy is called electric motor., (ii) A battery is used to pass a current through the, armature coil., (iii) In motor, a current carrying coil is set in rotation, due to the magnetic field., (iv) In motor, the split ring commutator is used., , (i), , The device which converts mechanical energy, into electrical energy is called electric generator., (ii) A galvanometer is used to detect the presence of, electric current., (iii) In generator, a current is produced when the coil, rotates in the magnetic field., (iv) In the AC generator, slip ring commutator is used., , (v) Electric motor runs on electricity., , (v) Electric generator generates electricity.

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Master Key Science and Technology Part-1 (Std. X), , 84, , (3), , Short circuiting and Overloading, , Ans., , Short circuiting, (i), , (4), , I f a live wire and neutral wire come in contact, or touch each other, short circuiting takes place., , Overloading, (i), , I f electrical appliances of high power rating are, switched on simultaneously, overloading occurs., , (ii) During a short circuit, the resistance becomes, very small., , (ii) In overloading, the power consumption of a, circuit becomes large., , (iii) In this, large amount of current flows which, produces heat and can cause fire., , (iii) I n this, large amount of current flows through, appliance beyond its permissible limit., , (iv) Can be avoided by periodic checking., , (iv) Can be avoided by not using many appliances at, a time., , Direct current and Alternating current, , Ans., , Direct current, (i), ,  he electric current that flows in one direction,, T, i.e. from higher potential to lower potential is, called as direct current (D.C.)., , Alternating current, (i), , The electric current that reverses its direction, periodically with time is called an alternating, current (A.C.)., , (ii) The frequency of DC is zero., , (ii) The frequency of AC in India is, 50 Hz., , (iii) DC cannot be used on a large scale for, household purpose., , (iii) AC can be used on a large scale for household, purpose., , Q.2.4. Give Scientific Reasons:, *(1) Tungsten metal is used to make a solenoid type, coil in an electric bulb., Ans., (i), The electric bulb works on the principle of heating, effect of electric current., (ii) The intensity of light emitted by the filament of, an electric bulb depends on the temperature of, the filament., (iii) When electric current is passed through the, filament, a large amount of heat is generated and, filament becomes hot, then emits light., (iv) The melting point of Tungsten is very high, (3380°C), so that filament can be heated to a high, temperature without melting., (v) Hence, the filament of electric bulb is made up of, Tungsten., *(2) In the electric equipments producing heat, e.g. iron, electric heater, boilers, toaster, etc an, alloy such as Nichrome is used, not pure metal., Ans., (i),  he appliances like iron, electric toaster etc., T, works on the principle of heating effect of an, electric current., (ii) At high temperature pure metal can get oxidised, and also pure metals have low resistivity., (iii) An alloy such as Nichrome, has high resistivity, , (iv), , and can be heated to a high temperature without, undergoing oxidation., Thus, Nichrome is preferred over other pure, metals and are used for making coil for devices, working on heating effect of electric current., , *(3) For electric power transmission, copper or, aluminium wire is used., Ans., (i),  opper and Aluminium are good conductors of, C, electricity., (ii) Copper and aluminium have large number of, free electrons and are highly ductile metals., (iii) They have less resistivity, due to which these, metals have high conductivity., (iv), ,  ence, connecting wires in the electric circuit are, H, generally made of copper and aluminium., , *(4) In practice the unit kWh is used for the, measurement of electrical energy rather than, joule., Ans., (i),  he unit of electric power (W) is a very small unit,, T, hence 1000W or 1kW is used in practice., (ii), (iii), ,  ow if 1kW power is used for 1 hour, it means, N, 1kW × 1h i.e. 1kWh energy is consumed., joule is the SI unit for energy which is very, small, hence on commercial scale kWh is used to, measure electrical energy.

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Effects of Electric Current, , *(5) Why are carbon brushes used? How do these, work? (Find out; Textbook Page 65), Ans. , Brushes make contact with the split ring and are, responsible for the make and break of the circuit., As the brushes are made up of carbon they offer, less resistance and act as a good conductor and, are also cheaper to replace., *(6) Iron is a conductor of electricity, but when we, pick up a piece of iron resting on the ground,, why don’t we get electric shock? (Can you recall?;, Textbook Page 47), Ans. , In a piece of iron, the net flow of electrons is zero, and hence current is zero. Hence we don’t get, electric shock. If we want an electric current to, flow through it, then we need to connect it to a, source of potential difference i.e. cell., (7) The material used for making fuse wire has low, melting point., Ans., (i),  he fuse wire works on the principle of heating, T, effect of electric current., (ii) The fuse is made up of an alloy with low melting, point., (iii) This wire can allow certain maximum current to, flow through it. If current exceeds this value, the, heat generated in the coil is sufficient to melt it., (iv) Due to this electric current stops flowing through, the circuit and the valuable electrical appliances, can be saved from possible damage due to, excessive electric current., (v) Due to this, the material used for making fuse, wire should have low melting point., , 85, , (ii), , I t is caused if the insulating coating of a wire is, spoilt or lost or if faulty electric appliances are used., , Effects of shortcircuiting:, (a) During a shortcircuit, the resistance of the circuit, becomes very small and hence huge amount of, current flows through it., (b) This produces a large amount of heat and raises, the temperature. As a result, the circuit catches, fire and can cause damage., *(2) What is Solenoid ? Compare the magnetic field, produced by a solenoid with the magnetic field, of a bar magnet. Draw neat figure and name, various components., Ans. , Solenoid:, (i), When a copper wire with a resistive coating is, wound in a chain of loops (like a spring) it is, called solenoid., (ii) When an electric current is passed through a, solenoid, magnetic field is produced in it., (iii) The magnetic lines of force are similar to that of, the lines of a bar magnet., (iv) Solenoid has all properties of the field produced, by a bar magnet., Magnetic, lines of, force, , Solenoid, , S, , N, , Q.2.5. Answer the following:, (1) If lines of force passing through the coil are, increased, will current be induced? Why?, Ans. Yes current will be induced. As the lines of force, passing through the coil are increased, there is a, change in intensity of magnetic field and thus,, current is induced in the conductor., , (v), , (2), , , (a) Copper wire with Resistive coating,, , What is overloading?, , Ans. A flow of large amount of current in a circuit,, beyond a permissible value of current, is called, overloading. It occurs when many electrical, appliances of high power rating, such as geyser,, heater, oven etc., are switched on simultaneously., , Fig 4.1: Magnetic lines of force of a magnetic field, produced by a current passing through a solenoid coil.,  ne end of the coil acts as the south pole, while, O, the other end as the north pole., Various components are:, (b) Battery, (c) Plug key, *(3) Which device is used to produce electricity?, Describe with a neat diagram., , Q.3.1. Explain the following:, , (a), , Electric motor, , *(1) How does short circuit form? What is its effect?, , (c), , Electric generator (DC), , Ans. Shortcircuiting:, (i), If a live wire (phase wire) and a neutral wire, come in direct contact or touch each other, shortcircuiting takes place., , (b), (d), , Galvanometer, Voltmeter, , Ans. , (c) Electric generator (DC) - The device which, converts mechanical energy into electrical energy, is called electric generator.

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Master Key Science and Technology Part-1, 103(Std. X), , 86All About Electromagnetism, , B, , C, , N, , Iron core, , S, A, , Armature coil, , D, , B1, , B2, , Split ring, Axle, , Fig. 5.6 D. C. Generator, , Fig 4.2: DC generator, (i) Armature coil, (ii) Strong magnets, For description refer Q. 4.2 1 (c), (iii) Split rings or commutator (iv) Brushes, *(4) How do we decide that a given material is a, (v)good, Bulb., conductor of electricity or is an insulator?, (Can, youcomponents, recall?; Textbooksuch, Page 47), The, as armature coil,, strong, magnet,, brushes, arethrough, the same, which, Ans. , (a) If, the current, passes, easily, a material,, are, used, for, AC, generators., that means it has large number of free electrons, andSplit, it is rings, a good, of is, electricity., or conductor, commutator, the same as used, in, electric, motor., (b) If a material does not allow current to flow, through, it, that, means, it has less, Working, of DC, generator, : or practically no, free electrons. Hence, it is an insulator., (a) When the coil of DC generator rotates in the, *(5) How, can wefield,, write potential, mechanical, power inis a, magnetic, difference, manner, similar, to, the, electrical, power?, produced in the coil. This gives rise to(Think, the, about it; Textbook Page 48), flow of current. This is shown by glowing of, Electric Work, the bulb., Ans. Electrical Power =, Time, (b) In D.C. generator, the flow of current in the, Energy, consumed, circuit is in the same, direction, as long as the, , =, Time, coil rotates in the magnetic field., Work, Power = one, (c)Mechanical, This is because, Timebrush is always in, , contact with the arm of the armature moving, *(6) Derive, Joules, Law, with, help ofwith, a neat, up and, other, brush, is the, in contact, thecircuit, arm, diagram., of the armature moving downward in the, magnetic field. I, , (.), , Describe the safety, that you, + measures, A, would adopt while, electrical, VAB Vhandling, R, +, –, appliances., B, + handling of, Ans. The measures, adopted, for, safe, –, A, electrical appliances are given, – below :, (4), , (i), , The electrical appliances should be earthed, , Fig 4.3, Work, Ans. , Power = Time, , properly to, leakage of current which, VABprevent, Q, ∴, P, =, [∴, W = VQ], can cause electric, shock., t, (ii) Rubber gloves made from insulated, Q, materials, be(i)worn, with rubber, ∴ P = VABshould, I ..........., ∴ I along, =, t, sole shoes., Now, H = P × t ........... (ii), (iii) During lightning, electric circuits should be, Substituting, switched, off. (i) in (ii),, , [, , ], , (iv), To avoid, many appliances, , H = VABoverloading,, It, should not be connected at the same point., ∴ H = IR × I × t, (v) Insulation of the wires should be checked, ∴ H = I2Rt, regularly, to avoid short circuit., This is called Joules Law of Heating Effect., , Q.13. Read the following paragraph and, *(7) , In the the, above, circuit, ifgiven, the resistor, answer, questions, below:is replaced, , by a motor, in which form will the energy given, In our homes we receive supply of electric, by the cell get transformed into? (Use your brain, power through a main supply, either, power?; Textbook Page 47), supported through overhead electric poles, Ans. , above circuit,, if theOne, resistor, replaced by, orInbythe, underground, cables., of theiswires, motor,, then usually, the energy, given, the cell will get, inathis, supply, with, red by, insulation, transformed, intolive, mechanical, cover, is called, wire orenergy., positive., Another, wire, with, black, insulation, (8) How will you increase the strength of is, magnetic, called, neutral, wire, (or, negative)., In, our, field in an electromagnet?, country, the potential difference between, Ans. , We can increase the strength of magnetic field by, the two is 220V., three ways:, At the meter board in the house, these, (i), Increase the magnitude of current in the circuit, wires, pass, into, an, electricity, (ii) meter, Increase, the number, turnsThrough, of the coil, through, a main of, fuse., the, main, switch, they, are, connected, to, the, live, (iii) Reduce the radius of the coil., wires in the house. Often two separate, (9), Write a note on Galvanometer., circuits are used, one of 15A current rating, Ans. , for appliances with higher power rating, as geysers, air, coolers, etc.The, other, (i) such, A galvanometer, is an, instrument, which, is used to, circuit, of 5A, currentofrating, forcurrent, bulbs, fans, detectisthe, presence, electric, in a circuit., etc., The, earth, wire,, which, has, insulation, (ii) If the current in the circuit is zero, the galvanometer, of green colour is usually connected to a, will show zero or no deflection., metal plate deep in the earth near the, (iii) house., The deflection, is onaseither, side measure,, of the zero mark, This is used, a safety, depending, directionthat, of the, current., especially, forupon, thosethe, appliances, have, a, body, electric, press, toaster,, (10) metallic, What do, you e.g., mean, by earthing?, table, fan, etc., The, metallic, body is, Ans., connected to the earth wire which provides, (i) a M, anyresistance, electrical conducting, appliances path, have for, metallic, low, the body., If by chance insulation of the wire melts, the, current., live wire may touch the metallic body of the, When insulation of wires is damaged or, appliance. If we touch this body with bare hands,, there is fault in the appliances, the neutral, we may receive severe electric shock., (ii), ,  o avoid such accidents, metal casings of all, T, electrical appliances must be earthed. One end of, the copper wire is connected to the metal case of, the appliance and the other end is connected to

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Effects of Electric Current, , the copper plate which is burried deep inside the, earth., (iii), ,  his system of connecting the metal casing of an, T, electrical appliance with a copper plate burried, deep inside the earth is called as earthing., , (11) State the factors on which the strength of the, magnetic field of a circular loop depends?, Ans., (i), , Current flowing through the loop, , (ii), , Number of turns in the loop, , (iii), , Radius of the loop, , Q.3.2. Open Ended Questions, (1) Which of the statement given below correctly, describes, *(i) The magnetic field near a long, straight current, carrying conductor?, (a) The magnetic lines of force are in a plane, perpendicular to the conductor in the form of, straight lines., , 87, , (2), , State the rule you will use:, , (a) To find the direction of the motion of a conductor,, if direction of the current and magnetic field are, known to you., (b) To find the direction of the induced current,, if the direction of motion of conductor and, magnetic field are known to you., Ans., (a) If the direction of current and directions of, magnetic field is known, then we will use, Fleming’s Left Hand Rule to know the direction, of motion of the conductor., (b) If the direction of motion and magnetic field is, known we will use Fleming’s Right Hand Rule to, know the direction of the induced current., Q.3.3. Figure/Application Based Questions, *(1) Identify the figure and explain the uses of, following:, (a), , (b) The magnetic lines of force are parallel to the, conductor on all the sides of a conductor., (c) The magnetic lines of force are perpendicular to, the conductor going radially outward., (d) The magnetic lines of force are in concentric, circles with the wire at the centre, in a plane, perpendicular to the conductor., Ans: The correct statement is, (d) The magnetic lines of, force are in concentric circles with the wire as the, centre, in a plane perpendicular to the conductor., , Fig 4.4, , Ans., (i), , Cartridge Fuse, , (ii), ,  use is a device which protects the appliances, F, from excess current., , (b) , , *(ii) Electromagnetic induction, (a), , Charging of an electric conductor., , (b) Production of Magnetic field due to a current, flowing through the coil., (c) Generation of a current in a coil due to relative, motion between the coil and the magnet., (d) Motion of coil around the axle in an electric, motor., , Ans., (i), , Fig 4.5, Miniature circuit breaker (MCB), , (ii) When the current in the circuit suddenly, increases, this switch opens and current stops., (c), , Ans. , The correct statement is (c), Generation of a, current in a coil due to relative motion between, the coil and the magnet., The process by which a changing magnetic field, in a conductor induces a current in another, conductor is called electromagnetic induction., This was first studied by the scientist Michael, Faraday., , A, , Fig 4.6

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Master Key Science and Technology Part-1 (Std. X), , 88, , (i), , DC generator, , (ii), , It is used to generate direct current., , (2) Name the following diagrams and explain the, concept behind them., *(a), , Ans., , Fig 4.7, , (i), , Fleming’s right hand rule, , (ii), ,  his rule is used to find the direction of the, T, induced current., , (iii), , Stretch the thumb, the index finger and the middle, finger of the right hand in such a way that they, are perpendicular to each other. In this position,, the thumb indicates the direction of motion of, the conductor, the index finger indicates the, direction of the magnetic field, and the middle, finger shows the direction of the induced current., This rule is known as Fleming’s right hand rule., , *(b), , Ans., , Fig 4.8, , (i), , Fleming’s left hand Rule., , (ii), ,  his rule is used to find the direction of the force, T, on the conductor., , (iii), ,  ccording to this rule, the left hand thumb, index, A, finger, and the middle finger are stretched such, that they are perpendicular to each other. If the, index finger is in the direction of the magnetic, field, and the middle finger points in the direction, of the current, then the direction of the thumb is, the direction of the force on the conductor., , Ans. , (i), Right Hand Thumb Rule, (ii) This rule is used to find out the direction of the, magnetic field produced by current flowing, through an electrical conductor., (iii) Imagine that you have held the conductor in your, right hand in such a way that your thumb points, in the direction of the current. Then turn your, fingers around the conductor, the direction of the, fingers is the direction of the magnetic lines of, force., (3) Answer the following based on the graph given, below., 2, Current Ampere, , Ans., , Direct current (DC), Alternating current (AC), , 1.5, 1, 0.5, 0, –0.5, –1, –1.5, –2, , 0, , 0.5, , 1, 1.5, 2, time (s), Fig 4.10, (i), After how many seconds does AC change its, direction?, Ans: , 0.5 sec, (ii) What is the maximum value of AC?, Ans: , 1A, (iii) What is the maximum value of DC?, Ans: , 1A, (iv) What is the time interval for 1 cycle of AC?, Ans: , 1 sec, (v), , What is the frequency of AC and DC?, , Ans: , f=, , 1, ∴ AC = 1 Hz and DC = 0 Hz, T, , (4) , Answer the following based on the diagram, given below., , (c), , Coil, , Solenoid coil, , B, , Fig 4.9, , Galvanometer, Battery, Fig 4.11

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Effects of Electric Current, , (i), , 89, ,  hat change do we observe in the Galvanometer, W, when the current in the Solenoid coil is, switched off?, , Ans. If we open the plugkey and turn the current to, zero in the coil, the pointer of the Galvanometer, deflects to a side and quickly comes back to zero., If the current in the coil is started again, the, pointer again deflects to the other side and then, returns quickly to zero., (ii), ,  hat change do we observe in the galvanometer, W, when a current is passing through the solenoid, coil and the coil is displaced laterally with, respect to the coil?, , Ans. If the coil is displaced laterally, then current will, be induced in the coil and the Galvanometer will, show a deflection., (iii), , Ans. When the current is downward, the force, experienced by the conductor will be outward., (ii), , I f the conductor experiences force inwards, then, what would be the direction of current?, , Ans. If the conductor experiences force inwards, then, the direction of current would be upward., (iii), ,  hich rule helps us to find the force experienced, W, by a current carrying conductor in the above, diagram?, , Ans. Fleming’s Left Hand Rule., (6), , Answer the following based on the diagram, given below., Solenoid, , (i), ,  hat change do we observe in the galvanometer, W, if the current in the Solenoid coil is increased?, , (5) Answer the following based on the diagram, given below., , I, , B, , Define : Solenoid, , Ans. When a copper wire with a resistive coating is, wound in a chain of loops (like a spring) it is, called solenoid., (iii), , (7), , H, N, , (ii), , N, , What is the nature of the magnetic field inside, the solenoid?, , Ans. The magnetic lines of force inside a solenoid are, parallel to each other., , I, S, , Stand, , Solenoid, , S, ,  ame and state the law responsible for the, N, above phenomena., , Ans. Faraday’s law of induction: Whenever the, number of magnetic lines of force passing, through the coil changes, current is induced, in the coil. The current produced in the coil is, called the induced current., , Complete the diagram of magnetic lines of force, passing through a solenoid., Magnetic, lines of, force, , Ans. If the current is increased, the strength of the, magnetic field increases and the deflection in the, galvanometer also increases., (v), ,  hat is the direction of the force experienced by, W, the conductor when the current is downward?, ,  hat change do we observe in the galvanometer, W, if the displacement of the Solenoid is faster?, , Ans. The faster is the displacement of the solenoid,, larger is the deflection of the Galvanometer, pointer., (iv), , (i), , A, , Answer the following based on the diagram, given below., Conductor loop, Magnetic lines, N, , C, , Copper wire, Magnet, , S, , Cardboard, R, A, , Fig 4.12, , A

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Master Key Science and Technology Part-1 (Std. X), , 90, , (i), , Which rule helps to find the direction of, magnetic field?, , B, , Ans. Right Hand Thumb Rule., (ii), , N, , State any 2 factors on which the strength of, magnetic field, for a circular loop depends?, , Ans. The strength of the magnetic field, for a circular, loop, depends on the current flowing through the, wire and the number of turns., , Ans., , Ans., , Fig 4.15: Electric Motor, (i), , Construction:, Armature coil: Rectangular loop of copper wire, having resistive coating., , (ii), ,  trong Magnet: Armature coil is placed between, S, the north pole and south pole of the magnet (such, as horse shoe magnet)., , (iii), ,  plit ring: The two halves of the ring (X and Y), S, have a resistive coating on the inner side and, conducting surface on the outer side., , (iv), ,  arbon Brushes: Two stationary carbon brushes, C, (E and F) are connected to the outer side of the, split ring., , (v), ,  attery: The battery supplies the electric current, B, to the armature coil., , I, , I, , Carbon, brushes, (E and F), , K, , Circuit for Alternating Current, (AC), , F, , Axle, , I, , (2), , D, Y, , E, , (DC), , Fig 4.13, , X, , Split rings (X and Y), , Circuit for Direct Current, I, , S, A, , Q.4.1. Draw a neat and labelled diagram:, (1), , C, , Working:, (i), , When the circuit is completed, the current flows, in the branch AB of the loop from A to B through, the carbon brushes E and F., , (ii), , (1) Explain the construction and working of the, following. Draw a neat diagram and label it., ,  ince the direction of the magnetic field is, S, from north pole to south pole, according to the, Fleming’s left hand rule, a force is exerted on the, branch AB in downward direction and CD in, upward direction., , (iii), ,  hus, the loop and the axle start rotating in an, T, anticlockwise direction., , *(a), , (iv), ,  fter half rotation, the current in a loop starts, A, flowing in the direction DCBA., , (v), ,  herefore, a force is exerted on the branch CD, T, in downward direction and on the branch AB in, the upward direction, and the loop continues to, rotate in the anticlockwise direction., , (vi), ,  hus, the current in the loop is reversed after each, T, half rotation and the loop and the axle continue, to rotate in the anticlockwise direction., , Fig 4.14, (3), DC generator, Ans. For diagram refer fig 4.2., (4), Magnetic lines of force through a Solenoid, Ans. For diagram refer fig 4.1., Q.4.2. Answer in brief, , Electric Motor, , Ans. A,  device which converts electrical energy into, mechanical energy is called electric motor., Principle of working:, Electric motor works on the principle that a, current carrying conductor placed in a magnetic, field experiences a force.

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Effects of Electric Current, , 91, , *(b) AC Generator:, Magnetic field, N, , coil C, , B, , Conducting, ring, , B1, R2, ,  urrent flows from B2 to B1 in the first half of the, C, revolution in the external circuit., , (iv), ,  fter half revolution, arms AB and CD exchange, A, their positions. Due to this, the induced current, flows in the direction DCBA. But, arm AB is in, contact with B1, through slip ring and arm CD is, in contact with B2., , (v), , I n the second half cycle, in the external circuit,, electric current flows from B1 to B2, i.e. opposite, to the previous half rotation., , (vi), ,  his repeats after every half rotation, and, T, alternating current is produced., DC Generator:, , S, D, R1, , A, , (iii), , B2, , Carbon, brushes, , Axle, , (c), , (For diagram, refer fig 4.2), , Fig 4.16: Electric generator, Ans: A generator which converts mechanical energy, into electrical energy in the form of alternating, current is called AC Generator., , Ans. The device that converts mechanical energy into, the electrical energy in the form of direct current, is called DC generator., Principle of working:, , Principle of working:,  lectric generators works on the principle of, E, electromagnetic induction., , (i), , Construction:, , Construction:, , (i), ,  rmature coil: Rectangular loop of copper wire, A, having resistive coating., , (i), ,  rmature coil: Rectangular loop of copper wire, A, having resistive coating., , (ii), ,  trong Magnet: Armature coil is placed between, S, the north pole and south pole of the magnet (such, as horse shoe magnet)., , (ii), , (iii), ,  lip rings: The ends of the armature are connected, S, to conducting rings R1 and R2 which are fixed to, the axle with the resistive coating in between the, ring and the axle., ,  trong Magnet: Armature coil is placed between, S, the north pole and south pole of magnet (such as, horse shoe magnet)., , (iii), ,  plit ring: The two halves of the ring (X and Y), S, have a resistive coating on the inner side and, conducting surface on the outer side., , (iv), ,  arbon Brushes: The stationary carbon brushes, C, B1 and B2 are connected to a galvanometer., , (iv), , (v), ,  alvanometer: It detects the presence of current, G, in the circuit., ,  arbon Brushes: Two stationary carbon brushes, C, (E and F) are connected to the outer side of the, split ring., , (v), , Electric bulb or Ammeter :, , (i), , (a) T,  he output is shown by the glowing bulb, connected across the carbon brushes., , Working:, (i), , (ii), ,  hen the armature coil ABCD is rotated by, W, external force in the magnetic field provided, by strong magnets, it cuts the magnetic lines of, forces. Thus, the changing magnetic field induces, current in the coil.,  fter rotation of axle, arm AB moves upwards,, A, while arm CD moves downwards. Therefore,, ABCD rotates in the clockwise direction., As per the Fleming’s right hand rule, the, current is induced in the Coil and flows from, A, B and C, D. Current flows in the, direction ABCD., ,  C generator works on the principle of, D, electromagnetic induction., , (b) Ammeter is connected in the circuit to detect, the presence of electric current in the circuit., Working:, (i), ,  hen the coil of DC generator rotates in the, W, Magnetic field, potential difference is produced, in the coil., , (ii), ,  s the coil rotates in the magnetic field, the, A, potential difference produces induced current, according to Fleming’s right hand rule., , (iii), ,  he presence of current can be detected either by, T, electric bulb or galvanometer.

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Master Key Science and Technology Part-1 (Std. X), , 92, , (iv), , I n DC generator, the flow of current in the, external circuit is in one direction as long as the, coil rotates in the magnetic field. This is because, one brush is always in contact with the arm, moving up and other brush is in contact with the, arm moving downward in the magnetic field., , (v), , Thus, this generator is called DC generator., , (ii) Which effects of electric current do you, find? (Use your brain power?; Textbook Page 47), , Q.4.3. Paragraph Based Question, , (a), , (1) Read the paragraph and answer the following, question., Galvanometer is a sensitive device which, works on the same principle as that of an, electric motor that we have studied earlier., We can make some electrical measurements, with it. A coil is positioned between the pole, pieces of a magnet in such a way that the, pointer on the galvanometer dial is connected, to it. When a small current (for example 1 mA), flows through the coil, the coil will rotate. The, rotation will be proportional to the current., Voltmeter and Ammeter also work on the, same principle. In galvanometer, the pointer, deflects on both the sides of the zero mark, depending on the direction of the current., (i), , State the principle on which Galvanometer, works., , Ans. It works on the principle that a current carrying, conductor placed in a magnetic field experiences, a force., (ii), ,  tate the relation between the strength of the, S, current and the deflection of Galvanometer?, , (b), , (c), Fig 4.17, , Ans., , All About Electromagnetism, , (i), , I n the above pictures, we observe different, household: 5.2, appliances, which work on electricity., ACTIVITY, , (ii) In the picture (a), The electric heater and the 2, (2) lamps, Takework, two on, coils, of about, heating, effect 50, of turns., electric Insert, current, them, over a(b), non-conducting, and, in picture, and (c), the fancylindrical, and the bell, roll,, showneffect, in the, following, figure, work, onas, magnetic, of electric, current., (thick, can be50, used)., Connect, (2) Take, twopaper, coils roll, of about, turns., Insert coil, them, over, non-conducting, roll,, 1 to aa battery, with a plugcylindrical, key. Connect, coilas, shown, the following figure, paper, roll, 2 to in, a galvanometer., Plug(thick, the key, and, canobserve, be used)., Connect, coil, 1, to, a, battery, with, the deflection in the galvanometer., a plug key. Connect coil 2 to a galvanometer., Unplug the key and again observe the, Plug the key and observe the deflection in, deflection. (HOTS), the galvanometer. Unplug the key and again, observe the deflection. (HOTS), Coil 1, , (iv), ,  hen the current in the Galvanometer is, W, switched on and later switched off, how will the, deflection in the Galvanometer change?, , Ans. The galvanometer will deflect on one side of the, zero mark when the current is switched on and on, the opposite size, when the current is switched off., Q.4.4. Activity Based Question., *(1), , (i) What do you observe in the following, pictures?, , D, fo, , (i), , D, , Earth, Wire, Live, Wire, Neutral, Wire, , B, , (ii) M, , +, , Name any three devices working on the same, principle as Galvanometer., , Ans. Electric motor, Voltmeter and Ammeter., , (3), , Coil 2, , Ans. , As the strength of the current increases, the, deflection of galvanometer also increases., (iii), , ACTIV, , _, , ( ), , G, , Key, K, , Fig 4.18, Fig. 5.8, (i) What do you observe when you switch on the, current in coil 1, i.e. when you connect coil 1 to, (i) theWhat, do you observe when you switch on, battery?, the coil, current, in coil 1, to, i.e.the, when, youwe, connect, Ans: When, 1 is connected, battery,, observe, coil 1 tointhe, deflection, thebattery?, galvanometer connected to coil 2., , (ii) , dodo, you, (ii)What, What, youobserve, observewhen, whencurrent, currentin, in coil, coil 1, reaches, a steady, valuevalue, or zero?, 1 reaches, a steady, or zero?, Ans: When the current in coil 1 reaches a steady value, (iii) Why is current induced in coil 2 when we, or becomes zero, there is no deflection in the, connect coil 1 to a battery?, galvanometer., Ans., (i), , When coil 1 is connected to the battery, we, , (iii) El, , Ans. R, , (iv) A, , Ans. R, (v), , D, , Ans. R

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(iii) Electric motor, Effects, of Electric, Ans., Refer, Fig.Current, 5.6, , 93, , (iv)(iii) , ACWhy, generator, is current:, , induced in coil 2 when we, connect coil 1 to a battery?, , (i) Observe the pattern of iron filings., , Ans: The iron filings which are sprinkled get, Ans: , magnetized and on tapping the cardboard sheet,, (i), When, we, connect, the, coil, to, a, battery,, current, (v) DC generator :, they arrange themselves in concentric circles, starts flowing, hence there is change in the, around the wire., magnetic, Ans. Refer, Fig. field., 5.8, (ii) This change in magnetic field induces a current, (ii) What changes are seen in the magnetic field, in, coil, 2., lines when the current is increased?, ACTIVITY : 5.4, (iii) The process by which changing magnetic field, Ans: When current is increased the magnetic field, induces a current in another conductor, is called, (4) Place the white paper on the drawing board, lines are formed more closer and more distinctly, electromagnetic induction., indicating that the intensity of the magnetic field, and, place, a barpaper, magnet, the middle, of the, (3) , Place, the white, on thein, drawing, board and, increases., place Sprinkle, a bar magnetsome, in the iron, middlefilings, of the paper., paper., on it, Sprinkle some iron filings on it, (iii) What happens to the magnetic needle of the, compass when the compass is brought near, the wire, and moved away from the wire? Also, observe what happens to the magnetic needle, when strength of current is increased., , Ans. Refer Fig. 5.7, , N, , S, , Ans: , (i), , The magnetic needle in the compass deflects more, when we moved towards the wire and it starts, deflecting less when we move it away from the, wire., , (i) Tap the board gently, what do you observe?, , (ii), , Ans: The iron filings arrange themselves in the form of, concentric circle, which indicates the direction of, magnetic field of a bar magnet., , Similarly, on increasing the strength of the, current, it is observed that the magnetic needle, deflects more., , (iii), , This indicates that the strength of magnetic field, increases with increase in strength of the current., , Fig 4.19, Fig., 5.11, , (ii) Why do the iron filings arrange themselves in a, pattern as shown in the figure above?, Ans: The magnetic field of the bar magnet influence, the iron filings due to which they behave like tiny, magnets and hence align themselves along the, direction of magnetic lines., (4) Take a battery, (12V) a variable resistance (or a, rheostat), an ammeter (0.5 A), a plug key, needle, and a long straight thick copper wire. Take a, rectangular cardboard, pass this wire through, center. Fix the cardboard and see that it does, not slide up or down. Sprinkle iron filings, uniformly on the cardboard. Close the key and, tap the cardboard gently., R, Cardboard, , Magnetic needle, , Thick copperwire, , K, , Fig 4.21, , (iv) What happens to the magnetic needle of the, compass when direction of the current is, reversed in the straight conductor? State the, reasons for your observation., Ans:, (i) , The direction in which the magnetic needle gets, deflected changes, when direction of the current, is reversed., (ii), , This shows that if direction of current is reversed,, the direction of magnetic field lines is also, reversed., , (6) Take a rectangular cardboard with two holes., Take a coil of large numbers of turns. Insert the, coil through the holes such that the turns are, normal to the plane of the cardboard. Connect, two ends of the coil in series with battery, a key, and a rheostat as shown in figure. Sprinkle iron, filings uniformly on the cardboard. Plug the key,, tap the cardboard gently. Note your observations.

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when we move away from the loop the, Master Key Science & Technology (Std. X), concentric circles representing magnetic, field lines become bigger and bigger. At the, (10) Take three ring shaped magnets, insulated, centre of, theScience, circular, loop, the magnetic, Master, Key, and, Technology, Part-1field, (Std. X), copper, wire, (1.5m), battery, and paper clips., lines are straight., , 108, 94, , tap the cardboard gently. Note your, observations., , (9), State, what, isthe, the into, direction, ofofthe, magnetic, Cut, theiswire, three, pieces., One piece, (7) , State, what, direction, the, magnetic, field, lines, with, respect, to, each, other, fieldshould, lines with, respect1m, to long, each other, in in, every, be about, and the, other two, every, segment, of, the, loop, and, the, centre, segment of the loop and the centre of the loop., 0.25m, longWhere, each. Take, 1m piecefield, of wire and, of, the, loop., the magnetic, Where, is the, magneticisfield, strongest? (Refer, following, wrap this(Refer, wirethe, around, twofigure)., fingers to form, the strongest?, following, figure)., , B, , A, , oppo, , and p, , three, , abov, , Take, , a coil. Take the loose end of this wire and them, -, , wrap it once along the diameter of the coil. 5.17., , ( ), K, , +, , (i), , The two arms of the coil should be directly, , Fig., 5.15, Fig 4.22, , N, , Ans: , (i), OnOn, tapping, the cardboard,, the iron, Ans. Observation:, Observation:, tapping, the cardboard,, the, filings, align, themselves, in, concentric, circles, near, iron filings align themselves in circles near, the, loop., thecurrent, currentcarrying, carrying, loop. Thus it can be seen, (ii) Thus, it can be seen that the magnetic field lines, that the magnetic field lines are circular near, are circular near the current carrying loop., the current carrying loop. It is observed that, (iii) It is observed that when we move away from, when, away, from, the loopthe, the, the, loopwe, themove, concentric, circles, representing, concentric, representing, magnetic, fieldcircles, lines become, bigger and magnetic, bigger., field, become, bigger loop,, and bigger., At the, (iv) At, thelines, centre, of the circular, the magnetic, field, lines, are, straight., centre of the circular loop the magnetic field, , lines are straight., (9), , S, , Ans. On gi, , curre, , rotati, (ii), , Ans: , Paper clip, (i) Ans., It can, be seen, that that, eacheach, segment, of the, It can, be seen, segment, of circular, the, Ring produces, magnets, loopcircular, carrying, current, loop, carrying, currentmagnetic, producesfield, linesmagnetic, in the same, thedirection, loop., fielddirection, lines in within, the same, Cell, (ii) At the, centre, of, the, circular, loop, all, the, magnetic, within the loop. At the centre of the circular, fieldloop, linesallare, in, the, same, direction,, due, tothe, which, the magnetic field lines are in, Fig., 5.17, the same, strength, of magnetic, field increases, and, direction,, due to which, the strength, of it is, the strongest., , magnetic field increases and it is the, strongest., opposite each other. Bend the paper, , Ans. On p, , deve, , field, , exert, rotat, , clip, , and prepare a holder for the coil. Pile the, three magnets on the table. Place the coil, above the magnet on the paper clip holder., Take free ends of the wire and connect, them to the battery as shown in the figure, 5.17. Give a single spin to the coil., (i), , S, , What do you observe?, , Ans. On giving a spin to the coil which is carrying, current we observe that the coil starts, rotating till the current is being passed., (ii), , Fig. 5.16, , What makes the coil rotate on passing the, current?, , Ans. On passing the current the magnetic field, Ans. It can be seen that each segment of the, circular loop carrying current produces, magnetic field lines in the same direction, within the loop. At the centre of the circular, loop all the magnetic field lines are in the, same direction, due to which the strength of, magnetic field increases and it is the, strongest., , develops around the loop. The magnetic, field of the three magnets placed below, exerts a force on the coil due to which it starts, rotating., , What, , curre, , Fig. 4.23, 5.16, Fig, Coil, , State what is the direction of the magnetic, vvv, field lines with respect to each other in, every segment of the loop and the centre, of the loop. Where is the magnetic field, strongest? (Refer the following figure)., , N, , What

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Effects of Electric Current, , 95, , ASSIGNMENT - 4, Time : 1 Hr., , , Marks : 20, , Q.1. (A) Fill in the blanks:, , (3), , (1), , The electric bulb consists of a filament made up of, , (2), , In the electric bill, the electricity usage is expressed in, , ., , (3), , When a live wire and a neutral wire come in contact,, , takes place., , , , ., , Q.1. (B) Choose and write the correct option:, , (2), , (1) The device which converts mechanical energy into electrical energy is called, (a) Electric charge, (2), , The deflection of the pointer of, (a) Voltmeter, , , , (b) Electric generator (c) Electric fuse, (b) Ammeter, , (d) Electric motor, , is on either side of zero mark., (c) Galvanometer, , (d) Thermometer, , Q.2. Answer the following: (Any 2), , (4), , (1), ,  cell is connected to a 9 ohm resistance, because of which heat of 400 J is produced per second due, A, to current flowing through it. Obtain the potential difference applied across the resistance., , (2), , Distinguish between: Direct current and Alternating current, , (3), , State Maxwell’s cork - screw rule., , , (1), , , Q.3. Answer the following: (Any 2), , (6), , Name the following diagrams and explain the concept behind them., (a), , (b), , (2), , An electrical iron uses a power of 1100 W when set to higher temperature. If set to lower temperature,, it uses 330 W power. Find out the electric current and the respective resistance of 220 V., , (3), ,  n electric tungsten bulb is connected into a home circuit. The home electric supply runs at 220 V, A, potential difference. When switched on, a current of 0.45 A flows through the bulb. What must be power, (wattage) of the bulb? If it is kept on for 10 hours, how many units of electricity will be consumed?., , , , Q.4. Answer the following: (Any 1), , (5), , (1), , Explain the construction and working of Electric Motor. Draw a neat labelled diagram., , (2), , Derive Joules Law with the help of a neat circuit diagram., , vvv

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5, , Heat, , Points to Remember:, zz Heat is one of the forms of energy and can be obtained by transforming any other form of energy., zz Temperature is the degree of hotness or coldness; it is measured with the help of thermometer., zz Heat is one of the important factors on which the change of state depends., zz Generally substances expand on heating and contract on cooling., zz During change of the state of a substance, heat may be gained or lost but there is no change in temperature., zz The constant temperature at which substance changes from solid state to liquid state is called melting point, , of that solid., , zz The constant temperature at which substance changes from liquid state to gaseous state is called as boiling, , point of the liquid., , zz Latent heat: The amount of heat energy either absorbed or released during change of state is called ‘Latent, , Heat’. Latent heat is not recorded on the thermometer in the form of rise or fall in temperature., , zz Specific latent heat of fusion: The amount of heat energy absorbed at constant temperature by unit mass of a, , solid to convert into liquid phase is called the specific latent heat of fusion., , zz Latent heat of vapourization: The amount of heat energy absorbed at constant temperature during, , transformation of liquid into gas is called Latent heat of vapourization., , zz Specific latent heat of vapourization: The amount of heat energy absorbed at constant temperature by unit, , mass of a liquid to convert into gaseous phase is called specific latent heat of vapourization., , zz Different substances have different melting points and boiling points. The values of melting point, boiling, , point and latent heat depend on atmospheric pressure., , zz The phenomenon in which ice converts to liquid due to applied pressure and then re-converts into ice, once, , the pressure is removed is called regelation., , zz The behaviour of water between its temperature from 0°C to 4°C is called anomalous behaviour of water., zz At 4°C, the volume of water is minimum, hence its density is maximum., zz Effects of anomalous behaviour of water in nature:, , - Survival of aquatic plants and animals., - Cracking of rocks in winter., - Pipelines carring water may burst in cold region during winter., zz Atmosphere contains water vapour. The amount of moisture in air due to the presence of water vapour in it, , is called 'humidity'., , zz The temperature at which air becomes saturated with water vapour is called 'dew point'., zz The mass of water vapour per unit volume of air is called absolute humidity. It is measured in kg/ m3., zz The ratio of the mass of water vapour in a given volume of air at given temperature to the mass of water, , vapour required to saturate the same volume of air at the same temperature is called relative humidity., , zz At dew point, relative humidity is 100%. If the relative humidity is greater than 60%, the air is called 'humid', , (moist) and if the relative humidity is less than 60%, the air is 'dry'., , zz In SI system, heat is measured in joule (J) and in CGS system, it is measured in calorie (cal)., zz The amount of heat required to raise the temperature of 1 g of water by 1°C (from 14.5°C to 15.5°C) is called, , one calorie of heat., , (96)

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Heat, , 97, , zz The amount of heat required to raise the temperature of 1 kg of water by 1°C (from 14.5°C to 15.5°C) is called, , one kilocalorie of heat., , zz The amount of heat required to raise the temperature of unit mass of substance through 1°C is called the, , specific heat capacity of a substance., , zz Specific heat capacity is denoted by letter, ‘c’. Its SI unit is J/g°C and its CGS unit is cal/ g°C., zz Specific heat capacity can be measured by mixing method., zz If the system of both the objects is isolated from the environment:, , - Heat lost by hot object = Heat gained by cold object, zz Heat Q = mL (m – mass, L – latent heat) : (change of state), zz Heat Q = mc∆t (m – mass, c – specific heat capacity, ∆t – change in temperature) : (change of temperature), zz Heat lost by hot object = heat gained by calorimeter + Heat gained by water, , (20) CGS unit of specific heat capacity is, , MASTER KEY QUESTION SET - 5, Q.1. (A) 1. Fill in the blanks:, (1) Temperature is a degree of, , of a substance., , (2), , The melting point of ice is, , °C., , *(3), ,  uring transformation of liquid phase to solid, D, phase, the latent heat is, ., , ., , *(21) If objects of equal masses are given equal heat,, their final temperature will be different, this is, due to difference in their, ., (22), , Heat lost by hot object =, , ., , (23) The specific heat capacity of a substance can be, measured by, method., , (6) Melting of ice due to applied pressure, getting, converted back into ice after the pressure is, removed is called, ., , Ans. , (1) hotness or coldness (2) 0 (zero) (3) given out, (released) (4) Latent heat of vaporisation (5) lost, (6) Regelation (7) 4 (8) dew point (9) kg/m3, (10) Humidity (11) Absolute humidity (12) 100%, (13) humid (14) less than (15) joules (J) (16) calorie, (17) 4.18 (18) J/kg°C (19) specific heat capacity, (20) cal/g0c (21) specific heat capacities (22) Heat, gained by cold object (23) Mixing., , (7) The density of water is maximum at, , Q.1. (A) 2. Complete the analogy:, , (4) The quantity of heat absorbed by liquid at constant, temperature to convert into gas is, ., (5) When substance is converted from liquid to solid, state, the heat energy is, ., , °C., , (8) The temperature at which the air gets saturated, with water vapour is called, temperature., (9), , Absolute humidity is measured in, , ., , (10) The presence of water vapour in the atmosphere, is called, ., *(11) The amount of water vapour in air is determined, in terms of its, ., (12), , At dew point, the relative humidity is, , ., , (13) If the relative humidity is more than 60%, then, the air is, ., (14) If the relative humidity is, the air is dry., (15), , In SI system, heat is measured in, , (16), , In CGS, heat is measured in, , (17), , 1 calorie =, , (18), , SI unit of specific heat capacity is, , 60%, then, ., ., , joule., ., , (19) The heat energy required to raise the temperature, of unit mass of a substance through 1°C is called, of the substance., , (1), , Boiling point of water : 100°C :: Melting point of, ice:, , Ans. 0°C - Ice melts at 0°C, (2), , Latent heat of fusion of ice : 80 cal/g :: Steam :, , Ans. 540 cal/g - Latent heat of fusion of steam is 540, cal/g, (3) CGS unit of heat : calorie :: CGS unit of heat, capacity:, Ans. cal/g°C - Heat capacity of substance is measured, per gram of the substance per degree, increase in, temperature, (4), , Water : 1 cal/g°C :: Kerosene:, , Ans. 0.1 cal/g°C - The specific heat of Karosene is 0.52, cal/g°C., (5) Dry air : less than 60% Relative humidity :: Humid, air:, Ans. More than 60% Relative humidity - When the, relative humidity is more than 60%, we feel that, the air is humid.

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Master Key Science and Technology Part-1 (Std. X), , 98, , Q.1, , (A) 3. Answer in one sentence., , (1), , What is melting point of a solid?, , Ans. , The constant temperature at which a solid, converts to liquid is called melting point., (2), , What do you mean by latent heat of fusion?, , mass of water vapour required to saturate the, same volume of air at the same temperature is, called relative humidity., (13) Write the formula of percentage relative humidity., Ans. % relative humidity, Actual mass of water vapour, in the air in a given volume, =, × 100, Mass of water vapour, required to saturate the same, volume at same temperature., , Ans. The amount of heat energy absorbed at constant, temperature during transformation of solid into, liquid is called latent heat of fusion., (3), , Define boiling point of liquid., , Ans. , The constant temperature at which a liquid, transforms into gaseous state, is called boiling, point of that substance., (4) What do you mean by Latent heat of vapourization?, Ans. The heat energy absorbed at constant temperature, during transformation of liquid into gas is called, Latent heat of vapourization., *(5) , What are the different ways of heat transfer?, Ans. Three different ways of heat transfer are, Conduction, Convection and Radiation., *(6) Is the concept of latent heat applicable during, transformation of gaseous phase to liquid phase, and from liquid phase to solid phase? (Use your, Brain Power! Textbook page no. 63), Ans. Yes, the concept of Latent heat is applicable in, both the cases. The heat energy will be evolved, from the substance., , (14), , What is kilocalorie?, , Ans. T,  he amount of heat energy required to raise the, temperature of 1 kg of water by 1°C, i.e. 14.5°C to, 15.5°C is called kilocalorie., (15), , What is calorie?, , Ans. The amount of heat energy required to raise the, temperature of 1 g of water by 1°C, i.e. 14.5°C to, 15.5°C is called calorie., (16), , Write the units of heat in CGS and SI systems., , Ans. In CGS system, heat is measured in calorie and in, MKS, is joule (J)., (17), , What is specific heat capacity?, , Ans. The amount of heat energy required to raise the, temperature of a unit mass of a substance by 1°C, is called specific heat capacity., (18), , What are the units of specific heat capacity?, , *(7) , Where does the latent heat go during these, transformation? (Use your Brain Power! Textbook, page no. 63), , Ans. The units of specific heat capacity, in CGS is, cal/g°C and in SI, it is J/kg°C., , Ans. The latent heat is used for the transformation of, state and is released into the atmosphere., , Ans. Heat always gets transferred from a hot body, to the cold body till the temperature of both the, bodies becomes equal., , (8) , How anomalous behaviour of water can be, experimented in the laboratory?, Ans. Anomalous behaviour of water can be experimented, by using Hope's apparatus., (9), , What do you understand by the term 'Humidity'?, , Ans. T,  he moisture of air due to the presence of water, vapour in it is called Humidity., (10), , What is dew point temperature?, , Ans. T,  he temperature at which the air gets saturated, with water vapour is called dew point temperature., (11), , What is absolute humidity?, , (19), , (20), , How does heat get transfered?, , What is the principle of heat exchange?, , Ans. The principle of heat exchange is, when a hot, object is kept in contact with a cold object then,, Heat lost by hot object = Heat gained by cold object., [Isolated from the environment], Q.1. (A) 4. Match the columns:, (1), , Column A, (1) Melting point of ice, , Column B, (a) 100°C, , (2) Boiling point of water (b) 540 cal/g, , Ans. The mass of water vapour in a unit volume of air, is called absolute humidity., , (3) Specific Latent heat of, vaporisation of water, , (12), , (4) Specific Latent heat of (d) 0°C, fusion of ice, , What is relative humidity?, , Ans. The ratio of actual mass of water vapour in a, given volume of air at a given temperature to the, , Ans. (1 – d), (2 – a), (3 – b), (4 – c), , (c) 80 cal/g

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Heat, , (2), , 99, , Column A, (1) Humidity, , (a) Mass of water vapour, , (2) Absolute, humidity, , (b) Moistness or dampness, in air, , (3) Relative, humidity, , (c) Proportion of vapour, content of air, , Ans. (1 – b), (2 – a), (3 – c), (3), , relative humidity is 100% (11) True (12) False., In CGS, heat is measured in calorie. (13) False., 1 cal = 4.18 J. (14) False. Specific heat is denoted by, letter 'c'. (15) True (16) False. Specific heat capacity in, CGS is measured in cal/g°C (17) False. Calorimeter is, used to measure specific heat capacity. (18) True, , Column B, , Column A, (1) Heat, , (1) The specific heat capacity of, (a) Mercury, (c) Water, , Column B, (a) cal/g°C, , (2) Specific heat capacity (b) kg/m3, (3) Absolute humidity, , Q.1 (B) Choose and write the correct option:, , (c) calorie, , Ans. (1 – c), (2 – a), (3 – b), Q.1 , (A) 5. State whether following statement are, True or False. If false; rectify the statement., (1), , Different substances have different melting points., , (2), ,  ifferent substances have same specific heat, D, capacities., , (3), , At 4°C the volume of water is maximum., , (4), , Density of water is maximum at 4°C., , (5), , Presence of moisture in air is called humidity., , (6), , Heat is necessary to saturate the air with vapour., , (7), ,  ater vapour content of air is measured by, W, relative humidity., , (8), , Absolute humidity is measured in kg/m3., , (9), , Humidity is measured by relative humidity., , (10), , At dew point, the relative humidity is 60%., , (11), , S.I. unit of heat is joule., , (12), , In CGS system, heat is measured in joules., , (13), , 1 J = 4.18 cal., , (14), , Specific heat is denoted by letter 'h'., , (15), ,  pecific heat capacity is measured in SI system, S, in J/g°C., , (16), , I n CGS, specific heat capacity can be measured, in calorie., , (17), ,  hermometer is used to measure specific heat, T, capacity., , (18), , The specific heat capacity of water is 1.0 cal/g°C., , Ans. , (1) True (2) False. Different substances have different, specific heat capacities. (3) False. At 4°C volume is, minimum, density is maximum. (4) True (5) True, (6) False. Temperature has influence on saturation, of air. (7) False. It is measured by absolute humidity., (8) True (9) True (10) False. At dew point the, , is maximum., , (b) Copper, (d) Iron, , (2) If temperature of water increases from 1°C to, 3°C, the density of water, ., (a) remains the same (b) decreases, (c) increases, (d) fluctuates, (3) Aquatic plants and animals can survive in cold, region because of, ., (a), (b), (c), (d), (4), , humidity, dew point, heat capacity, Anomalous behaviour of water, , Water is such a substance which, upto 4°C, (a), (b), (c), (d), , at 0°C, , expands on heating, contracts on heating, contracts on cooling, remains unchanged, , (5) If temperature of water is lowered from 4°C to, 3°C; its volume, ., (a) increases, (b) decreases, (c) remains the same (d) fluctuates, (6) The specific heat capacity is measured in C.G.S., system in, ., (a) cal, (c) cal/g°C, (7), , (b) kcal, (d) joule/kg°C, , At dew point, the relative humidity is, (a) 100%, , (b) 10%, , (c) 60%, , (d) 50%, , (8) If the pressure is applied on ice, its melting point, ., (a) decreases, (b) increases, (c) remains the same (d) fluctuates, (9) The specific heat capacity of water is, kcal/kg°C, (a) 10, , (b) 1.0, , (c) 1.5, , (d) 0.5

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Master Key Science and Technology Part-1 (Std. X), , 100, , (10) When water boils and is converted into steam,, then, ., , (d) temperature at which the density of water is, maximum, , (a) heat is taken in and temperature remains, constant., (b) heat is taken in and temperatures rises., (c) heat is given out and temperatures lowers., (d) heat is given out and temperature remains, constant., , (15) Why do street ice cream vendors add salt their, freezing mixture intermittently?, (a) to increase the melting point of the freezing, mixture., (b) to maintain the melting point of the freezing, mixture as constant., (c) to reduce the melting point of the freezing, mixture, (d) to reduce the weight of the freezing mixture., , (11) In Hope’s a apparatus when the temperature of, water drops from 4°C to 3°C water, ., (a) Remains still, (b) moves upwards, (c) moves downwards (d) freezes solid, (12), , (16), , Wax layer, , Lead, , Copper, , (17), , With reference to the above diagram, which of, the following relation is incorrect?, , (13), , (14), , (b) Ccopper > Clead, , (c) CIron > Clead, , (d) Clead > Ccopper, , (a) 4180 J, , (b) 103 J, , (c) 1 cal, , (d) 4180 cal, , Heat is absorbed when, (a), (b), (c), (d), , Iron, , (a) CIron > Ccopper, , Heat needed to raise the temperature of 1 kg of, water from 14.5°C to 15.5°C is, ., , (18), , ., , water is converted into vapour, water is converted into ice, steam is converted into water, all of these, , Heat absorbed when 1 g of ice melts at 0°C to form, 1g of water at the same temperature is, cal, , Which of the following is a correct relation?, , (a) 80, , (b) 800, , (a), (b), (c), (d), , (c) 540, , (d) 54, , 4.18 calories = 1 joule, 1 calories = 4.18 joule, 107 calories = 1 joule, 1 calories = 107 joule, , (19), , In the experiment of Hope’s apparatus, the point, of intersection of the two curves in the graph, corresponds to the, ., , Temperature °C, , 12°, , T2, , 8°, 4°, 0°, , 5, , 10 15 20 25 30, Time (Minutes), , 35, , 40 45, , (a) melting point of ice, (b) freezing point of water, (c) temperature at which the density of water is, minimum, , A sample of air containing certain amount of water, vapour is saturated at a particular temperature. If, the temperature of air sample is raised further,, then, ., (a), (b), (c), (d), , The sample becomes oversaturated., The sample remains saturated., The sample reaches dewpoint., The sample becomes unsaturated., , Ans. ( 1) (c) water (2) (c) increases (3) (d) Anomalous, behaviour of water (4) (b) contracts on heating (5) (a), increases (6) (c) cal/g°C (7) (a) 100% (8) (a) decreases, (9) (b) 1.0 (10) (a) heat is taken in and temperature, remains constant. (11) (b) moves upwards (12) (d), Clead > Ccopper (13) (b) 1 calories = 4.18 joule (14) (d), temperature at which the density of water is maximum, (15) (c) to reduce the melting point of the freezing, mixture (16) (a) 4180 J (17) (a) water is converted, into vapour (18) (a) 80 (19) (d) The sample becomes, unsaturated.

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Heat, , 101, , Q.2.1. Solve the following:, , Q = mL (Heat energy during change of state), , 3cA = 5cB, 5, cA =, c, 3 B, → Specific heat of A is more., 5, → It is greater by a factor of, 3, , Note: Specific heat capacity (c) = 1 cal/g°C = 1, kcal/kg°C, , NUMERICALS FOR PRACTICE, , Type: A, Q = mc∆T , (Heat energy during change in, temperature), , ∆T = Higher temperature – Lower temperature, , , , *(1) (a) How much heat energy is necessary to raise, the temperature of 5 kg of water from 20°C to, 100 °C., Ans. Given:, , Mass (m) = 5 kg, , , , Temp. difference (∆T) = 100 - 20 = 80°C, , , , Specific heat capacity (c) = 1 kcal/kg°C, , To find: Heat energy (Q) = ?, Formula: Q = mc∆T, Solution: Q = 5 × 1 × 80, , (1) How much heat is required to raise the temperature, of 500 g of mercury from 20°C to 100°C. (Sp. heat, cap. of Hg is 0.033 kcal/kg°C), (1.32 kcal), (2) The initial temperature of 40 g copper ball is 24°C., When 200 cal heat is supplied, the temperature, rises to 74°C. Find the specific heat capacity of, copper?, (0.1 cal/g°C), (3) Calculate the heat gained by 50 g of ice at 0°C to, change into water of the same temperature. (Latent, heat of fusion of ice is 80 cal/g), (4000 cal), (4) Calculate the amount of heat absorbed by 100 g, of ice to change into water of 30°C. (Specific heat, capacity of water is 1 cal/g°C), (11000 cal), , = 400 kcal, Heat energy needed is 400 kcal., , Type: B, , (1) (b) How much energy needs to be supplied to, 150g of water at 100°C to convert it to steam at, 100°C. (Lvap = 540 cal/g), Ans. Given:, , m = 150g, , , , L = 540 cal/g, , Principle of Heat Exchange, Heat energy lost by, Heat energy gained, hot object, = by cold object, Heat lost by hot, Heat gained by, body (QA), = cold body (QB), Q = mc∆T, Q = mL, Note: cwater/Ice/Steam = 1 cal/g°C = 1 kcal/kg°C, , To find: Q = ?, Formula: Q = mL, Solution: Q = 150 × 540, , Lfusion for water = 80 cal/g = 80 kcal/kg, , = 81000 cal, , , Heat energy needed is 81000 cal., *(2) Equal heat is given to two objects A and B of, mass 1g. Temperature of A increases by 3°C and, B by 5°C. Which object has more specific heat?, And by what factor?, Ans. Given:, Mass (m), Temperature, difference (∆T), , , For A, , For B, , 1g, , 1g, , 3°C, , 5°C, , Q A = QB, , To find: cA = ?, , cB = ?, , Formula: Q = mc∆T, Solution: QA = QB, , Lvap for water = 540 cal/g = 540 kcal/kg, , *(1) Liquid ammonia is used in ice factory for making, ice from water. If water at 20 °C is to be converted, into 2 kg ice at 0 °C, how many grams of ammonia, is to be evaporated? (Given: The latent heat of, vaporization of ammonia= 341 cal/g), QA, QB, Water/Ice, , Ammonia, , 2kg, , m1, , (c), , 1 kcal/kg°C, , -, , t1, t2, L, , 20°C, 0°C, 80 kcal/kg, , 341 kcal/kg, , Ans. Given: (m), , To find: m1 = ?, , , , mAcA∆TA = mBcB∆TB, , Formulae: QA = QB, , , , 1 × cA × 3 = 1 × cB × 5, , Q = mc∆T, Q = mL

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Master Key Science and Technology Part-1 (Std. X), , 102, , Solution:, , (150 × 80) + {150 × 1 × (50 - 0)} = { m1 × 540} + {m1 × 1 ×, (100 - 50)}, , (1) Water, , mL, mc∆T, Water (20°C), → Water (0°C) → Ice (0°C), , QA = mc∆T + mL ............... (i), , 12000 + 7500 = m1 (540 + 50), , , 19500 = m1 × 590, , , , (2) Ammonia, , 590, m1 = 33.05 = 33, , , , QB= m1L1 ............... (ii), , Hence, 33g steam has to be mixed., , from (i) and (ii), QA = QB, mc∆T + mL = m1L1, 2 × 1 × (20 - 0) + (2 × 80) = m1 × 341, , , 40 + 160 = m1 × 341, , , , 200 = m1 × 341, , , m1 = 0.5865 kg, or, m1 = 586.5 g, , m1 =, , 200, 341, , *(3) If 80 g steam of temperature 97 °C is released on, an ice slab of temperature 0 °C, how much ice will, melt? How much energy will be transferred to the, ice when the steam will be transformed to water?, QA, Steam, , QB, Ice, , 80g, , m1, , (c), , 1 cal/g°C, , 1 cal/g°C, , t1, , 97°C, , 0°C, , t2, , 0°C, , 0°C, , L, , 540 cal/g, , 80 cal/g, , Ans. Given: (m), , Hence, 586.5g of ammonia has to be evaporated., *(2) A thermally insulated pot has 150 g ice at, temperature 0 °C. How much steam of 100 °C has, to be mixed to it, so that water of temperature, 50 °C will be obtained?, QA, QB, Ans. Given:, , (m), (c), t1, t2, L, , 19500, , m1 =, , Steam, m1, 1 cal/g°C, 100°C, 50°C, 540 cal/g, , Ice, 150g, 1 cal/g°C, 0°C, 50°C, 80 cal/g, , To find: m1 = ?, Formulae: QA = QB, Q = mc∆T, Q = mL, Solution:, (1) Steam, , mL, mc∆T, Steam(100°C) → Water (100°C), → Water (50°C), , ∴ QA= m1L1 + m1c1∆T1 ............... (i), (2) Ice, , mL, mc∆T, Ice, (0°C), →, Water, (0°C), → Water (50°C), , , ∴ QB = mL + mc∆T .................... (ii), from (i) and (ii), QB = QA, mL + mc∆T = m1L1 + m1c1∆T, , To find: QA = ?, m1 = ?, Formulae: QA = QB, Q = mc∆T, Q = mL, Solution:, (1) Steam, , mc∆T, mL, → Water (0°C), (97°C), Steam(97°C) → Water, , , ∴ QA = mL + mc∆T, , , = (80 × 540) + {80 × 1 × (97 - 0)}, , , , = 43200 + 7760, , , , QA = 50960 cal ............... (i), , (2) Ice, mL, , Ice (0°C) → Water (0°C), QB= m1L1 ............... (ii), from (i) and (ii), QA = QB, 50960 = m1L1, 50960 = m1 × 80, , , , 50960, 80, m1 = 637g, m1 =, , Hence, 637g of ice will melt and 50960 cal of, heat energy is transfered to ice.

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Heat, , 103, , Type: C, Calorimeter, Heat gained, Heat gained, Heat lost by, +, =, by liquid, by calorimeter, hot object, Q = Q1 + Q2, Q = mc∆T, Q = mL, Note: C(copper) = 0.1 cal/g°C = 0.1 kcal/kg°C, Final temperature becomes equal to, 'T' for all substances, *(1) A copper sphere of 100g mass is heated to raise, its temperature to 100°C and is released in water, of mass 195g and temperature 20°C in a copper, calorimeter. If the mass of calorimeter is 50g,, what will be the maximum temperature of water?, Ans. Given:, (m), (c), t1, t2, , Q, Copper, , Q1, Water, , Q2, Calorimeter, , 100g, 0.1 cal/, g°C, 100°C, T, , 195g, 1 cal/, g°C, 20°C, T, , 50g, 0.1 cal/, g°C, 20°C, T, , To find: T = ?, Formulae: Q = Q1 + Q2, Q = mc∆T, , , , Solution:, , The maximum temperature of water will be, 23.8°C., *(2) , A calorimeter has mass 100g and specific heat, 0.1 kcal/ kg°C. It contains 250 gm of liquid at, 30 °C having specific heat of 0.4 kcal/kg°C. If we, drop a piece of ice of mass 10 g at 0°C, what will, be the temperature of the mixture?, , Ans. Given:, , Q, , Q1, , Q2, , Ice, , Water, , Calorimeter, , (m), , 10g, , 250g, , 100g, , (c), , 1 cal/, g°C, , 0.4 cal/, g°C, , 0.1 cal/, g°C, , t1, , 0°C, , 30°C, , 30°C, , t2, , T, , T, , T, , L, , 80 cal/g, , –, , –, , To find: T = ?, Formulae: Q = Q1 + Q2, Q = mc∆T, Q = mL, , , , Solution:, (1) Ice, , mL, mc∆T, Ice, (0°C), →, Water, (0°C), → Water (T°C), , , ∴ Q = mL + mc∆T ............... (i), , (1) Copper, Q = mc∆T ............... (i), (2) Water, Q1= m1c1∆T1 ............... (i), (3) Calorimeter, Q2= m2c2∆T2 ............... (iii), from (i), (ii) and (iii), , (2) Water, Q1= m1c1 ∆T1 ............... (ii), (3) Calorimeter, Q2= m2c2 ∆T2 ............... (iii), from (i), (ii) and (iii), Q = Q1 + Q2, , mL + mc∆T = m1c1∆T1 + m2c2∆T2, , Q = Q1 + Q2, mc∆T = m1c1∆T1 + m2c2∆T2, , (10 × 80) + {10 × 1 × (T - 0)} = {250 × 0.4 × (30 - T)}, , {100 × 0.1 × (100 - T)} = {195 × 1 × (T - 20)} + {50 ×, 0.1 × (T - 20)}, , 800 + 10 T = 100 (30 - T) + 10 (30 - T), , , , , + {100 × 0.1 × (30 - T)}, , , , 10 (100 - T) = 195 (T - 20) + 5 (T - 20), , , , , , 1000 - 10T = (T - 20) (195 + 5), , 10T + 110T = 3300 - 800, , , , 1000 - 10T = (T - 20) 200, , , , , , 1000 - 10T = 200 T - 4000, , , , 1000 + 4000 = 200 T + 10 T, , , , , 5000 = 210 T, 5000, T=, = 23.8°C, 210, , 800 + 10T = 110 (30 - T), 120T = 2500, , T =, , 2500, = 20.8°C, 120, , The temperature of the mixture will be 20.8°C.

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Master Key Science and Technology Part-1 (Std. X), , 104, , due to presence of water vapour in it is called, humidity., , NUMERICALS FOR PRACTICE, (1) A solid of mass 250 g at 90°C is dropped in a, calorimeter containing 100 g of water at 15°C. The, resulting temperature of mixture becomes 40°C., The mass of calorimeter is 100 g and its specific, heat capacity is 0.1 cal/g°C. Find the specific heat, capacity of the solid., (0.2 cal/g°C), , (9), , (2) A copper sphere of mass 500 g is heated to 100°C, and then introduced into a copper calorimeter, containing 100 g of water at 20°C. Find the, maximum temperature of the mixture, if the mass, of calorimeter is 100g and specific heat capacity is, 0.1 cal/g°C., (45°C), , (11), , Dew point temperature, , Ans. , The temperature at which the air becomes, saturated with water vapour is called dew point., (10), , Absolute humidity, , Ans. The mass of water vapour in unit volume of air is, called absolute humidity. It is measured in kg/m3, Relative humidity, , Ans. The ratio of the mass of water vapour in given, volume of air at given temperature to the mass, of water vapour required to saturate the same, volume of air at the same temperature is called, relative humidity., , Q.2.2. Define/ write the laws:, , (12), , (1), , Ans. , The amount of heat required to raise the, temperature of 1 g of water by 1°C (from 14.5°C to, 15.5°C) is called one calorie of heat., , Latent heat of fusion, , Ans. The heat energy absorbed at a constant temperature, during transformation of solid into liquid is called, latent heat of fusion., (2), , Specific latent heat of fusion, , Ans. The amount of heat energy absorbed at a constant, temperature by unit mass of a solid to convert, into liquid phase is called the specific latent heat, of fusion., (3), , Melting point, , Ans. The constant temperature at which solid changes, to liquid state is called melting point of that solid., (4), , Latent heat of vaporisation, , (13), , One calorie of heat, , One kcal heat, , Ans. , The quantity of heat required to raise the, temperature of 1 kg of water through 1°C (from, 14.5°C to 15.5°C) is called kilocalorie heat., (14), , Specific heat capacity, , Ans. , The quantity of heat required to raise the, temperature of unit mass of substance through, 1°C is called specific heat capacity., (15), , Principle of heat exchange:, , Ans. The heat energy absorbed at constant temperature, during transformation of liquid into gas is called, Latent heat of vapourization., , Ans. When a hot body is kept in contact with a cold, body, isolated from the environment, then the, quantity of heat lost by the hot body is equal to, the quantity of heat gained by the cold body., , (5), , Q.2.3. Write short notes:, , Specific latent heat of vapourization, , Ans. The amount of heat energy absorbed at constant, temperature by unit mass of a liquid to convert, into gaseous phase is called specific latent heat of, vapourization., (6), , Boiling point, , Ans. , The constant temperature at which a liquid, changes to gaseous state is called as boiling point, of the liquid., (7), , Regelation, , Ans. , The phenomenon in which the ice converts, to liquid due to applied pressure and then, re-converts to ice again once the pressure is, removed is called regelation., (8), , Humidity, , Ans. The presence of dampness or moistness in air, , *(1) Latent heat, Ans., (i), The quantity of heat absorbed or released, , by a substance to change its state at constant, temperature is called latent heat., (ii), ,  he heat energy absorbed at constant temperature, T, during transformation of solid into liquid is called, latent heat of fusion. Latent heat of fusion of ice is, 80 cal/g., , (iii), ,  he heat energy absorbed at constant temperature, T, during transformation of liquid into gas is called, Latent heat of vapourization. The latent heat of, vaporisation of water is 540 cal/g., , (iv), ,  uring change of state there is no rise or fall of, D, temperature.

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Heat, , (2), , 105, , Regelation, , Ans., (i), If two small ice cubes are pressed upon each, other, the pressure at the surfaces of contact, increases and melting point is reduced., , container. At the middle of the cylinder, there is, a flat bowl., (iii), ,  wo thermometers, one at the upper region and, T, other at the lower region are fixed to record the, temperature., , (iv), ,  he freezing mixture of salt and ice is taken in the, T, flat bowl. Temperatures are recorded after every, 30 min., , (ii), , Due to this, ice melts and forms water., , (iii), ,  hen pressure is removed, then the melting point, W, comes back to normal and water freezes again to, form ice, sticking the two pieces of ice together., , (iv), , Ice melts when subjected to pressure and then, when pressure irreleased, it reconverts into ice, again. This is called regelation., , (v), ,  ater of 10°C is poured into the cylindrical, W, container. Initially, both the thermometers show, the temperature of 10°C., , (3), , Anomalous behaviour of water, , (vi), ,  s the temperature of water decreases due to, A, freezing mixture, it contracts, density increases, and goes to the bottom of the container till 4°C., , Ans., (i),  enerally, liquids expand on heating and contract, G, on cooling., (ii) Water, however, shows an exceptional behaviour., (iii) If water at room temperature is cooled, it contracts, till 4°C, but if it is cooled below 4°C (between 4°C, and 0°C), it expands instead of contracting., (iv), ,  n the contrary, if water at 0°C is heated, it, O, contracts instead of expanding upto 4°C., , (v), ,  his exceptional behaviour of water between 4°C, T, and 0°C is called anomalous behaviour of water., , (vi), , (viii) A,  s the temperature falls below 4°C, water in the, region expands instead of contracting, density, decreases and this water rises towards the surface., (ix), ,  fter sometime, the upper thermometer shows the, A, temperature of 0°C and the lower thermometer, shows 4°C., , Examples found in nature:, , (x), , Finally both thermometer show 0°C, , (a) Survival of aquatic plants and animals., , (xi), , In this way, the anomalous behaviour of water, can be demonstrated in the laboratory by using, Hope's apparatus., , (5), , Relative Humidity, , (b) Cracking of rocks in winter., (c) Pipes conducting water in cold regions burst, in a winter., (4), , (vii) D,  ue to this, for sometime the thermometer at the, bottom shows temperature of 4°C., , Hope's apparatus, , Ans., , Ans., Thermometer T2, Freezing, mixture, , (i), , 0°C, Cylindrical, trough, , % relative humidity, Actual mass of water vapour in the air, in a given volume, =, × 100, Mass of water vapour required to, saturate the same volume at same, temperature., , Thermometer T1, 4°C, (ii), ,  he relative humidity at the dew point is 100%., T, If the relative humidity is more than 60%, we feel, that the air is humid., , (iii), , If the relative humidity is less than 60%, we feel, that the air is dry., , Fig 5.1 Hope's apparatus, (i), (ii), ,  he anomalous behavior of water can be studied, T, with Hope's apparatus., In Hope's apparatus, there is a metallic cylindrical, ,  he ratio of actual mass of vapour content in the, T, air for a given volume and temperature to that, required to make the air saturated with vapour at, that temperature is called the relative humidity.

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Master Key Science and Technology Part-1 (Std. X), , 106, , Q.2.4. Distinguish between:, *(1) , What is the difference between heat and temperature?, Ans., , Heat, (i), , It is a form of energy., , (i), , (ii), (iii), , SI unit is joule (J), Two bodies of different materials having, same quantity of heat can be at different, temperatures., It can be measured by using Calorimeter., , (ii) SI unit is kelvin (K), (iii) Two bodies at same temperature can have, different quantities of heat., , (iv), (2), , It is a measure of degree of hotness or coldness of, an object., , (iv) It can be measured using Thermometer., , Absolute Humidity and Relative Humidity, , Ans., , (3), , Temperature, , Absolute Humidity, , Relative Humidity, , (i), , The mass of water vapour per unit volume of, air is called absolute humidity., , (i), , The ratio of the mass of water vapour in a given, volume of air at given temperature to the mass, of water vapour required to saturate the same, volume of air at the same temperature is called, relative humidity., , (ii), , It has a unit., , (ii) It has no unit., , (iii) It is measured in kg/m3., , (iii) It is measured in percentage., , (iv) Mass of water vapour present is measured., , (iv) Extent of humidity is measured., , Humidity and Dew point, , Ans., , Humidity, , Dew point, , (i), , The dampness or moistness of air due to, the presence of water vapour in it is called, humidity., , (i), , The temperature at which air becomes saturated, with water vapour is called dew point., , (ii), , It is usually measured in terms of relative, humidity which is a percentage., , (ii) It is measured in terms of temperature in degree, celcius., , (iii), , When air is saturated with water vapour, the, relative humidity is 100%., , (iii) When the temperature falls below the dew, point, then relative humidity is 100%, dew starts, forming., , Q.2.5. Give Scientific Reasons:, , (ii), , *(1) , You know that as we go higher than the sea level,, the boiling point of water decreases. What would, be effect on the melting point of solid? (Use your, Brain Power!; Textbook page no. 64), Ans., (i), As we go higher than sea level, the atmospheric, pressure decreases., (ii) So melting point of ice increases., ,  e feel that an object is hot, when our body, W, temperature is low as compared to that of the, object because heat is transferred from object to, body., , (iii), , imilarly when body temperature is high as, S, compared to that of object, we feel that the object, is cold because heat is transferred from body to, object., , *(2) , We feel that some objects are cold, and some are, hot. Is this feeling related in some way to our body, temperature? (Can you tell?; Textbook page no. 64), Ans., (i), If two objects are in contact with each other, there, will be an exchange of heat between them., , (3) , Why is a white trail observed at the back of a, flying plane in clear sky?, Ans., (i), In winter, as the plane flies the vapour released, by the aeroplane engine condenses and forms, clouds.

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Heat, , 107, , (ii), ,  his cloud appears in the form of a white trail, T, behind the flying aircraft., , (iii), ,  ence, a white trail is observed at the back of a, H, flying plane., , *(4) , How will you explain following statements with, the help of the anomalous behaviour of water?, (Use your Brain Power!; Textbook page no. 66), (a), In regions with cold climate, the aquatic, plants and animals can survive even when the, atmospheric temperature goes below 0°C., Ans., (i), In cold countries during winter, the temperature, of air falls to 0°C or even below., , (iii), , (iv), ,  s there is no place in the crevices for the, A, expansion of water, it exerts tremendous pressure, on the rocks which results in the crumbling of the, rocks., Hence, sometimes rocks crumble into pieces in, cold countries during winter., , Q.3.1. Explain the following / meaning of statement:, *(1) Role of Latent heat in the change of state of a, substances., Ans., (i), During change of state of a substance, object, , absorbs or release heat energy., (ii), , But the temperature does not change., , (ii), ,  s the temperature of air decreases, the water at, A, the surface of the lake begins to contract due to, which its density increases., , (iii), ,  he heat energy is utilized for weakening the, T, bonds between the atoms or molecules in the, substance., , (iii), ,  hen the temperature of the water falls below, W, 4°C, due to its anomalous behaviour, the water, expands instead of contracting., , (iv), ,  ence, this energy which causes a change in state, H, and does not change the temperature is called, Latent heat., , (iv), ,  s the temperature of water falls to 0°C, the water, A, on the surface of the lake gets converted to ice., , (v), , ince ice is a bad conductor of heat the water, S, below the ice layer continues to remain at 4°C even, though the temperature of air may fall below 0°C., , (vi), ,  ence, due to the anomalous behaviour of, H, water, aquatic animals stay alive even when the, temperature of air falls below 0°C, , (b), , In cold regions in winter the pipes for water, supply break and even rocks crack., , Ans., (i),  uring winter season, in cold countries the, D, temperature of atmosphere falls well below 0°C., (ii), ,  herefore, the temperature of water inside the, T, pipeline also decreases., , (iii), ,  hen the temperature of the water falls below, W, 4°C, it starts expanding instead of contracting, due to anomalous behaviour of water., , (iv), ,  ince there is no place for water to expand it exerts, S, tremendous pressure on the walls of the pipeline, causing them to burst. Hence, in winter the, pipeline carrying water burst in cold countries., , *(5) , In cold regions, the winter rocks crack due to, anomalous expansion of water., Ans., (i), (ii), ,  uring winter season, in cold countries the, D, temperature of atmosphere falls well below 0°C., If water is present within the crevices of the, , rocks, then due to its anomalous behaviour,, water begins to expand below 4°C., , *(2) , On what basis and how will you determine, whether air is saturated with vapour or not?, Ans., (i), We can determine whether air is saturated with, vapour or not on the basis of relative humidity., (ii), ,  elative humidity is the ratio of actual mass, R, of vapour content in air for given volume and, temperature to that required to make air saturated, with vapour at that temperature., , (iii), , If relative humidity is 100%, the air is saturated, with vapour. This is called dew point., , (iv), , If its value is less than 100%, then air is unsaturated., , *(3) , How can you relate the formation of water, droplets on the outer surface of a bottle taken, out of refrigerator with formation of dew., Ans., (i),  hen the bottle is taken out of the refrigerator,, W, the air around the bottle cools., (ii), ,  s the air cools, due to decrease in temperature it, A, becomes saturated with water vapour., , (iii), ,  s a result, the excess water vapour gets converted, A, into tiny droplets on the outer surface of the bottle, called dew., , (iv), ,  his is similar to the dew we see in the early, T, mornings on leaves of plants and window glass, of vehicle., , (4), , Specific heat capacity, , Ans., (i),  he amount of heat energy required to raise the, T, temperature of unit mass of substance through, 1°C is called specific heat capacity.

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Master Key Science and Technology Part-1 (Std. X), , 108, , 'm' is a mass of and object which has specific heat, capacity of 'c'. If its temperature is to be increased, by ∆T, then the heat absorbed by a substance can, be calculated as,, Heat absorbed = mc ∆T, , (iii), , ... ∆T is rise in, temperature, , I f the temperature of a substance is reduced, then, heat lost by a substance can be given as,, Heat lost = mc ∆T  ..... ∆T is fall in temperature., ∴ Specific heat capacity is necessary to find, amount of heat loss or gain., , (5), Principle of heat exchange, Ans., (i), When a hot and a cold object are kept in contact,, then, Heat lost by hot object = Heat gained by, cold object. [ In an isolated environment ], (ii), ,  hen a hot object and cold object are placed in, W, contact, then there is transfer of heat from hot, object to cold object till the temperature of both, become equal., , (iii), , I f there is no other loss or gain of heat energy from, the surrounding, then principle of heat exchange, is written as., Heat lost by hot body = Heat gained by cold body, , (6), ,  ow is specific heat measured? or Explain, H, 'mixing method' for measurement of specific, heat., , Ans., (i),  he specific heat of an object can be measured by, T, mixing method using a calorimeter., (ii), ,  hen a hot solid metallic ball (sphere) is put, W, in water in a calorimeter, heat exchange starts, between hot object, water and calorimeter., , (iii), ,  his continues till the temperature of all the three, T, becomes equal., , (iv), ,  eat lost by hot object (Q1) is equal to Heat gained, H, by water and calorimeter., , (v), , I f the specific heat of water and calorimeter are, known, then the specific heat of solid object can, be calculated., , What is meant by latent heat? How will the state, of matter transform if latent heat is given off?, Ans. , The quantity of heat absorbed or released, by a substance to change its state at constant, temperature is called latent heat. If Latent heat, is given off, then a gas will convert into a liquid,, and a liquid will convert into a solid., , *(8), , What is meant by ‘specific heat capacity’? How, will you prove experimentally that different, substances have different specific heat capacities?, , Ans., (i),  he quantity of heat energy required to raise a, T, temperature of unit mass of a substance through, 1°C is called specific heat capacity of a substance., (refer Fig 5.3), (ii) Three solid spheres of equal masses of material, iron, copper and lead are placed in boiling water for, same time. After removing them simultaneously,, they are placed on the wax tray together., (iii) Initially, all spheres have temperature of 100°C., The sphere which absorbs more heat can melt, more wax., (iv) Iron sphere melts more wax and goes deep into it, ray. The lead sphere goes less deep into the molten, wax. The copper sphere goes to intermediate depth., (v) This indicates that the heat absorbed by different, material differs. This is specific heat capacity of, metals., (vi) Specific heat capacity is denoted by letter 'c'. In, CGS, it is measured in cal/g°C, while in MKS, it is, measured in J/kg°C., *(9), , Which principle is used to measure the specific, heat capacity of a substance?, , Ans. The specific heat capacity of a substance is, measured using principle of heat exchange., It states that,, Heat lost by hot object = Heat gained by cold, object., (10), , Explain the given graph:, , *(a) , Observe the graph below. Temperature of, water is raised from 0°C, observe the change in, volume at different temperatures and explain, the anomalous behaviour of water., , Specific Volume (cm3/g), , (ii), , *(7), , 1000.35, 1000.30, 1000.25, 1000.20, 1000.15, 1000.10, 1000.05, 1000.00, 0, , 1 2 3 4 5 6 7 8 9 10, Temperature °C, , Fig : 5.2 Graph between the volume, and temperature of water, Ans., (i) As water at 0°C is heated upto 4°C, it contracts, instead of expanding. The volume of water at

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Heat, , 109, , 4°C is minimum. If water at 4°C is further heated,, then it expands and its volume increases., (ii) This exceptional behaviour which water shows, between 0°C to 4°C is called anomalous behaviour, of water., , *(2) In the below experiment, the wire moves through, the ice slab. However, the ice slab does not break., Why? (Use your Brain Power!; Textbook page no. 64, Ice, , (iii) From graph, we come to know that if water is, heated from 0°C to 4°C, it contracts instead of, expanding. Since volume at 4°C is minimum,, hence, density is maximum., Boiling water + vapour, C, D, Liquid -, , 100, 90, 80, 70, 60, 50, 40, 30, 20, 10, , W, at, , e, qu r, id, , state, , (0, 0), , A, , Ice + Water, (Solid + liquid), , B, 2 4 6 8 10 12 14 16 18 20 22 24 26, Time (Minutes), , Fig 5.3: Temperature vs Time Graph, Ans., (i), , The graph shown is a temperature - time graph of, Ice transforming into water and steam., , (ii), ,  B shows that, at constant temperature of 0°C, A, ice changes to water., , (iii) The temperature at which ice changes to water, is, called melting point of ice i.e. 0°C., (iv) , Now, water at 0°C starts heating and its, temperature gradually increases to 100°C., (v), , Weight, , Gaseous, , Li, , °C, , *(11) Explain the following graph:, , BC shows increase in temperature from 0°C to, 100°C without a change in state., , (vi) , CD shows that, at constant temperature of 100°C,, water changes to vapour., (vii) , The temperature at which water changes to, vapour is called boiling point of water i.e. 100°C., Q.3.2. Open Ended Question:, *(1) , While deciding the unit for heat, which, temperatures interval is chosen? Why?, Ans. , The temperature range of 14.5°C to 15.5°C is, chosen while deciding the unit for heat. If we, heat 1 kg of water by 1°C in different temperature, range other than 14.5°C to 15.5°C, the amount of, heat required will be slightly different., , Fig 5.4, , Ans., (i), (ii), (iii), ,  he melting point of ice becomes lower than 0°C, T, due to the applied pressure by the wire., So the Ice gets converted into water and the wire, moves through the ice slab., Once the pressure is removed, the water becomes, ice again. In this way the wire moves through the, ice slab without breaking it., , *(3) , Is there any relationship of latent heat with the, regelation? (Use your Brain Power! Textbook page, no. 64), Ans. Yes there is relationship between latent heat and, regelation, (i), The amount of heat energy absorbed or released, during change of state is called Latent heat., (ii) The phenomenon in which ice converts to liquid, due to applied pressure and then reconverts to ice, once the pressure is removed is called Regelation., (iii) Since there are change of states during Regelation, process latent heat is one of the factors affecting, Regelation., Q.3.3. Draw a neat and labelled diagram:, *(1) Hope's apparatus, Ans., Freezing, mixture, Cylindrical, trough, , Thermometer T2, 0°C, , Thermometer T1, 4°C, , Fig 5.5

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Master Key Science and Technology Part-1 (Std. X), , 110, , Experiment of specific heat capacity with three, solid spheres., , What is the state of the substance in the region, DE of the graph?, , Ans: , Liquid and gas, , Wax layer, , (ii) What is the heat energy absorbed in the region, EF called?, Ans: , Latent heat of vaporization, , Lead, , (2) Answer the following questions based on the, diagram given below., , Iron, °C, , *(3) Graph of time against temperature for Hope's, Apparatus., Ans., , Temperature °C, , 12°, , Upper Thermometer (T2), , 8°, , Lower Thermometer (T1), , (0, 0), , 4°, 0°, , 5, , 10 15 20 25 30, Time (Minutes), , 35, , 40 45, , D, , A, , B, 2 4 6 8 10 12 14 16 18 20 22 24 26, , (i), , What part of the graph shows a change from ice, to water at a constant temperature?, Ans. Part AB,  hich part of the graph shows a change in the, W, temperature without a change in the state?, Ans. Part BC, , Ice at 0°C, , (iii) D,  efine : Latent heat of Vapourization., Ans. The heat energy absorbed at constant temperature, during transformation of liquid to gas is called, latent heat of vaporization., , Water at, 4°C, , (iv), ,  hat is the heat energy absorbed from A to B, W, called?, Ans. Latent heat of fusion., , Fish, , (v), , What is the heat energy absorbed from C to D, called?, Ans. Latent heat of vaporization., , Fig 5.7: Aquatic animals in cold regions, Q.3.4. Figure/Application Based Questions., ,  nswer the following based on the graph given, A, below., , (3) , Answer the following based on the diagram, given below., Thermometer T2, Freezing, mixture, , 150, , Temperature (°C), , Temperature vs Time Graph, , (ii), , Survival of aquatic life., , Ans., , (1), , C, , Time (Minutes), , Fig 5.6: Time - Temperature Graph, *(4), , 100, 90, 80, 70, 60, 50, 40, 30, 20, 10, , e, qu r, id, , Copper, , W, at, , Ans., , (i), , Li, , *(2), , D, , Thermometer T1, , E, , 50, , -50, , Cylindrical, trough, , F, , 100, , 0, , °C, , B, , °C, , C, 10 20 30 40 50 60 70 80 90 100 110 120, , Time (s)

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Heat, , 111, , (i), ,  hen the temperature of water decreases from, W, 4°C to 2°C, will the level of water rise or fall in, the cylinder?, Ans. When the temperature of water decreases from, 4°C to 2°C, water expands and therefore it will, rise in the cylinder., , (i), , (ii), , (ii) Define the phenomenon in minimum words., Ans. The phenomena in which the ice converts to water, due to applied pressure and then re-converts to, ice again once the pressure is removed is called, regelation., , Which of the two thermometers, T1 or T2, will, show a temperature of 0°C first?, Ans. Thermometer T2 will show temperature of 0°C, first., (iii), , ipe lines crack in winter season in cold, P, countries due to which phenomena?, Ans. Pipelines crack in winter season due to anomalous, behaviour of water., , Specific Volume (cm3/g), , (4) , Answer the following based on the diagram, given below., 1000.35, 1000.30, 1000.25, 1000.20, 1000.15, 1000.10, 1000.05, 1000.00, 0, , 1 2 3 4 5 6 7 8 9 10, Temperature °C, , (i), Which behavior of water is shown in the graph?, Ans. Anomalous behavior of water., (ii), ,  t what temperature is the volume of water, A, minimum?, Ans. 4°C, ,  bove experiment is meant to understand, A, which phenomenon of heat?, Ans. Regelation, , (iii), ,  hat is the effect of the increase in pressure on, W, the melting point of ice?, Ans. As the pressure increases, the melting point of ice, decreases., (iv), , The wire does not breaks the ice slab while, passing through it. Why?, Ans. On applying pressure, the melting point of ice, reduces and it melts. As the wire moves down,, the pressure is removed the melting point comes, back to normal and water freezes back to ice, again. Thus, the ice slab doesnot breaks., (v) Will latent heat affect the process?, Ans. Yes, latent heat will affect the process since there, is a change in state involved., (6) , The graph given in daigram represents a, cooling curve for a substance being cooled from, its vapour to solid phase by releasing heat at a, constant rate of 100 J s–1., , 200, , (iii), ,  n cooling from 0°C to 4°C, what is the effect on, O, (a) Volume (b) Density of water, Ans. (a) The volume of water decreases on cooling, from0°C to 4°C, , Temperature (°C), , 150, , (b) The density of water increases on cooling, from 0°C to 4°C., (5), , Answer the following based on the diagram, given below., Ice, , (i), , A, B, , C, D, , 100, , E, F, , 50, 0, 5, , 10, , 15, , 20, , 25, , 30, , 35, , Time (s), , What is the boiling point of the substance?, , Ans. The boiling point of the substance is 150°C., (ii), , What happens in the region DE?, , Ans. The region DE represents freezing of the, substance where the liquid changes into solid at, a constant temperature (= 100°C)., (iii), Weight, , What is the melting point of the substance?, , Ans. The melting point of the substance is 100°C.

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Master Key Science and Technology Part-1 (Std. X), , 112, , Q.4.1. Paragraph Based Question:, , *(i) Heat is transferred from where to where?, Ans. Heat gets transferred from hot object to the cold, object., , *(1) , Read the following paragraph carefully and, answer the following questions:, , *(ii) Which principle do we learn about from this, process?, Ans. Principle of Heat exchange., , If heat is exchanged between a hot and cold, object, the temperature of the cold object, goes on increasing due to gain of energy and, the temperature of the hot object goes on, decreasing due to loss of energy., , *(iii) How will you state the principle briefly?, Ans. The principle of Heat exchange states that, if heat, is exchanged between a hot and cold object, then, heat energy lost by hot object is equal to heat, energy gained by the cold object., , The change in temperature continues till the, temperatures of both the objects attain the, same value. In this process, the cold object, gains heat energy and the hot object loses, heat energy. If the system of both the objects, is isolated from the environment by keeping, it inside a heat resistant box (meaning that the, energy exchange takes place between the two, objects only), then no energy can flow from, inside the box or come into the box., , *(iv) Which property of the substance is measured, using this principle., Ans. Using this principle, the specific heat capacity of, a substance can be measured., , vvv, , ASSIGNMENT - 5, Time : 1 Hr., , , Marks : 20, , Q.1. (A) Answer the following:, , (3), , (1), , What is Specific latent heat of fusion?, , (2), , Name the apparatus used to experiment anomalous behavior of water., , (3), , What is the Principle of Heat exchange?, , , (1), , Q.2. (B) Choose and write the correct option., The specific heat capacity of, (a) Mercury, , (2), , (b) Copper, , (2), , is maximum., (c) Water, , (d) Iron, , Aquatic plants and animals can survive in cold regions because of, , ., , (a) Humidity (b) Dew point (c) Heat capacity (d) Anomalous behaviour of water, , , Q.2. Answer the following: (Any 2), , (4), , (1), , In cold regions in winter, rocks crack due to anomalous expansion of water., , (2), , Distinguish between heat and temperature., , (3), , Equal heat is given to two objects A and B of mass 1g. Temperature of A increases by 3°C and B by 5°C., Which object has more specific heat? and by what factor?, , , (1), , , Q.3. Answer the following: (Any 2), , (6), , Answer the following based on the diagram given below., Ice, , Weight

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Heat, , 113, , (i), , Above experiment is meant to understand which phenomenon of heat?, , (ii), , Define the phenomenon in minimum words., , (iii), , What is the effect of the increase in pressure on the melting point of ice?, , (2), , A calorimeter has mass 100 g and specific heat 0.1 kcal/kg°C. It contains 250 gm of liquid at 30°C having, specific heat of 0.4 kcal/kg°C. If we drop a piece of ice of mass 10g at 0°C, what will be the temerature of, the mixture?, , (3), , If 80 g steam of temperature 97°C is released on an ice slab of temperature 0°C, how much ice will melt?, How much energy will be transferred to the ice when the steam will be transformed to water?, , , , Q.4. Answer the following: (Any 1), , (5), , Explain Hope's apparatus with a neat and labelled diagram., , (2), , Answer the following questions based on the diagram given below., , (0, 0), , id, , qu, , A, , B, 2 4 6 8 10 12 14 16 18 20 22 24 26, Time (Minutes), , , (i), , D, , er, , C, , W, at, , 100, 90, 80, 70, 60, 50, 40, 30, 20, 10, , Li, , °C, , (1), , vvv, , Temperature vs Time Graph, What part of the graph shows a change from ice to water at a constant temperature?, , (ii) Which part of the graph shows a change in the temperature without a change in the state?, (iii) Define : Latent heat of Vapourization., (iv) What is the heat energy absorbed from A to B called?, (v), , What is the heat energy absorbed from C to D called?below., , vvv

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6, , Refraction of Light, CONCEPT MAP, Refraction of Light, Phenomena, , Total Internal Reflection, e.g. mirage, , Refraction, Refractive, index, sin i, , , 2 η1, sin r, , l, l, l, , When light is , incident, , Rainbow, , V1, V2, , Effects of, Atmospheric, Refraction., Twinkling of stars, Advanced sunrise, Delayed sunset., , Dispresion, , Combined Effect, Leads to the, formation of, l, , Denser to, Rarer, Away from, the normal, , Rarer to, Danser, Towards the, normal, , Normal at, the boundary, of 2 medium, Goes, undeviated, , l, l, l, , splitting of, light into seven, colours., Band of 7 colours, is called spectrum, Violet deviates, the most, Red deviates the, least, , Points to Remember:, �, , Refraction of light: The phenomenon of change in the direction of light when it passes from one transparent, medium to another obliquely, is called refraction of light., , �, , Laws of Refraction:, (a) The incident ray and the refracted ray are on the opposite sides of the normal at the point of incidence, and all the three lie in the same plane., (b) For a given pair of media,, , , , If ‘i’ is angle of incidence and ‘r’ is the angle of refraction; then,, , sin i, = constant = η, sin r, , This constant is called as the ‘refractive index’ of second medium with respect to the first medium. It, is denoted by 2n1., sin i, ∴ 2n1 =, = n., sin r, This is also called Snell’s law. A ray incident along the normal (i = O) goes forward in the same direction (r = O)., , , The refractive index of the second medium with respect to the first medium is given by the ratio of magnitude, of velocity of light in the first to that in second medium. It is denoted by 2n1., �, , The refractive index depends upon the relative speed of propagation of light in different media., , �, , Suppose v1 is the velocity of light in medium 1 and v2 is the velocity in medium 2., (114)

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Refraction of Light, , 115, , ∴ Refractive index =, , , , n1 =, , 2, , Velocity of light in first medium, Velocity of light in second medium, v1, v2, , In the same way, the refractive index of the first medium with respect to second medium is denoted as,, , , �, , n2 =, , 1, , v2, v1, , Refraction of light across the medium:, (a) When the light ray moves from an optically rarer medium to optically denser medium, it bends, towards the normal., (b) When a ray of light travels from an optically denser medium to an optically rarer medium, it bends, away from the normal., (c) When the ray is incident normally at the boundary between the two media, it does not change its, direction and hence does not get refracted., , �, ,  irage is due to the apparent random wavering of the hot air rising above the heated roads. It is the, M, apparent image of an object originating from the ground below the object., , �, , Twinkling of stars:, (a) Stars emit their own light and shine at night. They are at infinite distances from us, hence appear as, the point source., (b) Because of mobility of atmospheric air and change in air density and the temperature, the atmosphere, is not steady. Hence, refractive index of air goes on changing continuously and randomly., (c) When atmosphere refracts more light towards us, the star is seen bright. When atmosphere refracts, less light towards us, the star is seen dim. Thus, due to change in the refractive index of atmosphere,, stars appear twinkling at night. Different colour lights have different frequencies. In vacuum, the, velocity of light rays of all frequencies is the same. In a medium like glass, the angle of refraction are, different for different colours., , �, , Dispersion of light:, (a) The phenomenon of splitting of white light or sunlight into its component colours while passing, through a medium is called the dispersion of light., , (b), , The band of coloured components of a light beam is called its ‘spectrum’., , (c), , Various colours obtained after splitting are in a sequence - Violet, Indigo, Blue, Green, Yellow, Orange , and Red. The acronym is VIBGYOR and can be used to remember the colours., , �, , The angle of incidence for which the angle of refraction is 90° is called as critical angle., , �, ,  hen the angle of incidence is larger than the critical angle, the angle of refraction is more than 90°, then the, W, ray comes back in the denser medium. Hence, total internal reflection takes place. When angle of incidence, is more than critical angle, total internal reflection takes place., , sin i, sin r, For total internal reflection, i = Critical angle, r = 90°, sin i, ∴ 1n2 =, = sin i , ( sin 90° = 1), sin 90°, When light enters a rarer medium from a denser medium, it gets partially reflected. This is called partial, reflection., ∴ 1n2 =, , �, �, ,  ormation of rainbow: Rainbow is a natural phenomena with a combined effect of dispersion, refraction, F, and total internal reflection of light.

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Master Key Science and Technology Part-1 (Std. X), , 116, , MASTER KEY QUESTION SET - 6, , (2) , Critical angle, angle of incidence, angle of, refraction, alternate angle, , Q.1. (A) 1. Fill in the blanks and rewrite the, completed statements:, , Ans. Alternate angle. Others are concerned with, refraction., , *(1) Refractive index depends on the, of, light., *(2) The change in the, of light rays while, going from one medium to another is called, refraction., (3) The refractive index of air goes on, while coming to the earth surface., (4) The splitting of sunlight into its component, colours is called, of light., (5) In the dispersion of sunlight through glass prism, the, light deviates the most., (6) , When the light ray is incident normal to the, interface between any two media, the angle of, incidence is, ., (7) When the ray of light passes from denser medium, to rarer medium obliquely, it bends, the normal., (8) When the value of ‘r’ is 90°, then the ‘i’ for it is, called as, ., (9) When light travels from denser medium to rarer,, then the partial, takes place., v1, (10) The refractive index 2n1 =, , (3) Reflection, dispersion, refractive index, refraction, , (11) The refractive index of diamond is, , Ans. Violet : The wavelength of violet is the smallest, and therefore it deviates the maximum., , ., , (12) When the light passes through glass prism, the, light deviates the least., (13) The scientist, used the glass prism, for obtaining spectrum from the sunlight., , Ans. Refractive index. Refractive index is a value while, the rest are natural phenomena of light., (4), , Brown, blue, green, red, , Ans. Brown. It is not a component of white light. Blue,, red and green are components of white light., Q.1. (A) 3. Complete the analogy:, (1) Refractive index of water : 1.33 : : Refractive index, of air :, Ans. 1.0003 : The value of refractive index is different, for different media., (2) , Stars : Point source of light : : Planets :, Ans. Collection of point sources of light : Planets are, much closer to us as compared to stars, therefore,, they appear as a collection of point sources., (3) Diamond : Denser medium : : Air :, Ans. Rarer medium : The speed of light is more in air, than in diamond., (4) Least deviation : Red : : Maximum deviation :, , (5) , Spectrum : Dispersion : : Twinkling of stars :, , (14) The wavelength of visible light range for human, is between, ., , Ans. Refraction : Spectrum is formed by dispersion of, white light or sunlight, whereas the changes in, the refractive index of air stars appear to twinkle., , (15) The wavelength of, which is, , Ans. 1.33 : Liquid water is denser than solid ice., , (16), , 1 nm =, , ., , light is the least, , m., , (17) The wavelength of violet light is, , ., , (18) When the angle of incidence is more than critical, angle, takes place., Ans. , (1) velocity (2) direction (3) increasing (4) dispersion, (5) violet (6) 0° (7) away from (8) critical angle, (9) reflection (10) v2 (11) 2.42 (12) red (13) Isaac, Newton (14) 400 nm and 700 nm (15) violet, 400 nm, (16) 10-9 (17) 400 nm (18) Total internal reflection, Q.1. (A) 2. Find the odd word out:, (1) Mirage, twinkling of stars, scattering of light,, apparent position of stars, Ans. Scattering of light. It is reflection, rest are refraction., , (6), , Ice : 1.31 : : Water :, , (7) Maximum refractive index : Violet : : Minimum, refractive index :, Ans. Red : Red light has the maximum wavelength, and its refractive index is minimum compared to, violet light which has the smallest wavelength., v, (8), n = 1 : : 1n2 :, 2 1, v2, v, Ans. 2 : Velocity of light in a medium depends on the, v1, refractive index of the medium., (9) , ∠ i < critical angle : Partial reflection : : ∠ i >, critical angle :, Ans. Total internal reflection : If the angle of incidence, is greater than the critical angle, then all the light, gets reflected back into the dense medium

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Refraction of Light, , (10), , Lens : Transparent : : Mirror :, , Ans. O,  paque : Light can pass through the lens whereas, in a mirror, light cannot pass through because it, is covered by a mirror coating on the side., Q.1. (A) 4. Answer in one sentence:, (1), , What is the refractive index of water?, , Ans. The refractive index of water is 1.33., (2), ,  rite the ratio of velocity of light in first medium, W, to the velocity of light in the second medium., , Ans. Refractive index, =, ∴, (3), , Velocity of light in the first medium, , *(11) W,  hat is meant by reflection of light?(Can you, recall?; Textbook page no. 73), Ans. , When rays of light fall on a smooth surface they, turn back. This is called reflection of light., *(12) Have you seen a mirage which is an illusion of, the appearance of water on a hot road or in a, desert? (Can you tell?; Textbook page no. 76)., Ans. , Yes. I have seen mirage on hot road during, summer days., Q.1. (A) 5. Match the pairs:, (1), , Velocity of light in the second medium, v, n = 1, 2 1, v2, , What do you mean by dispersion of light?, , Ans. , The process of separation of light into its, component colours while passing through a, medium is called dispersion of light., (4), , 117, , What is critical angle?, , (a) 1.5, , (2) Air, , (b) 2.42, , (3) Alcohol, , (c) 1.77, , (4) Benzene, , (d) 1.36, , (2), , Column A, (1) Formation of, rainbow, , Column B, (a) Total internal, reflection, , (2) Twinkling of stars (b) Dispersion, , Ans. , Light changes its direction when passing from, one transparent medium to another transparent, medium. This is called the refraction of light., , Ans. , If the angle of incidence is zero, then the angle of, refraction is also zero., , (1) Diamond, , Ans. (1 – b), (2 – e), (3 – d), (4 – a), , What is refraction of light?, , (6) What is the value of angle of refraction if angle of, incidence is 0°?, , Column B, , (e) 1.0003, , Ans. T,  he angle of incidence for which the angle of, refraction is 90° is called critical angle., (5), , Column A, , (3) Mirage, , (c) Dispersion,, refraction,, reflection, , (4) Spectrum, , (d) Atmospheric, refraction, , Ans. (1 – c), (2 – d), (3 – a), (4 – b), (3), , Column A, , Column B, , (7) When sunlight passes through glass prism which, light deviates the least?, , (1) Refraction, , (a), , With respect to, vacuum, , Ans. When sunlight passes through glass prism, the, red light deviates the least., , (2) Refractive index, , (b), , Change in, direction of light, , (8), , (3), , What is total internal reflection?, , Ans. When the angle of incidence in a denser medium, is more than its critical angle, then the light, reflects back into the denser medium. This is, called total internal reflection., (9), , (10), , (4) Dispersion, , What is absolute refractive index?, , Ans. , The refractive index of medium with respect to, vacuum is called its absolute refractive index., , (c) Spectrum, (d), , sin i, sin r, , (e) Mirage, , What is partial reflection?, , Ans. When light enters a rarer medium from denser, medium, a part of it gets reflected and comes, back into the denser medium as per laws of, reflection. This is called partial reflection., , Absolute, refractive index, , Ans. (1 - b), (2 - d), (3 - a), (4 - c), Q.1. (A) 6. State whether the following statements, are 'True' or 'False'., (1), , Light has different velocities in different media., , (2) The speed of violet light is less than that of red, light.

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Master Key Science and Technology Part-1 (Std. X), , 118, , (3) Incident ray and refracted rays are parallel to the, normal., , (d), , (4) The refractive index of the medium depends on, the wavelength of light., , *(2) If the refractive index of glass with respect to air, , (5) , When light rays travel from rarer to denser, medium, it bends towards the normal., (6) When light ray travels from denser medium to, rarer medium, it travels parallel to the normal., (7) If angle of incidence is zero, then the angle of, refraction is 90°., (8) When light travels obliquely from glass to air, it, bends away from the normal., (9) The splitting of white light into its component, colours is called dispersion., Ans. , (1) True (2) True (3) False. Incident ray and refracted, ray lie on the opposite sides of normal. (4) True, (5) True (6) False. Light ray bends away from the normal., (7) False. The angle of refraction would also be zero., (8) True (9) True, Q.1. (A) 7. Name the following:, (1) The phenomenon of splitting of sunlight into, spectrum., , 3, , then the refractive index of air with respect, 2, to glass, ., 1, 2, 1, (a), (b) 3, (c) 3, (d), 3, 2, *(3) We can see the sun even when it is little below the, horizon because of, ., is, , (4) The ratio of sine of angle of incidence to the sine, of angle of refraction., Ans. , Refractive index, (5) The light ray which gets deviated the least while, passing through glass prism., , (c) Dispersion of light, , (d) Absorption of light, , (c) does not bend, , (d) turn back, , (5) When light passes through glass slab, the incident, ray and emergent ray are, ., (a) intersecting , , (b) parallel, , (c) making an angle, , (d) touching, , (6) When light ray travels from denser medium to, rarer, then it, ., (a) bends towards normal, (b) bends away from normal, (c) goes undeviated, , Ans. , Diamond, Ans. , Atmospheric refraction, , (b) Refraction of light, , (a) bends towards normal (b) bends away, , A material whose refractive index is 2.42., , (3) The reason for stars twinkling at night., , (a) Reflection of light, , (4) When light ray passes from air to glass slab then, it, ., , Ans. Dispersion of light, (2), , Changing refractive index of the atmospheric, gases, , (d) deviates randomly, (7), , The velocity of light in air is, (a) 3 × 10, , 10 , , (c) 1.5 × 10, , 8 , , (b) 3 × 10, , m/s., 8, , (d) 0.3 × 108, , (8) Rakesh performs the experiments on tracing the, path of a ray of light passing through a rectangular, glass slab for different angles of incidence. He, observes that in all cases, , Ans. , Red light, , (a) ∠i > ∠r but ∠i = ∠e (b) ∠i < ∠r but ∠i = ∠e, , (6) Phenomenon behind formation of rainbow., , (c) ∠i > ∠e but ∠i = ∠r (d) ∠i < ∠e but ∠i = ∠r, , Ans. , Dispersion, total internal reflection and refraction, (7), , The unit of refractive index., , Ans. , No unit, (8), , Refractive index of kerosene., , Ans. 1.39, Q.1. (B) Choose and write the correct option:, *(1), , What is the reason for the twinkling of stars?, , (a), , Explosions occuring in the stars from time to time, , (b), , Absorption of light in the earth’s atmosphere, , (c), , Motion of stars, , (9) When a ray of light travels from air to glass and, strikes the surface of separation at 90°, then it, ., (a) bends towards the normal, (b) bends away from the normal, (c) passes without bending, (d) reflects to air, (10) A ray of light incident from a denser medium, passes through a rarer medium in a straight line., What should be angle of incidence ?, (a) 0°, , (b) 30°, , (c) 60°, , (d) 120°

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Refraction of Light, , 119, , (11) The ray of light gets deviated when it passes, from one medium to another medium because, ., , The correctly marked angles are:, , (a) the colour of light changes, , r, , i, , (b) the frequency of light changes, , δ, , e, , (c) the velocity of light changes, (d) None of these, (12) A ray of light strikes the glass slab at an angle of, 50°. What is the angle of incidence ?, (a) 50°, , (b) 40°, , (c) 60°, , (d) 120°, , (13) A glass-slab is placed in the path of convergent, light. The point of convergence of light, ., , (18), , (b) ∠i and ∠e, , (c) ∠i, ∠e and ∠δ, , (d) ∠i, ∠r and ∠e, , Will the light travel through soapy water with the, same velocity as it travels through air?, (a) Yes , , (c) Depends on the conditions, , (b) moves away from slab, , (d) None of these, (19), , (d) undergoes lateral shift, (14) , What is the speed of light in a transparent, medium having absolute refractive index 1.25?, (a) 1.25 × 108 m/s, , (b) 2.4 × 108 m/s, , (c) 3.0 × 108 m/s, , (d) 1.5 × 108 m/s, , (15) Which colour of light deviates the least in the, spectrum obtained with a prism ?, (a) Red, , (b) Yellow, , (c) Violet (d) Blue, , The figure shows the path of ray of light, propagating from medium 1 to medium 2. The, refractive index of medium 1 with respect to, medium 2 is, ., M, , In an experiment to trace the path of a ray of light, passing through a rectangular glass slab, four, students tabulated their observations as given, below. Which student is correct., (a) S.N. ∠i ∠r ∠e, 1., 2., 3., , 300, 450, 600, , 200, 290, 360, , 300, 430, 600, , (b), , S.N., 1., 2., 3., , ∠i, 300, 450, 600, , ∠r, 150, 200, 280, , ∠e, 380, 530, 670, , (c), , S.N., 1., 2., 3., , ∠i, 300, 450, 600, , ∠r, 300, 450, 600, , ∠e, 200, 290, 360, , (d), , S.N., 1., 2., 3., , ∠i, 300, 450, 600, , ∠r, 200, 290, 360, , ∠e, 200, 290, 360, , Medium 1, , 30o, , 45o, , Medium 2, , N, (a) sin 450 , sin 600, , (b) sin 600, sin 450, , sin 450 , sin 300, , 0, (d) sin 30, sin 450, , (c), (17), , (b) No, , (a) moves towards the slab, (c) remains at the same point, , (16), , (a) ∠i and ∠r , , After tracing the path of a ray of light through, a glass prism, a student marked the angle of, indicence (∠i), angle of emergence (∠e) and the, angle of deviation (∠D) as shown in the diagram., , (a) a, (20), , (21), , (b) b, , (c) c, , (b) d, , If nv, nr, nb are the refractive indices of violet, red, and blue light respectively, in a given medium, then which of the following is true?, (a) nv = nb = nr , , (b) nv > nb > nr, , (c) nv > nb < nr , , (b) nv > nr > nb, , vV = vR= vG are the velocities of violet, red and, green lights respectively, passing through a prism, after dispersion of white light. Which among the, following is a correct relation?

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Master Key Science and Technology Part-1 (Std. X), , 120, , (a) vV = vR= vG , , (b) vV < vG= vR, , (c) vV > vR > vG , , (b) vV < vR= vG, , Ans. , (1) (d) changing refractive index of atmospheric gases, 2, (2) (d), (3) (b) refraction of light (4) (a) bends, 3, towards normal (5) (b) parallel (6) (b) bends away, from normal (7) (b) 3 × 108 (8) (a) ∠ i > ∠ r but ∠ i = ∠ e, (9) (c) passes without bending (10) (a) 0° (11) (c) the, velocity of light changes (12) (b) 40° (13) (b) moves, away from the slab (14) (b) 2.4 × 108 m/s (15) (a) Red, sin 600, (16) (b), (17) (b) ∠i and ∠e (18) (b) No, sin 45 0, (19) (a) (20) (b) nv > nb > nr (21) (b) vV < vG= vR, Q.2.1 Solve the following:, Formula:, Refractive index of second medium with respect, to first medium,, v1, , n = v, 2 1, 2, v1 = velocity of light in first medium, v2 = velocity of light in second medium, sin i, sin r, = angle of incidence, , Also 2n1 =, i, , r = angle of refraction, *(1) The absolute refractive index of water is 1.36., What is the velocity of light in water? (velocity, of light in vacuum is 3 × 108 m/s), Ans. Given: wnvac = 1.36, , vvac = 3 × 108 m/s, To find: vw = ?, vvac, Formula: wnvac =, vw, 3 × 108, Solution: 1.36 =, vw, 3 × 108, 1.36, , vw =, , , , vw = 2.21 × 108 m/s, , The velocity of light in water is 2.21 × 108 m/s, (2) Light travels with a velocity 1.5 × 108 m/s in a, medium. On entering 2nd medium its velocity, becomes 0.75 × 108 m/s. What is the refractive, index of the 2nd medium with respect to the first, medium., Ans. Given: v1 = 1.5 × 108 m/s, v2 = 0.75 × 108 m/s, , To find: 2n1 = ?, , Refractive index of 2nd medium with respect, to 1st is 2., *(3) If the speed of light in a medium is 1.5 x 108, m/s, what is the absolute refractive index of the, medium?, Ans. Given: vm = 1.5 × 108 m/s, , , vvac = 3 × 108 m/s, , To find: mnvac = ?, , vv, Formula: mnvac = v, m, 8, Solution: mnvac = 3 × 10 8, 1.5 × 10, mnvac = 2, , Absolute refractive index of the medium is 2., *(4) If the absolute refractive indices of glass and, water are 3/2 and 4/3 respectively, what is the, refractive index of glass with respect to water?, 3 vv, Ans. Given: gnv = =, 2 vg, 4 vv, , n = =, w v, 3 vw, To find:, , n =?, , g w, , Formulae: gnw =, Solution : gnw =, , , , , , v1, Formula: 2n1 = v, 2, 8, Solution: 2n1 = 1.5 × 10 8, 0.75 × 10, , n =2, 2 1, , , , =, , , , =, , vg, vw, vg, vw, vg, vv, , ×, , vv, vv, vw, , ×, vg vv, 3 3, =, ×, 2 4, , , , , vw, , n = 9, 8, Refractive index of glass with respect to water, is 9, 8, g w, , *(5) If the refractive index of second medium with, respect to first medium is 2n1 and that of third, medium with respect to second medium is, n What and how much is 3n1? (Can you tell?;, 3 2., Textbook page no. 75), Ans. Refractive index of 2nd medium w.r.t. 1st.

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Refraction of Light, , n =, , 121, , v1, , ........(i), v2, Refractive index of 3rd medium w.r.t. 2nd., v2, n =, ........(ii), 3 2, v3, Similarly,, 2 1, , n1 = refractive index of 3rd medium w.r.t. 1st., v1, v2, n =, ×, 3 1, v3, v2, v2 v1, =, ×, v3 v2, , From (i) and (ii), 3, , n = 3n 2 × 2n 1, , 3 1, , *(6) If the angle of incidence and angle of emergence, of a light ray falling on a glass slab are i and e, respectively, then prove that, i = e., Ans., P, , Incident ray, , A, , i Normal, , r, Refracted, ray, , air B, glass, , r, C, , e, , Emergent ray, , Fig. 6.1 : Refraction of light passing through a glass slab, , Solve and fill in the blanks, , Sr. Velocity, No. of light in, the first, medium, v1, , Velocity, of light in, the second, medium, v2, , (1) 3 × 108 m/s, , 1.2 × 108 m/s, , (2), , 2.25 × 108 m/s, , Refractive Refractive, Index, Index, n1, , 2, , 1, , n2, , 4, 3, , (3) 2 × 108 m/s, , 1.5, 2, (0.67)], 3, , (1) Refraction of light, Ans. , The phenomenon of change in the direction of the, light ray passing from one transparent medium, to another obliquely is called refraction., (2) Dispersion, , 1, n = n, 2 1, 1 2, 1st medium be air, 2nd medium be glass, 1, n = n ........(i), g a, a g, At surface AB, sin i, n =, ........(ii), g a, sin r, At surface CD, sin r, ........(iii), sin e, , , , Substituting (ii) and (iii) in (i), sin i sin e, =, sin r sin r, , sin i = sin e, i=e, , (4), , Q.2.2. Define/write the laws:, , S, , n =, , (1) The speed of light in a transparent medium is, 2.5 × 108 m/s. Find the absolute refractive index of, that medium., (1.2), 3, (2) The refractive index of benzene is, and that of, 2, 4, water is, . Find the refractive index of water, 3,  8, with respect to benzene.,  9 , 4, (3) The refractive index of water is, and speed of, 3, 8, light in air is 3 × 10 m/s. Find the speed of light in, water., (2.25 × 108 m/s), , [(1) 2.5, 0.4; (2) 3 × 108 m/s, 0.75; (3) 3 × 108 m/s,, , D, , a g, , NUMERICALS FOR PRACTICE, , Ans. , The splitting of white light into its component, colours is called dispersion., (3) Spectrum, Ans. , The band of coloured components of a light beam, is called Spectrum., (4), , Angle of incidence, , Ans. , The angle made by an incident ray with the, normal is called angle of incidence. It is denoted, by ‘i’., (5), , Angle of refraction, , Ans. , The angle made by refracted ray with the normal, is called as angle of refraction. It is denoted by ‘r’., (6), , Laws of refraction, , Ans., (i) The incident ray and the refracted ray are on, the opposite sides of the normal at the point of, incidence and all three lie in the same plane.

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Master Key Science and Technology Part-1 (Std. X), , 122, , (ii) For a given pair of media, the ratio of the sine of, the angle of incidence to the sine of the angle of, refraction is constant., If ‘i’ is the angle of incidence and ‘r’ is the angle of, sin i, refraction, then, = constant, sin r, , This constant is called refractive index of a, medium., *(5) Laws of reflection (Can you recall?; Textbook page, no.73), Ans. , , (ii) The refractive index of the second medium with, respect to the first, Ray, medium is given by, Medium 1, v1, the ratio of the, Air, magnitude, of, Medium 2, velocity of light in, Glass, v, the first medium to, 2, that in the second, medium., It, is, Fig. 6.3: Light ray going, represented as, from medium 1 to medium 2, , (i) The incident ray, reflected ray and the normal are, all in the same plane., , , n = velocity of light in medium 1, 2 1, velocity of light in medium 2, v1, ∴ 2n1 =, v2, , (ii) The incident ray and reflected ray are on opposite, sides of the normal., (iii) The angle of incidence and angle of reflection are, of equal measure., , (iii) The refractive index depends upon the relative, speed of propagation of light in different media., , Q.2.3. Write the short notes on:, , (iv), , Refractive index has no unit., , (1), Dispersion of light, Ans., , (3), , Partial and Total internal reflection, , ight, , Sun l, , Glass Prism, Fig. 6.2 : Dispersion of light, , Ans. (For diagram; refer Fig. 6.8, Page No. 114), , R, O, Y, G, B, I, V, , (i) , The process of separation of light into its, component colours while passing through a, medium is called dispersion of light., (ii) Refractive index of a medium is different for, different colours., (iii) When white light enters a single medium like, glass, the angles of refraction are different for, different colours., (iv) When white light from the sun through air, enters, any refracting medium, it emerges as a spectrum, of seven colours., (v) Sir Issac Newton was the first person to use a, glass prism to obtain sun’s spectrum., (2), , Refractive index, , Ans., (i) The ratio of sine of angle of incidence to the sine, of angle of refraction is called refractive index., sin i, ∴ 1n 2 =, sin r, , (i) , When light ray travels from denser medium, to rarer, reflection takes place partially at the, interface, this is called partial reflection., (ii) Angle of incidence (i) is smaller than the angle, of refraction, (r). If ‘i’ increases gradually, then ‘r’, also increases as refractive index is a constant., (iii) The angle of incidence for which the angle of, refraction is 90°, is called critical angle., (iv) When the angle of incidence in a denser medium, is more than critical angle, the light reflects back, into the denser medium. This is called total, internal reflection., (4), , Mirage or Have you seen a mirage which is an, illusion of water on a hot road or desert. Explain., , Ans., (i) A mirage is observed on a hot day in the desert or, road., (ii) The air near the hot road or desert surface is hot, and hence rarer than the air above., (iii) , As height increases, the refractive index also, increases., (iv) The direction of light rays keeps changing due, to refraction and appears to be coming from the, image of the object on the ground., (v), , This is called a mirage.

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Refraction of Light, , (5), , Ans., , 123, , Rainbow formation or Prove that a rainbow is, the combined effect of the refraction, dispersion, and total internal reflection of light., Light ray, , (ii) They, therefore do not appear as point sources, but appear as a collection of point sources., (iii) Changes in atmospheric refractive index changes, the position as well as the brightness of individual, point source, but the average position and total, average brightness remains unchanged., (iv), , Water droplet, , (3) We can see the sun even when it is little below, the horizon., , red, , blue, , Ans., Internal reflection, , Fig. 6.4: Rainbow formation, (i) The beautiful phenomenon of the rainbow is a, combination of different phenomena namely, dispersion of light, internal reflection and, refraction of light., (ii) The rainbow appears in the sky after a rain, shower., (iii), , As a result of this planets do not appear to, twinkle., , The water droplets act as small prisms., , (iv) When sunlight enters the water droplets present, in the atmosphere, they refract and disperse the, incident sunlight. That results in a spectrum., (v) Then the light rays get reflected internally inside, the droplet and finally again get refracted., , (i) By sunrise we mean the appearance of the sun, above the horizon., (ii) When the Sun is somewhat below the horizon, its, light rays are able to reach us along a curved path, due to the refraction through Earth’s atmosphere., (iii) Thus we see the Sun even before it emerges above, the horizon., (iv) , Samething happens at the time of sunset, we, keep seeing the Sun for a short while even after it, goes below horizon., (4), , Red colour is used in danger signal., , Ans., (i) In a visible light range, the wavelength of red, light is maximum, i.e. 700 nm., , (vi) As a collective effect of all these phenomena, the, seven coloured rainbow is observed., , (ii) The maximum scattering of light is for blue light, and the least scattering of red light takes place., , (vii) In the morning, rainbow can be seen towards, west, while in the evening, it can be seen towards, the East., , (iv), , Q.2.4. Give scientific reasons, (1), , Stars twinkle at night., , Ans., (i) Stars are self luminous and can be seen at night in, the absence of sunlight., (ii) They appear to be point sources as they are at a, very large distance from us., (iii) The apparent position of the star keeps changing, a bit because of the motion of atmospheric air, and changing air density and temperature., (iv) Because of this, refractive index of air keeps, changing continuously., , (iii) Thus, red light can travel to longer distance and, can be visible from far off distance., Therefore, the danger signals are red coloured., , (5) It is possible to enjoy a rainbow at fountain in, any season., Ans., (i) The phenomenon of rainbow is a combined effect, of dispersion, internal reflection and refraction., (ii) The rainbow appears in the sky after a rainfall., (iii) The water droplets of a fountain also act as small, prisms as the rain drops., (iv) , When the light rays enter these droplets, it first, gets dispersed, then internally reflected and then, refracted while coming out of the droplets. All these, three processes together produce the rainbow., , (v) So the position and brightness of the stars keeps, changing and so the stars appear to be twinkling., , *(6) Will light travel through a glass slab with the, same velocity as it travels in air? (Use your brain;, Textbook page no. 74), , (2), , Planets do not twinkle., , Ans., , Planets are much closer to us as compared to, stars., , (ii) As refractive index increases, velocity of light, decreases., , Ans., (i), , (i), , Refractive index depends on the velocity of light.

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Master Key Science and Technology Part-1 (Std. X), , 124, , (iii) Since refractive index of glass is greater than air,, the velocity of light in glass is lesser than that in, air., , , Light travels along a straight line path in a, transparent medium. But, it does not travel in the, same direction in all media., , *(7) Will the velocity of light be same in all media?, (Use your brain; Textbook page no. 74), , The direction of light rays change when it enters, from one transparent to another as the velocity of, light is different in different media., , Ans., (i), , Refractive index depends on the velocity of light., , (ii) Refractive index is different for different media., (iii) Velocity of light will also be different for different, media., Q.2.5. Label the diagram., (1), (ii), (i), , (3) Light rays when travelling from air to water, bends, ., Ans. Light rays when travelling from air to water bend, towards normal., , When light rays travel from rarer medium to, denser medium obliquely, it bend towards, normal as the velocity of light decreases after, entering into denser medium, i.e. water., (4) , Light rays when travelling obliquely from, Benzene to Air bend, ., Ans. , Light rays when travelling obliquely from, Benzene to Air, bend away from the normal., , (iii), , Generally, when light rays travel obliquely from, denser medium to rarer medium, it bends away, from the normal as the velocity increases after, entering air., , (iv), Ans. (i) Atmospheric layers (ii) Apparent position, (iii) Horizon , , (iv) Real position, , (2), , Benzene is a denser medium, while air is the rarer, medium. Velocity of light in benzene is less than, that in air., (5) The splitting of Sunlight into its component, colours is called, ., , (iii), , Ans. The splitting of Sunlight into its component, colours is called dispersion., , (ii), , The band of coloured components of a light beam is, called its spectrum. The various colours are Violet,, Indigo, Blue, Green, Yellow, Orange and Red., Q.3.2 , Draw neat and labelled diagram for the, following:, , (i), Ans. (i) Internal Reflection (ii), , (1) Refraction of light passing through a glass slab., , Water droplet, , (iii) Incident Sunlight, Q.3.1. Complete the following statements and explain., *(1) , The, , refractive index, of light., , depends, , on, , the, , Ans. The refractive index depends upon the relative, speed of propagation of light in different media., This is because light has different velocities in, different media., *(2) The change in, of light rays while, going from one medium to another is called, refraction., Ans. The change in the direction of light rays while, going from one medium to another obliquely is, called refraction., , A, P, , Incident ray, B, , i, , N, , Air, , Q, , Glass, r, Refraction i1, of light, S, R, M, C, e, D Air, , Refracted ray, , Emergent ray, , Fig. 6.5: Refraction of light passing through a glass slab, (2) Dispersion of light., Ans. (For diagram, refer Fig. 6.2)

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Refraction of Light, , 125, , (3) Partial and total internal reflection., Ans., Refracted rays, Air, , Water, , r, , i, , r1, , i1, , Light, source, , r = 90°, ic, , i > ic, , Total internal, reflection, , Partial reflection, Medium 2, Fig. 6.6: Partial and total internal reflection, , (4) Rainbow formation., , (ii), , Ans. (For diagram, refer Fig. 6.4), , Ans. , Violet, Indigo, Blue, Green, Yellow, Orange and, Red., , Q.4.1. Read the following paragraph and answer the, following questions., (1), , In, vacuum the velocity of light radiation, with different wavelength is the same. But in, a material medium, the velocity of different, colour radiations is not the same and they, travel with different paths. Due to this, the, refractive index of the medium is different, for different colours. If white beam of light is, incident on a glass prism, the angle of refraction, for different colours is different. When, Sunlight is incident on a refracting surface,, then it gets split ịnto seven coloured spectrum, and emerges out. The splitting of light into, its component colours is called dispersion., , (iii), , (i), , How do emergent rays appear?, , Ans. The emergent rays appear as a band of colours, called spectrum., , What is dispersion?, , Ans. , The splitting of Sunlight or white light into its, component colours is called dispersion of light., (iv), , What is spectrum?, , Ans. , The band of coloured components of a light beam, is called as Spectrum., (v) Which light deviates the least and which one, deviates the most?, Ans. , When light is incident on a surface of a glass, prism, the red light deviates the least, while, violet light deviates the most., (2), , Sir Isaac Newton was the first to use a glass, prism to obtain the spectrum of Sunlight. A, glass prism is a transparent medium bounded, by two plane surfaces inclined at an angle., When white light is dispersed into seven, colours by a prism, different colours of light, bend through different angles with respect to, incident ray. Out of these seven colours, red, light bends the least, while violet light bends the, most. So, the rays of each colour emerge along, different paths and becomes distinct. Hence,, we get spectrum of seven - different colours., , Write the colours of the Spectrum?, , he beautiful phenomenon of the rainbow, T, is a combination of different phenomena dispersion, internal reflection and refraction, of light. The rainbow appears in the sky after a, rainfall. The water droplets act as small prisms., When sunlight enters the water droplets present, in the atmosphere, they refract and disperse the, incident Sunlight. Then, they reflect internally, inside the droplet and finally again refract it., As a collective effect of all the phenomena, the, seven coloured rainbow is observed., , (i) Which three major phenomena are responsible, for rainbow?, Ans. , The three major phenomena responsible for, rainbow are dispersion, internal reflection and, refraction of sunlight., (ii), , How do refraction and dispersion take place?, , Ans. , When light rays of Sun enter the water droplets of, atmosphere, the refraction of light occurs as the

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Master Key Science and Technology Part-1 (Std. X), , 126, , medium changes from air to water. The direction, of light rays also change and at the sametime, due, to decrease in the velocity of light, the splitting of, light rays result in dispersion., (iii) Give a diagrammatic representation of Rainbow, Production., Ans. (For diagram, refer Fig. 6.4), Q.4.2. Activity Activity / Application based Question:, (1), , Activity:, (For diagram, refer Fig. 6.5), , (i) Keep a glass slab on a blank paper and draw its, outline PQRS as shown in the figure., , (iv) , It is found then, that a beam of white light, emerges from the other side of the second prism., *(3) You must have seen chandeliers having glass, prism. The light from a tungsten bulb gets, dispersed while passing through these prisms, and we see coloured spectrum. If we use on LED, light instead of tungsten bulb, will we be able, to see the same effect? (Use your brain; Textbook, page no. 77), Ans. , Yes. We can see the same effect for LED light also., But the spectrum obtained is not continuous as in, the case of sunlight., (4), , (ii) , Draw an inclined straight line on the side of, PQ so that it intersects PQ at N. Pierce two pins, vertically at two points A and B along the line., (iii) Look at the pins A and B from the opposite side of, the slab and pierce pins C and D vertically so that, the images of A and B are in line with C and D., (iv) Now remove the chip and the pins and draw a, straight line going through points C and D so, that it intersects SR at M., , Objects beyond and above a holy fire appear to, be shaking. Give scientific reasons., Ans. , Yes., (i), , The air near the holi flame is hotter than the air, above, hence it is a rarer medium than the air, above., , (ii), , Because of this change of medium the refractive, index keeps on changing as we go higher and, higher above the flame and so the refraction of, light also changes., , (iii), , This gives a shaking effect to objects beyond and, above holy fire., , (5), , (a) Complete the diagram., , (b), , What is the effect on the velocity, , (v) Join points M and N. Observe the incident ray, AN and emergent ray MD., Questions:, (a) What is your observation about the emergent, ray and incident ray?, Ans. , Incident ray and emergent ray lie parallel to each, other., *(2) From incident white light how will you obtain, white emergent light by making use of two, prisms? (Use your brain; Textbook page no. 77), , Have you seen that objects beyond and above, a holi fire appear to be shaking? Why does this, happen? (Can you tell?; Textbook page no. 76). or, , of light when it enters water?, (c), , Ans., , If the light was incident on the, interface at an angle greater than, critical angle; which phenomen, on occurs?, , Ans. (a), , N, glass, Water, N1, , N, glass, Water, N1, , (b), , When light enters water, its velocity increases., , Fig. 6.7: Dispersion of light, , (c), , (i) Take a glass prism, Allow white light to fall on it., Obtain a spectrum., , At an angle greater than critical angle, Total, internal Reflection occurs., , (6), , (a) Complete the following diagram:, , (ii) , Take a second identical glass prism. Place it, parallel to the first prism in an upside down, position with the first prism., (iii) Allow the colours of the spectrum to pass through, the second prism., , Wh, , ite, , ligh, , t

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Refraction of Light, , (b), , 127, , Name the colour that deviates the most and, least, , (c), , Among orange and violet, which colour has, highest velocity., Ans. (a), Wh, , ite, , (b), (c), , (a), , Which phenomenon does the diagram show?, , (b), , Which law obeyed by light makes this, phenomenon occur?, , (c), , Name one more phenomenon in nature where, the above law is observed., , Ans. (a) Mirage, ligh, , t, , Violet deviates the most and Red deviates the, least., The velocity of orange is more than the velocity, of violet., , (b), , Refraction of light or Total internal Reflection., , (c), , Formation of Rainbow., , (9), , Answer the following questions from the, diagram:, , Incident, ray, , A, , B, , P, , (7), , (a) Answer the following questions based on, the concept of total internal reflection., (a), What is the angle of incidence called, when angle, of refraction is 900?, (b) State any one phenomenon in nature which is, based on total internal reflection., (c), State any one condition required for total internal, reflection., Ans. (a) The angle of incidence is called as Critical, angle., (b), , Mirage or Rainbow formation., , (c), , Light should travel from denser medium to rarer, medium., OR, Angle of incidence should be greater than critical, angle., , (8), , i, , N, , Medium Air, Normal, Q, , R, , r, Refracted ray, N1, S, , M, e, , R, C, , D, Emergent ray, , (a), , How many times does the light undergo, refraction?, , (b), , Which two angles are equal?, , Ans. (a) The light undergoes refraction 2 times., (b), , vvv, , angle of incidence (i) is equal to angle of, emergence (e).

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Master Key Science and Technology Part-1 (Std. X), , 128, , ASSIGNMENT - 6, Time : 1 Hr., , (1), , (2), , (3), , Marks : 20, , Q.1. (A) Attempt the following questions:, , (3), , Fill in the blanks and rewrite statement., (i) The wavelength of visible light range for human is, Find the odd one out:, Brown, Blue, Green, Red, Complete the analogy., , , , Diamond : Denser medium : : Air :, , , , Q.1. (B) Choose and write the correct option:, , (1), , (2), , , , ., , The velocity of light in air is, (a) 3 × 10, , 10, , (2), , m/s., , (b) 3 × 10, , (c) 1.5 × 108, , 8, , (d) 0.3 × 108, , When light ray travels from denser medium to rarer, then it, , ., , (a) bends towards the normal, , (b) bends away from the normal, , (c), , (d) deviates randomly, , goes undeviated, , Q.2. Answer the following: (Any 2), , (4), , (1), , Write short note on: Laws of refraction., , (2), , Give scientific reasons: "Red colour is used in danger signal.", , (3), , Solve: If the speed of light in a medium is 1.5 × 108 m/s, what is the absolute refractive index of the, medium?, , , , Q.3. Answer the following: (Any 2), , (6), , (1), , Draw neat and labelled diagram for partial and total internal refraction., , (2), , Complete the following statement and explain., , , (3), , , (i) The change in, , of light while going from one medium to another is called refraction., , If the angle of incidence and angle of emergence of a light ray falling on a glass slab are i and e, respectively, prove that, i = e., Q.4. Answer in detail. (Any 1), , (5), , (1), , With a neat labelled diagram explain Rainbow Formation., , (2), , (A) Define refraction , , N, , , , (B) (i), , Complete the diagram., , , , (ii), , , What is the effect on the velocity of light , when it enters water?, 1, , glass, Water, , N, , , (iii) If the light was incident on the interface at an angle greater than critical angle; which phenomen, , on occurs?, , vvv

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7, , Lenses, CONCEPT MAP, , (1), Lenses, , Convex lens, (Converging), , Concave lens, (Diverging), Position of, object, At Infinity, Image, At F1, , Uses, , Simple microscope, Compound microscope, , Position of, object, Between 2F1 & F1, Image, , Reflecting telescope, Human Eye, , Between F1 & O, , Eye Defect, Myopia, Unable to see, distant objects, , Hypermetropia, , Presbyopia, , Unable to see, nearby objects, , In old-age,, unable to see, nearby objects, , Corrected by, Concave lens, , At Infinity, Image, , At focus (F1), Image, , At Focus (F2 ), , At Infinity, , Beyond 2F1, Between F1 & 2F1, Image, Image, Between F2 & 2F2, , Refracting telescope, , Telescope Types, , Between 2F2, , (129), , Corrected by, Convex lens, , At 2F1, Image, At 2F2, , Corrected by, Convex lens, , Between F1 & O, Image, On the same side, as that of object

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Master Key Science and Technology Part - 1 (Std. X), , 130, , Points to Remember:, zz A lens is a transparent medium bounded by two surfaces; out of which atleast one surface is spherical., zz A lens having both surfaces spherical,bulging outward, is a convex lens or double convex lens. It is thicker at, , the centre as compared to its edges. This lens can converge the light incident on it. So, it is a converging lens., , zz A lens having both surfaces spherical,bulging inward, is a concave lens or double concave lens. It is thinner, , at the centre as compared to its edges. This lens can diverge the light incident on it. So, it is a diverging lens., , zz Double convex, plano – convex, concavo convex, positive meniscus, negative meniscus are the types of, , convex lenses., , zz Center of curvature (C): The centres of sphere whose parts form surfaces of the lenses are called centre of, , curvature of lenses. Each lens has two centre of curvature C1 and C2., , zz Radius of curvature (R): The radii (R1 and R2) of sphere whose parts form surfaces of the lenses are called the, , radii of curvature of the lens., , zz Principal axis: The imaginary straight line passing through the two centres of curvatures of lens (C) is called, , principal axis of the lens., , zz Optical centre (O) :- The central point of a lens on the principal axis through which light rays pass without, , changing their path is called optical centre., , zz Principal Focus (F): The point on the principal axis where all the incident rays are parallel to the principal axis, , actually meet or appear to meet after refraction from spherical lens is called Principal focus of a spherical, lens. Every lens has two principal foci., , zz Focal length (f) : The distance between the optical centre and the principal focus is called focal length of that lens., zz Images formed by spherical lenses can be studied with the help of ray diagrams. Ray diagrams are useful, , to study the position, relative size of the object and the image. It also helps to know the nature of the image, formed by that lens., , zz Following are the rules for obtaining images by convex lens :, , i) Rule 1: If the incident ray is parallel to the principal axis, then the refracted ray passes through the, principal focus (F)., ii) Rule 2: I f the incident ray is passing through the principal focus, then the refracted ray is parallel to, the principal axis., iii) Rule 3 : I f the incident ray of light passes through the optical centre of lens, it passes without changing, its direction., , zz Position, relative size and nature of image formed by a convex lens for various positions of the object., , Sr. No., 1, , Position of the object, At Infinity, , Position of the image, At Focus F2, , 2, 3, 4, 5, 6, , Beyond 2F1, At 2F1, Between F1 and 2F1, At Focus F1, Between O and F1, , Between F2 and 2F2, At 2F2, Beyond 2F2, At Infinity, On the same side of the, lens as the object, , Size of the image, Highly diminished,, Point - Sized, Smaller, Same size, Larger, Very Large, Very Large, , Nature of the image, Real and inverted, Real and inverted, Real and inverted, Real and inverted, Real and inverted, Virtual and erect, , zz Following are the rules for obtaining images by Concave lens., , Rule 1 : When the incident ray is parallel to the principal axis, the refracted ray when extended backwards,, passes through the principal focus., Rule 2: W,  hen the incident ray passes through the focus, the refracted ray is parallel to the principal axis., Rule 3: When the incident ray passes through optical centre, it passes undeviated.

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Lenses, , 131, , zz Position, relative size and nature of image formed by a Concave lens for various positions of the object., , Sr. No, , Position of the object, , Position of the image, , Size of the image, , Nature of the image, , 1, , At Infinity, , On the 1st Focus F1, , Point Image, , Virtual and erect, , 2, , Anywhere between optical, centre O and Infinity, , Between O and first focus F1, , Small, , Virtual and erect, , zz Lens formula : The formula showing relation between distance of the object (u), the distance of the image (v), , and the focal length (f) is called lens formula. 1 – 1 = 1, v u f, zz Linear magnification : Magnification is the ratio of height of the image (h2) to the height of the object (h1), Height of Image, h2, Magnification =, Height of object ∴ M =, h1, zz Magnification is also related to object distance (u) and Image distance (v)., , ∴, , Magnification =, , Distance of Image, v, Distance of object ∴ M =, u, , zz Power of lens (P) : The capacity of a lens to converge or diverge incident rays is called its power (P). The power, , of a lens depends upon its focal length (f)., , zz If focal length is expressed in metre, the power of lens is expressed in dioptre. ∴ P =, zz One dioptre is the power of a lens whose focal length is 1 metre., , 1, f(m), , 1, 1 metre, zz Combination of lenses : If two lenses of focal lengths f1 and f2 are placed together then their combined focal, length can be given by,, 1 1 1, = +, f f1 f2, 1 dioptre =, , zz Combined power or power of combination of two lenses is given by, P = P1 +P2, zz Human eye and function of its lens :-, , (a) The outer most thin, transparent membrane of the human eye is called cornea., (b) Light enters the eye through it and maximum amount of incident light is refracted inside the eye at the outer, surface of the cornea., (c) There is a dark, fleshy screen behind the cornea called the Iris., (d) There is a small hole of changing diameter at the centre of iris called the pupil. It controls the amount of light, entering the eye., (e) If the light falling on eye is too bright it contracts and if the light falling on eye is too dim it widens., (f) Behind the pupil there is a double convex transparent crystalline lens, which provides small adjustment of the, focal length to focus the image. This lens also creates real and inverted image of an object on the screen inside, the eye., (g) The screen consist of light sensitive cells called retina., (h) These cells gets excited when light falls on them and they generate electric signals., (i) Signals are conveyed to the brain through optic nerve., (j) The brain thus analyses these signals and converts them in such a way that we perceive the objects as they, actually are., zz The power of accommodation: The capacity of the eye lens to change its focal length as per need is known as, , power of accommodation.

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Master Key Science and Technology Part - 1 (Std. X), , 132, , zz Myopia (near sightedness): The type of eye defect in which person can see nearby objects clearly but cannot, , see distant objects clearly is called Myopia or nearsightedness. Myopia can be corrected by using a spectacle, of concave lens of suitable focal length. A concave lens with negative power is required for correcting myopia., , zz Hypermetropia (longsightedness): The type of eye defect in which a person can see distant objects clearly, , but cannot see nearby objects clearly is called Hypermetropia or farsightedness. A convex lens with positive, power is required for correcting Hypermetropia., , zz Presbyopia (old age hypermetropia): The power of accommodation of eye usually decreases with increasing, , age. The nearpoint of aged people recedes and they find it difficult to see nearby objects comfortably and, clearly without spectacle., , zz Uses of concave lens: Medical apparatus scanner, CD player, spectacles, etc. use concave lens., zz Uses of convex lenses: Optical instruments like Simple microscope, Compound microscope, Astronomical, , telescope, etc. make use of convex lens., , 1 th, of a second even after the image, 16, is vanished from the eye. This effect is called persistence of vision. Motion pictures and TV work on this, principle., , zz Persistence of vision: The sensation of the image lasts on the retina for, , zz Colour blindness: Some people cannot distinguish certain colours as they lack conical cells responding to, , certain colours. They can not even identify certain colours. These people are called ‘colour blind’ even though, their eye sight is normal., , MASTER KEY QUESTION SET - 7, Q.1. (A) 1. Fill in the blanks:, , (13), ,  he focal length of a healthy eye in relaxed, T, position is, ., , (14), , The, cells are responsible for the, perception of colours., , (15), ,  he power of convex lens is 0.5D. Hence, its focal, T, length is, ., , (16), ,  he image of the object at, T, principal focus., , (1), , The focal length of, , lens is positive., , (2), , In hypermetropia, the image is formed, the retina., , (3), , The power of myopic eye is, , (4), , Myopia can be corrected by using spectacle of, lens., , (17), , (5), ,  he, T, is the distance between optical, centre and principal focus., ,  he image formed by, T, virtual., , (18), , (6), ,  he ability of lens to converge or diverge light, T, rays depends on, of the lens., ,  he tendency of pupil to adjust the opening for, T, light is called, ., , (7), ,  he power of convex lens with focal length 40 cm, T, is, ., , (8), , he lens with, T, correct farsightedness., , (9), , The distance of distinct vision for human eye is, cm., , ., , power is used to, , (10), , The impression of an image lasts on retina for, of second., , (11), , The light enter the eye through, , (12), , An optical device used by watch repairers is, ., , ., , is on the, is always, , (19) The diameter of human eyeball is, ., (1) convex (2) behind (3) negative (4) concave, Ans. , (5) focal length (6) power (7) 2.5 D (8) positive (9) 25, (10) 1 th (11) cornea (12) simple microscope (13) 2 cm, 16, (14) conical (15) 2 m (16) infinity (17) concave lens, (18) adaptation (19) 2.4 cm, Q.1. (A) 2. Find the odd word out:, (1), , Cornea, iris, pupil, cerebrum., , Ans. Cerebrum; it is a part of brain while others are the, parts of human eye.

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Lenses, , (2), , Spectacle, microscope, torch, camera., , Ans. Torch; Concave lens is used in it while in others, mainly convex lens is used., (3) Longsightedness, myopia, short sightedness,, near sightedness., , 133, , (7), , Q.1. (A) 4. Match the columns:, (1), , (2) Telescope, , (2), , (2), , Column B, (a) 2.4 cm, 1 th, 16 of a second, (c) 2 cm, (b), , (d) 25 cm, , Ans. (1 – b), (2 – d), (3 – a), (4 – c), (3), , Image at infinity : In convex lens when the object, Ans. , is at the focus, a very large, real and inverted, image will be formed at infinity., (3), , Column A, (1) Persistence of, vision, (2) Distance of, distinct vision, (3) Diameter of eye, ball, (4) Focal length of, healthy eye lens, , Convex lens: Converging :: Concave lens:, , ., , Object at 2F1 : Image at 2F2 :: Object at F1 :, , ., , (a) To observe, celestial objects., (b) Watch repairers., , Ans. (1 – d), (2 – a), (3 – b), , Q.1. (A) 3. Complete the analogy:, , Diverging : Light rays parallel to the principal, Ans. , axis appear to be diverging from principal focus, in concave lens., , Column B, , (3) Simple microscope (c) Correction of eye, defect, (d) To observe tissue, , Ans. Flood light; It contains mirrors/reflectors while, others contain lens., (1), , Column A, (1) Microscope, , (4) Presbyopia, night blindness, myopia,, hypermetropia., , (5) Compound microscope, astronomical telescope,, simple microscope, flood light., , Mirror, , Uses, , (1) Plane mirror, , (a) Rear view mirror, , (2) Concave mirror, , (b) At laughing, gallery, (c) At a hair dresser, , (3) Convex mirror, , Myopia : Concave lens :: Farsightedness :, ., , (4) Irregular curved, mirror, , (d) At a dentist, , Convex lens : Convex or converging lens is used, Ans. , to correct the Farsightedness., , Ans. (1 – c), (2 – d), (3 – a), (4 – b), , (4), , (4), ,  oncave lens : Negative focal length :: convex, C, lens :, ., , ., , Ans. Infinity, , Ans. Longsightedness; it is corrected by convex lens, while others are corrected by concave lens., , Ans. Night blindness; it is a deficiency disease while, others are eye defects., , Near point : 25cm :: Far point:, , Column A, , Column B, , (1) Microscope, , (a) Prism, , Ans. Positive focal length : Focal length of a convex, lens is positive according to cartesian sign, convention., , (2) Telescope, , (5), , (3) Presbyopia, , (b) Spectacles, having convex, lenses, (c) To observe, distant objects, , Convex lens : Positive power :: Concave lens:, , ., , Ans. Negative power : The power of concave lens is, negative as the power depends on focal length., (6), , Rod cells : Sensitive to dim light :: Cone cells:, ., , Ans. Respond to colours : The conical cells respond to, colour and give information about the colour of, the object to the brain., , (4) Dispersion of light (d) Observation of, plant and animal, cells, (e) Weakness of, ciliary muscles, Ans. (1 – d), (2 – c), (3 – e), (4 – a)

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Master Key Science and Technology Part - 1 (Std. X), , 134, , (5), , Column A, , (2) True (3) True. (4) False. Pigment in iris gives, colour to the eye. (5) False. The light is first and, maximum refracted as it passes through cornea., (6)True (7) False. Optic nerve takes the impulse, from eye to brain. (8) False. Impression an the image, lasts on the retina for 1 th of a second. (9) True, 16, (10) False. In compound microscope, two convex, lenses are used., , Column B, , (1) Microscope, , (a) Torches and, headlights, (2) Telescope, (b) Observation of, plant and animal, cells, (3) Simple microscope (c) To observe, distant objects, (4) Concave mirror, (d) Watch repairers, (e) Correction of eye, defect, , Ans. (1 – b), (2 – c), (3 – d), (4 – a), *(6), , Column A, , Column B, , Column, C, , (i) Bifocal, (a) Near by, lens, object can, be seen, clearly, (2) Presbyopia, (b) Faraway (ii) Concave, lens, object can, be seen, clearly, (3) Nearsightedness (c) Problem of (iii)Convex, old age, lens, (1) Farsightedness, , Q.1. (A) 6. Answer the following in one sentence:, (1) If the object is at infinity from a convex lens, what, would be its size?, If the object is at infinity, then the image will be, Ans. , highly diminished point image., (2), , Write the lens formula., 1, 1, 1, Ans. , The lens formula is, –, =, v, u, f, where u - object distance, v - image distance and, f is focal length., (3), , Name the types of telescopes., , Ans. , Refracting telescope and reflecting telescope are, the two types of telescopes., (4), , How is hypermetropia corrected?, , Ans. (1 – b – iii), (2 – c – i), (3 – a – ii), , Ans. , Hypermetropia is corrected by using spectacles, of convex lens of suitable focal length., , Q.1. (A) 5. State whether the following statements, are ‘True’ of ‘False’. If false; write the correct, statement:, , (5), How is myopia corrected?, Ans. , Myopia is corrected by using spectacles of, concave lens of suitable focal length., , (1), , A lens is an opaque object., , (2), ,  yopia can be corrected by using a spectacle of, M, concave lens., , (3), ,  ypermetropia can be corrected by using a, H, spectacle of convex lens., , (6) By which lens the matchstick can catch fire in the, sunlight?, By using convex lens, a matchstick can catch fire, Ans. , in the sunlight., , (4), , Cornea gives colour to the eye., , (5), ,  he light is first and maximum refracted as it, T, passes through eye lens., , (6), ,  he perception of colours is concerned with cone, T, cells., , (7), , Auditory nerve take the impulse from eye to brain., , (8), , Impression of an image lasts on the retina for, th of a second., , (9), (10), , 1, 10, , The focal length of a convex lens is positive., , I n the compound microscope, two concave lenses, are used., (1) False. A Lens is a transparent object., Ans. , , (7), , What is persistence of vision?, , Ans. , The impression of the image lasts on the retina for, 1, , th of a second. The sensation on retina persists, 16, for a while. This is called persistence of vision., (8), , What is principal focus?, , Ans. T,  he point on the principal axis where all the, incident rays parallel to the principal axis actually, meet or appear to meet after refraction is called, principal focus., (9), , What is focal length?, , Ans. , The distance between optical centre and principal, focus of a lens is called focal length.

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Lenses, , (10), , 135, , What is the distance of distinct vision?, , Ans. , The minimum distance at which an object should, be placed from a normal eye so that there won’t, be any strain on the eye is called distance of, distinct vision. For normal human eye it is 25 cm., (11), , What do you mean by power of accommodation?, , Ans. T,  he capacity of the lens to adjust its focal length, as per need is called its power of accommodation., *(12) Why do we have to bring a small object near, the eyes in order to see it clearly. (Use your Brain, Power; Textbook Page 89), Ans. , If we bring small objects near the eyes, the angle, subtended by the object at the eye becomes large, which makes the eye to see the object clearly., Q.1. (B) Choose and write the correct option:, (1) For the normal human eye, the distance of distinct, vision is, , (9) Convex lens of power +5 D and concave lens, with power -3 D are placed together, then the, ., combined power is, , , (a) 5 D, , (b) +3 D, , (c) 2 D, , (10) In simple microscope,, , (d) –2 D, lens is used., , , , (a) Concave, , (b) Cylindrical, , , , (c), , (d) Convex, , Diverging, , (11) The perception of dim light is concerned with, cells., , , (a) Rod , , (b) Cone, , , , (c), , (d) Squamous, , Amoeboid, , (12) The impression of an image lasts on the retina fo, 1 th of a second, is called, , ., 16, , , (a) dispersion, , , , (b) refraction, , (2) The power of a convex lens of focal length 50 cm, is, ., , , , (c), , , , , , (d) internal reflection, , (13) The second focal point is located at, of a human eye., , (a) Retina , (b) Optic nerve, , , , (a) 15 cm, , (a) 2 D, , (b) 20 cm, , (b) 0.2 D, , (c) 25 m, , (c) 50 D, , (d) 25 cm, , (d) 0.5 D, , (3) The focal length of a concave lens with power, -4 D is, ., , , (a) –0.5 m (b) 0.5 m, , (c) –0.25 m (d) 0.25 m, , (4) If the incident ray passes through focus, then the, refracted ray is, to the principal axis., , , (a) parallel, , (b) opposite, , , , (c), , (d) intersecting, , perpendicular, , (5) The image is formed on the, human eye., , (a) Cornea , (b) Retina, , , (c), , Pupil , , of the, , (d) Ciliary muscle, , (6) If an object is placed between F1 and 2F1 of a, convex lens, then nature of the image formed is, ., , , (a) Real and inverted, , , , (b) Real and diminished, , , , (c), , , , (d) Virtual, inverted, , , , (c), , persistence of vision, , Cornea , , (d) Pupil, , (14) For a Convex lens, if the incident ray is parallel, to the principle axis, then the refracted ray passes, ., through the,, , , (a) Centre , , (b) Pole, , , , (c), , (d) Principal focus, , Optical centre, , (15) Inside water, an air bubble behaves, , , (a) like a flat plate, , , , (b) like a concave lens, , , , (c), , , , (d) like a concave mirror, , ., , like a convex lens, , (16) To obtain an image of the same size as the object, with the help of a convex lens, the object should, ., be placed, , Virtual, erect, , , , (a) at infinity, , (b) beyond F1, , , , (c), , (d) at 2F1, , (17), , , At what distance should a watchmaker hold his, lens from the watch?, , (8) Longsightedness can be corrected by using, lens., , , , (a) At the focal length, , , , (b) At less than focal length, , , , (a) cylindrical, , (b) concave, , , , (c), , , , (c), , (d) converging, , , , (d) At zero distance, , (7) In myopia,, clearly., , objects can be seen, , , , (a) distant , , (b) nearby, , , , (c), , (d) big, , small , , diverging, , between F1 and 2F2, , At more than the focal length

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Master Key Science and Technology Part - 1 (Std. X), , 136, , (18), , , When we enter a dark room, pupil of our eye, ., , , , (a) contracts, , (b) expands, , , , (c), , (d) none of these, , remains same, , Ans. , (1) (d) 25 cm; (2) (a) 2 D; (3) (c) –0.25 m;, (4) (a) parallel ; (5) (b) retina (6) (a) Real and inverted ;, (7) (b) nearby ; (8) (d) converging ; (9) (c) 2 D ;, (10) (d) convex; (11) (a) Rod ; (12) (c) persistence, of vision (13) (a) retina ; (14) (b) principal focus ;, (15) (d) like a concave lens ; (16) (d) at 2F1 ; (17) (b) At, less than focal length (18) contracts, Q.2.1. Solve the following:, , , Type: A, Problems based on the formulae:, 1, v, , –, , 1, , =, , u, , 1, , Formula:, , ∴, v, , h2, ∴, , Fig 7.1, ∴, , h2, , ∴, h1, u, , Fig 7.2, Tips:, (1), (2), (3), (4), (5), (6), (7), , ∴, ∴, , v, , If h2 is negative, then image is Real and Inverted, If h2 is positive, then image is Virtual and Erect, h1 greater than h2, image is diminished, h1 smaller than h2, image is magnified, v is positive then image is on other side of lens, v is negative then image is on same side of lens., f is +ve, it is convex lens, corrects Hypermetropia, or Farsightedness, f is -ve, it is concave lens, corrects Myopia or Near, Sightedness., , f = focal length, , (1) An object is placed at a distance of 15 cm from, a convex lens. If the focal length of the lens is, 60 cm, find the image distance., Ans. Given: Object distance (u) = –15 cm, Focal length (f) = 60 cm, To find: Image distance (v) = ?, v, , Solution:, , u, , u = object distance, , f, , , , Use the following ray diagrams and tips for, solving the numericals., , h1, , v = image distance, , 1, , ∴, , 1, v, 1, v, 1, v, 1, v, 1, v, 1, v, , =, =, =, =, =, =, , 1, , +, , f, 1, , 60, 1, 60, , 1, , 1, , –, , u, 1, , –, , v, , 1, , =, , f, 1, , =, , u, , f, , 1, u, , +, , 1, (–15), , –, , 1, 15, , 15 – 60, 900, – 45, 900, –1, 20, , v = –20 cm, The image of the object is formed at a distance, of 20 cm on the same side of the lens., , (2) An object is placed at a distance of 10 cm from, a convex lens of focal length 12 cm. Find the, position and nature of the image., Ans. Given: Object distance (u) = –10 cm, , Focal length, (f) = 12 cm, To find: Image Position (v) = ?, , Nature of image = ?, Formulae:, Solution:, , 1, v, , 1, v, , –, –, , 1, u, , 1, u, , =, =, , 1, f, 1, f

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Lenses, , ∴, ∴, ∴, , ∴, ∴, ∴, ∴, , 137, , 1, v, 1, v, 1, v, 1, v, 1, v, 1, v, , =, =, =, , =, =, =, , 1, , +, , f, 1, , 12, 1, 12, , f = – 30cm, Since, f is –ve,, , u, , Lens is concave i.e. diverging lens., , 1, , +, , Type: B, , –10, , Problems based on the formulae:, , 1, , –, , 1, , 10, , v, , 10 – 12, 12 × 10, 12 × 10, 60, , v = –60 cm, The image is formed at a distance of 60 cm on, the same side of the lens, it is a virtual and, erect image., , To find: Focal length (f) = ?, 1, , –, , v, , Solution:, , ∴, ∴, , ∴, ∴, ∴, , 1, f, 1, f, 1, f, 1, f, 1, f, , =, =, =, , =, =, =, , v, , –, , 1, –20, –1, 20, , –, , +, , 1, –60, , 60, , 20 × 60, , u, , Solution:, h2 =, , 1, u, , =, , 1, f, , ∴, , h2 =, , ∴, , h2 =, , h2, h1, , =, , v, u, , v × h1, u, 9×6, –18, 6, , –2, h2 = –3 cm, The height of the image is 3 cm and it is, inverted., , 1, , –40, , v, , Formula: M =, , u, , 20 × 60, , 30, , h1, , =, , To find: Height of the image (h2) = ?, , ∴, , 1, , –60 + 20, , –1, , h2, , f, , Distance of the image (v) = 9 cm, , Image distance (v) = –20 cm, , f, , =, , Distance of the object (u) = –18 cm, , Ans. Given: Object distance (u) = –60 cm, , 1, , u, , v = distance of the image, u = distance of the object, f = focal length, M = magnification, h2 = height of image, h1 = height of object, , 1, , Ans. Given: Height of the object (h1) = 6 cm, , *(3) An object kept 60 cm from a lens gives a virtual, image 20 cm in front of the lens. What is the, focal length of the lens? Is it a converging lens, or diverging lens?, , 1, , 1, , (1) An object 6 cm tall is placed in front of a convex, lens at a distance of 18 cm. If the image is formed, at a distance of 9 cm on the other side of lens,, find the height of the image., , –1, , Formulae:, , –, , M=, , –2, , , , f = –30 cm, , ∴, , 1, , *(2) An object is placed vertically at a distance of, 20 cm from a convex lens. If the height of the, object is 5 cm and the focal length of the lens is, 10 cm, what will be the position, size and nature, of the image? How much bigger will the image, be as compared to the object?, Ans. , Given: Height of the object (h1) = 5 cm, , , Focal length (f) = 10 cm, , , , Distance of the object (u) = –20 cm, , To find: Image distance (v) = ?, , , Height of the image (h2) = ?, , , , Magnification (M) = ?

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Master Key Science and Technology Part - 1 (Std. X), , 138, , Formulae: (i), , , (ii), , Solution:, (i), ∴, ∴, ∴, , 1, v, , v, , –, , h2, h1, , 1, u, =, , =, , 1, f, , v, u, , f, 1, , ∴, , 1, , ∴, , v, , 10 + 20, 20 x10, , =, , 1, , ∴, , 10, =, v 200, , h1, , =, , ∴, , h2 =, , (ii), , v, u, v × h1, , ∴, , u, 20 × 5, , ∴, , –20, , h = –5 cm, , ∴, , M=, , v, u, , =, , 20, –20, , v, , 1, v, 1, v, 1, v, 1, v, , 1, , +, , f, , 1, u, , =, , =, , 1, 10, , 1, 25, , 25 – 10, 250, 15, , =, =, , –, , 250, 3, 50, , ,v=, , 50, 3, , = 16.666, , v = 16.7 cm, h2, h1, h2, 5, , =, , v, u, 50, , =, , 3, –25, , h2 = – 50 × 5 =, 3 × 25, , –10, 3, , = – 3.333, , h2 = – 3.3 cm, , ∴, = –1, , Position of image = 16.7 cm, Size of the image = 3.3 cm, , M = –1, , ∴, , =, , 1, 1, 1, =, +, – 25, v, 10, , ∴, , v = 20 cm, h2, , 1, , ∴, , 1, 1, =, +, =–, +, 20 10, v, –20, 10, , h2 =, , Nature = Real and inverted, , The object and the image are of the same, size. Image is real and inverted, *(3) , 5 cm high object is placed at a distance of, 25 cm from a converging lens of focal length, 10 cm. Determine the position, size and type of, the image., Ans. Given: Height of the object (h1) = 5 cm, , , Focal length (f) = 10 cm, , , , Distance of the object (u) = –25 cm, , To find: Image distance (v) = ?, , , Height of the image (h2) = ?, 1, 1, 1, Formula:, –, =, u, v, f, , (i), , ∴, , ∴, , 1, , 1, , ∴, , (iii), , =, , u, , 1, , ∴, (ii), , –, , 1, , 1, , Solution:, 1 1 1, – =, v u f, , Type: C, Problems based on the formulae:, P=, , 1, f (m), , Dioptre, , Note:, When we have numerical in power and focal length:, , •, •, , Focal length should be in metre., Power should be in dioptre., , *(1) The focal length of a convex lens is 20 cm. What, is the power?, 20, Ans. , Given: Focal length (f) = 20 cm =, m, 100, , = 0.2 m, To find: Power of lens (P) = ?, Formula: P =, , 1, f (m)

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Lenses, , 139, , Solution: P =, 1, , ∴, , P=, , =, , ∴, , P=5D, , 0.2, , 10, , ∴, , 1, f (m), , 2, , Power of lens is 5 dioptre., *(2) , Doctor has prescribed a lens having power, +1.5D. What will be the focal length of the lens?, What is the type of the lens and what must be, the defect of vision?, , The focal length of the lens is 0.4 m., *(4) Three lenses having power 2, 2.5 and 1.7 D are, kept touching in a row. What is the total power, of the lens combination?, Ans. , Given: P1 = 2 D, P2 = 2.5 D, P3 = 1.7 D, To find: Ptotal = ?, Formula: Ptotal = P1 + P2 + P3, , Ans. Given: Power of lens (P) = 1.5 D, , Solution: Ptotal = P1 + P2 + P3, , To find: Focal length (f) = ?, Formula: P =, Solution: P =, ∴, , 1, f=, P, , ∴, , 1, f=, 1.5, , ∴, , f=, , ∴, , f=, , f = 0.4 m, , 1, , ∴, , Ptotal = 2 + 2.5 + 1.7, , f (m), , ∴, , Ptotal = 6.2 D, , 1, , Total Power of combination is 6.2 D., , f (m), , Q.2.2. Define the following:, (1), , Centres of curvatures (C), , Ans. , The centres of the spheres whose parts form, surfaces of the lenses are called centres of, curvature of the lens., , 10, , (2), , 15, , Ans. The radii of the spheres whose parts form surfaces, of the lenses are called radii of curvatures., , 2, 3, , (3), , = 0.67 m, , Since f is +ve, lens used is convex, f = 0.67 m, The defect is Hypermetropia (farsightedness), , Radii of curvatures (R), , Principal axis, , Ans. The imaginary line passing through the centres, of curvature is called principal axis., (4), , Optical centre (O), , (3) Calculate the focal length of a corrective lens, having power + 2.5 D., , Ans. The imaginary point on the lens through which the, light ray passes undeviated is called optical centre., , Ans. Given: Power of lens (P) = 2.5 D, , (5), , To find: Focal length (f) = ?, Formula: P =, Solution: : P =, ∴, , f=, , ∴, , f=, , ∴, , f=, , ∴, , f=, , 1, P, , 1, f, , dioptre, , (6), , Focal length (f), , Ans. , The distance between optical centre and principal, focus is called focal length of a lens., (7), , (8), , 2.5, 1 × 10, 2.5 × 10, 25, , f (m), , Ans. , The point on the principal axis where all the, incident rays parallel to the principal axis actually, meet or appear to meet is called principal focus., , Convex lens, , Ans. A lens having both the spherical surfaces bulging, outward is a convex lens., , 1, , 10, , 1, , Principal Focus (F), , =, , 2, 5, , Concave lens, , Ans. A lens having both the spherical surfaces curved, inwards is a concave lens., (9), , Real image, , Ans. The image obtained when the refracted rays, actually meet each other is called a real image.

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Master Key Science and Technology Part - 1 (Std. X), , 140, , (10), , Virtual image, , (15), , Adaptation of pupil., , Ans. The image obtained when the refracted rays, appear to meet is called a virtual image., , Ans. The tendency of the pupil to adjust the opening, for light is called adaptation., , (11), , (16), , Converging lens, , Ans. The lens which converges the light incident on it, is called converging lens., (12), , Diverging lens, , Ans. The lens which diverges the light incident on it is, called diverging lens., (13), , Lens, , Iris, , Ans. The dark muscular diaphragm behind cornea is, called iris., (17), , Pupil, , Ans. The small opening of variable diameter at the, centre of iris is called the pupil., (18), , Power of accommodation, , Ans. A lens is a transparent medium bounded by, two surfaces, out of which at least one surface is, spherical., , Ans. The ability of the lens of adjusting focal length is, known as power of accommodation., , (14), , Ans. The minimum distance from the normal eye,, at which the objects can be seen clearly and, distinctly without any strain to the eye is known, as the distance of distinct vision., , Retina, , Ans. The light sensitive screen in the human eye is, called retina., , (19), , Distance of distinct vision, , Q.2.3. Distinguish between:, *(1), , Farsightedness and Nearsightedness., , Ans., , Farsightedness/ Hypermetropia, , Nearsightedness/ Myopia, , (i), , Curvature of cornea and eye lens decreases, , (i), , Curvature of cornea and eye lens increases, , (ii), , Eyeball flattens slightly., , (ii), , Eyeball elongates slightly., , (iii) Image of the nearby object falls behind the (iii) Image of the nearby object falls on the retina., retina., (iv) Image of the distant object falls on the retina., , (iv) Image of the distant object falls in front of the, retina., , (v), , (v), , This defect is corrected by using a spectacle of, convex lens of suitable focal length., , *(2) Concave lens and Convex lens., Ans., Concave lens, , This defect is corrected by using a spectacle of, concave lens of suitable focal length., , Convex lens, , (i), , It is also called as diverging lens., , (i), , (ii), , This lens is thinner in the centre than at its edges. (ii), , It is also called as converging lens., This lens is thicker at the centre than at its edges., , (iii) The principal focus is virtual., , (iii) The principal focus is real as well as virtual., , (iv) The focal length is negative., , (iv) The focal length is positive., , (v), , (v), , This lens is used to correct myopia., , (3), Real image and Virtual image., Ans., Real Image, (i), , When the refracted rays actually meet, the real, image is formed., , This lens can be used to correct hypermetropia., , Virtual Image, (i), , When the refracted rays appear to meet, the, virtual image is formed., , (ii) Real image can be taken on the screen., , (ii) Virtual image can not be taken on the screen., , (iii) Real image is always inverted., , (iii) Virtual image is erect (upright)., , (iv) In a ray diagram, real image is formed on (iv) In a ray diagram, virtual image is formed on, opposite side of lens., same side of lens.

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Lenses, , 141, , Q.2.4. Give scientific reasons:, *(1), , (v), , (i), (ii), , (4), A magnifying glass works on the principle of, , Ans., , simple microscope., , (i), , of a convex lens we get a virtual, erect and, , point on the principal axis., , This principle is used by the watch repairer to see, use, , a, , magnifying, , glass, , while, , (ii), , This point is called the principal focus., , (iii), , Light ray parallel to the principal axis falling on, a convex lens come together, i.e. get focused at, , repairing, , principal focus. So this type of lens is called a, , wristwatches., , converging lens., , *(2) We cannot clearly see an object kept at a distance, less than 25 cm from the eye., Or, If we bring an object closer than 25 cm from the, , (5), (i), , Ans., The minimum distance at which an object is, placed from the normal eye so that it can be seen, , way that they appear to be coming out of a point, on the principal axis., (ii), (iii), , (iii), , If the object is placed at a distance less than 25, cm, ciliary muscles are unable to adjust the lens, there is a strain on the eye and it cannot be seen, clearly., Hence an object should be placed at 25 cm or, more to have a clear and sharp image of it on the, retina., , *(3), , One can sense colours only in bright light., , (diverge) after refraction. So, this type of lens is, called diverging lens., (6) You can not enjoy watching a movie or television, from a very short distance from the screen., Ans., (i), , The light sensitive screen of the eye called retina, has two types of light sensitive cells., , (ii), , Rod cells are concerned with intensity of light, and cone cells are responsible for the vision in, , (iii), , (ii), , The ability of a normal eye or pupil to adjust the, opening for light is called adaptation., , (iii), , As the light is bright, the pupil contracts and, exposure in this condition for longer period of, time can give an additional strain on the eye., , (iv), , Due to this, person can not enjoy watching movie, or television program sitting close to the screen., , (7), , presence of light., , Ans., , Cone cells are also concerned with colour, , (i), , A concave lens is used to correct myopia., In Myopia, the curvature of the cornea and eye, , perception. They do not respond to dim light, , lens increases. Due to this, converging power of, , or no light. Rod cell respond to faint light but, , lens remains large., , conical cells do not., (iv), , If a person is very close to the screen, then the, intensity of light falling on the eye is more., , Ans., (i), , Light rays parallel to the principal axis falling, on a concave lens go away from one another, , called the distance of the distinct vision., This distance is 25 cm for a normal human eye., , This point is called the principal focus of the, concave lens., , clearly without giving any strain on the eye is, (ii), , Rays travelling parallel to the principal axis of, a concave lens diverge after refraction in such a, , it subtends a bigger angle at the eye? (Use your, brain power; Textbook Page 89), , Concave lens is also called as diverging lens., , Ans., , eyes, why can we not see it clearly even though, , (iv), , When light rays parallel to the principal axis are, incident on a convex lens, they converge to a, , the small parts more clearly. Hence, watchmakers, , (i), , Convex lens is also called a converging lens., , When an object is placed within the focal length, magnified image on the same side of the lens., , (iii), , in presence of light., , Simple microscope is used for watch repairs., , Ans., , Hence, we can identify and differentiate colours, , Cone cells can function only in bright light., , (ii), , Also the eyeball elongates so that the distance, between the lens and retina increases.

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Master Key Science and Technology Part - 1 (Std. X), , 142, , (iii), , (8), ,  concave lens of suitable focal length can create, A, required divergence, hence, after convergence, by eye lens sharp image of the distant object is, formed on the retina and the defect is be corrected., Old people sometimes use bifocal glasses., , Ans., (i), ,  ew old people, suffer from both myopia and, F, hypermetropia., , (ii), , I n such cases a bifocal lens is required to correct, the defect. The upper part is concave lens to, correct myopia and lower part is convex lens to, correct hypermetropia., , (iii), ,  he concave lens is useful for distant object vision, T, and convex lens for nearby object vision., , (iv), , Hence, old people sometimes use bifocal glasses., , (9) , Hypermetropia can be corrected by using, convex lens., Ans., (i), , (ii), (iii), , In hypermeteropia curvature of the cornea and, eye lens decreases so that the converging power, of the lens becomes less.,  ue to the flattening of the eye–ball the distance, D, between the lens and retina decreases.,  his defect can be corrected by using a convex, T, lens. This lens converges the incident rays which, are further converged by retina to get a sharp, image of nearby object on the retina., , (10) Why do we have to bring a small object near, the eyes in order to see it clearly. (Use your brain, power; Textbook Page 89), Ans., (i), , The Apparent size of an object depends on the, angle subtended by the object at the eye., , (ii), , When the object is closes to the eye. The angle, subtended is larger and it appears bigger, hence, we have to bring a small object near eye., , Q.3.1. Answer the following questions:, (1),  hich are the different types of lens?, W, Ans. In all, there are six different types of lenses,, namely, bi-convex, plano-convex lens, Positive, Meniscus, bi-concave lens, plano-concave lens, and Negative Meniscus., *(2), , , , What is magnification? Write its formula., Or, What is the relation between h1, h2, u and v., (Use your brain power; Textbook Page 85), , Ans. , The ratio of height of image (h2) to the height of, the object (h1) is called Magnification (M)., , ∴ Magnification =, , Height of the image, , Height of the object, h2, ∴, M=, , ... (i), h1, Magnification can also be calculated by the, ratio of the image distance (v) to the object, distance (u)., Image distance, ∴ Magnification =, object distance, v, ∴, M=, , ... (ii), u, From (i) and (ii),, h2, v, ∴, M=, =, u, h1, (3), , What is the Power of lens?, , Ans., (i) The capacity of a lens to converge or diverge, incident rays is called its power (P)., (ii) The power of a lens depends on its focal length, (f):, (iii), , Power of lens is the reciprocal of focal length in, meters., , ∴, , Power of lens =, , ∴, , P=, , 1, , 1, focal length (metre), , f (m), , (4), What do you mean by combined focal length?, Ans. The reciprocal of combined focal length is equal, to the sum of the reciprocals of focal lengths of, number of all lenses placed together., , ∴, , 1, f, , =, , 1, f1, , +, , 1, f2, , where f - combined focal length,, f1 - focal length of 1st lens,, , f2 - focal length of 2nd lens., (5), , What is the use of astronomical telescope?, , Ans. A,  stronomical telescope is used to observe the, celestial objects like stars and planets., (6), , Enlist the functions of Iris., , Ans. Functions of Iris are:, (i), , To decide the size of the pupil., , (ii), , To give colour to the human eye., , (7) Write the uses of concave lens., Ans. The uses of concave lens are as follows:, (i) Concave lens are used to correct Myopia., (ii), , In P,  eep hole of the outside door., , (iii) In flash lights to magnify light produced by, bulb.

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Lenses, , (8) Write the uses of convex lens., Ans., (i) , Watch repairers use simple microscope which, has convex lens., (ii) To correct hypermetropia, a spectacle of convex, lens is used., (iii) Astronomical telescope, compound microscope, are the optical instruments where two convex, lenses are used., (9), What is persistence of vision?, Ans., (i) We see an object because the eye lens creates its, image on the retina., (ii) The image is on the retina as long as the object is, in front of us. The image disappears as soon as, the object is taken away., (iii), , , However, this is not instantaneous and the image, remains imprinted on our retina for 1 th of a, 16, second after the object is removed., , (iv), , The sensation on retina persists for a while. This, is called persistence of vision., , (10), , What is colour blindness?, , Ans., (i) Some people are unable to identify or distinguish, certain colours due to the lack of cone cells in the, retina., (ii) This is a hereditary disorder and linked with X, chromosomes., (iii) This defect/disorder is called colour blindness, and person is called colour blind., (iv) It is mostly for the recognition of red and green, colour., *(11) What is the function of iris and the muscles, connected to the lens in human eye?, , 143, , (v), ,  rain processes all the information received and, B, we see the actual image of the object., , (vi), ,  hus conical cells are responsible for perceiving, T, different colours., , *(13) What are real and virtual images? How will you, find out whether an image is real or virtual? Can, a virtual image be obtained on a screen? (Can, you Recall; Textbook Page 82), Ans. , (i), , I f the reflected or refracted rays from the lens, actually meet at a point when light rays from the, object are incident on it, then the image formed is, Real and it can be seen on a screen., , If the reflected or refracted rays from the lens, appear to meet at a point when light rays from the, object are incident on it, then the image formed is, called virtual., (ii), , I f the images are inverted then they are real and, if they are erect then they are virtual., , (iii), , Virtual images cannot be obtained on a screen., , *(14) Indicate the following terms related to spherical, mirrors in figure: Poles, centre of curvature,, radius of curvature, principal focus. (Can you, Recall; Textbook Page 80), Ans. , Radius of curvature, Centre of, curvature, M, Principal axis, , P, , C, , Pole, of the, mirror, , Ans., (i), , An iris in human eye imparts colour to the eye., , (ii) The muscles connected to the lens in the human, eye adjust the focal length of eye lens., *(12) H,  ow do we perceive different colours? (Can you, tell; Textbook Page 91), Ans. , (i), (ii), (iii), , (iv), ,  he retina in our eyes is made up of many light, T, sensitive cells., These cells are shaped like a rod and like a cone., The rod like cells respond to the intensity of light, and give information about the brightness or, dimness of the object to the brain., The conical cells respond to the colour and give, information about the colour of the object to the, brain., , Fig 7.1, *(15) , How are concave and convex mirrors, constructed? (Can you Recall; Textbook Page 80), Ans., , A, B, , Fig 7.2

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Master Key Science and Technology Part - 1 (Std. X), , 144, , (i), (ii), (iii), ,  enerally, concave and convex mirror are, G, spherical mirrors., Spherical mirrors are parts of a hollow glass sphere, like the part B in the figure., The inner or outer surface of this part is coated with, a shiny substance to produce spherical mirror., , Q.3.2. Draw well labelled diagrams for the following., (1) A ray diagram for object position at infinity for, a convex lens or convergence of rays by convex, lens., A, O, 2F1, , 2F2, , F2, , F1, , *(4) A ray diagram for object position between 2F1, and F1 for a convex lens., A, , 2F1 B, , B 2F1, , F1, , O, , B’, , 2F2, , 2F2, , O, , A’, Fig 7.6, *(5) A ray diagram for object positioned at F1 for a, convex lens., A, , 2F1, , Image position: At F2, Nature: Real, inverted and highly diminished., Fig 7.3, , F2, , F1, , B’, , Image position: Beyond 2F2, Nature: Real, inverted and magnified., , B, , *(2) A ray diagram for object position beyond 2F1, for a convex lens., A, , F2, , B, F1, , F2, , O, , 2F2, , Image position: At infinity, Nature: Real, inverted and highly magnified., Fig 7.7, *(6) A ray diagram for object position between F1, and O for a convex lens., A’, , A’, Image position: Between F2 and 2F2, Nature: Real, inverted and diminished., Fig 7.4, *(3) At which position will you keep an object in, front of a convex lens so as to get a real image of, the same size as that of the object ? Draw a figure., Or, A ray diagram for object position at 2F1 for a, convex lens., , A, B’ 2F1, , F1 B, , F2, , 2F2, , Image position: On the same side of lens as the object., Nature: Virtual, erect and magnified., Fig 7.8, , A, B, 2F1, , O, , F2, F1, , O, , Image position: At 2F2, Nature: Real, inverted and of same size., Fig 7.5, , 2F2, B1, , *(7) Divergence of rays by concave lens., , A1, , O, 2F1, , F1, , Image position: At F1, , Nature: Virtual, erect, highly diminished, , Fig 7.9

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Lenses, , 145, , (8) , Ray diagrams showing images obtained by, concave lens., P, , P1, , 2F1 Q, , F1, , Q1, , D, , A, , Q.4.1. Answer the following in brief., *(1) Draw a figure explaining various terms related, to a lens., Ans. Concepts related to lens:, A, , S1, , O, , Image position: On the same side of lens, Nature: Virtual, erect, diminished, , 1, , C1, , R2, , The structure of Human eye., Muscles, , Ciliary muscles, , S2, , O, , Fig 7.10 Image formed by a concave lens, (9), , 2, , R1, , C2, , B, , Iris, , Fig 7.13a Cross section of convex lens, , Pupil, , Optic, nerve, Optic, disc, , Crystalline, lens, Cornea, Sclerotic, coat, Choroid, , S1, R1, , B, 2, , R2, C2, , D, , Fig 7.11 Structure of human eye, , Lens becomes flat, , 1, , O, , C1, , Retina, , (10) The change in the shape of the lens while seeing, distant and nearby objects., , A, , S2, , C, , b, Fig 7.13b Cross section of ccncave lens, , (i) Centre of curvature (C): The centres of spheres, whose parts form surfaces of the lenses are, called centres of curvatures of the lenses. A lens, with both surfaces spherical, has two centres of, curvature C1 and C2., , Light coming, from a distant, object, Fig 7.12a, Lens becomes, rounded, , Light coming, from a nearby, object, Fig 7.12b, , (ii) Radius of curvature (R): The radii (R1 and R2), of the spheres whose parts form surfaces of the, lenses are called the radii of curvature of the lens., (iii) Principal axis: The imaginary line passing, through both centres of curvature is called the, principal axis of the lens., (iv) Optical centre (O): The point inside a lens on, the principal axis, through which light rays, pass without changing their path is called the, optical centre of a lens. In figure 7.14, rays, P1Q1, P2Q2 passing through O are going along a, straight line. Thus, O is the optical centre of the, lens.

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Master Key Science and Technology Part - 1 (Std. X), , 146, P1, , Light rays parallel to the principal axis falling on, a convex lens come together, i.e. get focused at a, point on the principal axis. So, this type of lens is, called a converging lens., , Q2, , P3, , Rays travelling parallel to the principal axis of, a concave lens diverge after refraction in such a, way that they appear to be coming out of a point, on the principal axis. This point is called the, principal focus of the concave lens. As shown in, figure 7.15 (b) F1 and F2 are the principal foci of, the concave lens., , Q3, , O, , P2, , Q1, , P1, , Light rays parallel to the principal axis falling, on a concave lens go away from one another, (diverge) after refraction. So, this type of lens is, called a diverging lens., , Q2, , P3, , (vi) Focal length (f): The distance between the optical, centre and principal focus of a lens is called its, focal length., , Q3, , O, , (2) Enlist the rules for drawing ray diagrams of, convex lens., P2, , Ans. F,  ollowing are the rules for obtaining images by, convex lens., , Q1, Fig 7.14 Optical centre of a lens, , Incident ray, , (v) Principal focus (F): When light rays parallel to, the principal axis are incident on a convex lens,, they converge to a point on the principal axis., This point is called the principal focus of the, lens. As shown in figure 7.15a F1 and F2 are the, principal foci of the convex lens., , F1, , Refracted ray, O, , F2, , Fig 7.16, Rule (i): When the incident ray is parallel to the, principal axis, the refracted ray passes through, principal focus (F2)., , F1, , F2, , f, , Refracted ray, , F1, , Fig 7.15 (a) Principal focus of a convex lens, , O, , F2, , Incident ray, Fig 7.17, F1, , F2, f, , Fig 7.15 (b) Principal focus of concave lens, , Rule (ii): When the incident ray passes through, the principal focus (F1), the refracted ray is, parallel to the principal axis.

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Lenses, , 147, , Q.4.2. Explain the following with the help of a diagram:, (1) Myopia or Nearsightedness, , Incident ray, , F1, , O, , F2, Reflected ray, , Nearby objects can be seen clearly, , Fig 7.18,  ule (iiii): When the incident ray passes through, R, the optical centre of the lens it passes without, changing its direction., *(3) , What is the Cartesian sign convention used, for spherical mirrors? (Can you Recall; Textbook, Page 84), Ans. , According to the Cartesian sign convention,, the pole of the mirror is taken as the origin., The principal axis is taken as the X–axis of the, frame of reference. The sign conventions are as, follows., Y-axis, Direction of, incident rays, distances on, Distance distances on, the right (+ve), the left (–ve), upward, (+ve), Principal axis, P, X-axis, Distance, downward, (–ve), , Fig 7.19, (i), ,  he object is always kept on the left of the both, T, mirrors. All distances parallel to the Principal, axis are measured from the pole of the mirror., , (ii), ,  ll distances measured towards the right of, A, the pole are taken to be positive, while those, measured towards left are taken to be negative., , (iii), ,  istances measured vertically upwards from the, D, principal axis are taken to be positive., , (iv), ,  istances measured vertically downwards from, D, the principal axis are taken to be negative., , (v), ,  he focal length of concave mirror is positive, T, while that of a convex mirror is negative., , Myopic eye, Concave lens, , Correction of Nearsightedness, Fig 7.20 Nearsightedness, (i) Person having myopia can see nearby objects, clearly but the distant objects appear indistinct., (ii) In nearsightedness, the image of a distant object, forms in front of the retina. There are two reasons, for this defect., (a) The curvature of the cornea and the eye lens, increases, so that the converging power of the, lens remains large., (b) The eyeball elongates so that the distance between, the lens and the retina increases., (iii) This defect can be corrected by using spectacles, with concave lens of proper focal length., (iv) This lens diverges the incident rays and these, diverged rays can be converged by the lens in the, eye to form the image on the retina., (2) Hypermetropia or Farsightedness, , Faraway objects can be, seen clearly, , Hypermetropic eye, Convex lens, , Correction of Farsightedness, , Fig 7.19 Farsightedness

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Master Key Science and Technology Part - 1 (Std. X), , 148, , (i) In hypermetropia, a person can see distant objects, clearly but cannot see nearby objects distinctly., , (i) A compound microscope is made of two convex, lenses objective and eye piece., , (ii) The images of nearby objects are formed, behind the retina. There are two reasons for, farsightedness., , (ii) The axes of both lenses are along the same line., The lenses are fitted inside a metallic tube in such, a way that the distance between can be changed., , (a) Curvature of the cornea and the eye lens decreases, so that, the converging power of the lens becomes, less., , (iii) The objective has smaller cross-section and, smaller focal length. The eye piece has bigger, cross section, its focal length is also larger than, that of the objective., , (b) Due to the flattening of the eyeball the distance, between the lens and retina decreases., (iii) This defect can be corrected by using a convex, lens with proper focal length. This lens converges, the incident rays before they reach the lens., (iv) The lens then converges them to form the image, on the retina., (3) Presbyopia, (i) Presbyopia is an age related vision problem., The vision becomes blurred. Because of this old, people cannot see nearby objects., , (iv) Higher magnification can be obtained by the, combined effect of the two lenses., (v) The magnification occurs in two stages. The, image formed by the first lens acts as the object, for the second lens., *(5) , Explain the working of an astronomical, telescope using refraction of light., Ans., F0, , (ii) Generally the focusing power of the eye lens, decreases with age. The muscles near the lens, loose their ability to change the focal length of the, lens. The near point of the lens shifts farther from, the eye., (iii) Presbyopia can be corrected by using spectacle of, convex lens of suitable focal length., (iv) Sometimes people suffer from nearsightedness, as well as farsightedness. In such a case, bifocal, lenses are required to correct the defect., (v) In such lenses, the upper part is concave lens and, corrects nearsightedness while the lower part is a, convex lens which corrects the farsightedness., (4) Compound microscope, Object, , u, , Fe, , Fo, Objective lens, , u, Eyepiece, , Fe, , Fe, Objective lens, , Eyepiece, , Fig 7.21, (i) Telescope is used to see distant objects clearly, in their magnified form. The telescopes used to, observe astronomical sources like the stars and, the planets are called astronomical telescopes., (ii) Objective lens has large diameter and larger, focal length because of which maximum amount, of light coming from the distant object can be, collected., (iii) On the other hand the size of the eyepiece is, smaller and its focal length is also less. Both the, lenses are fitted inside a metallic tube in such, a way that the distance between them can be, changed., (iv) When rays of light enter the objective, they refract, and give a real inverted and diminished image., , Image, , Fig 7.20, , (v) The eye–piece is so adjusted that the image, becomes an object for the eyepiece and gives a

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Lenses, , 149, , virtual, enlarged and inverted image w.r.t to, object., Q.4.3. Application Based Questions:, (1), , Answer the questions based on the following, diagrams:, , (ii), , (c), , Due to the flattening of the eye ball the, distance between the lens and retina, decreases., , P = + 2D, 1, 1, f=, =, P, 2, = 0.5 m, ∴ focal length = 0.5 metre, , (1), , (2), , (d), , Convex lens., , (3), , Vijay is 16 yr old and wears a spectacle having a, power of -2D. Answer the following questions:, , (a), Name his eye defect., (b) State two reasons why this defect is caused?, (c), Compute the focal length of his lens., Ans. , (a), (b), , Myopia., (i), The curvature of the cornea and the eye lens, increases, so that the converging power of, the lens remains large., (ii) The eyeball elongates so that the distance, between the lens and the retina increases., , (c), , P = – 2D, , (3), Fig 7.22, (a), , Name the defect shown in figure (1)., , (b), , Give two reasons for this defect., , (c), , In which figure is the defect corrected, in figure, (2) or (3)?, , f=, , 1, p, , 1, 2, = – 0.5 m, , =–, , Ans. , (a), , Myopia., , (b), , (i), , The curvature of the cornea and the eye lens, increases, so that the converging power of, the lens remains large., , (4), , (ii), , The eyeball elongates so that the distance, between the lens and the retina increases., , Ankit developed an eye defect at the age of 65, yrs and had to wear a lens of power +3D. Answer, the following questions:, , (a), , Name the eye defect., , (b), , State two reasons why this defect is caused?, , (c), , Compute the focal length of his lens., , (c), , Figure (2), , (2), , Selwyn is 16 yr old and wears a spectacle, having a power of +2 D. Answer the following, questions:, , (a), , Name the eye defect he has., , (b), , State two reasons why this defect is caused?, , (c), , Compute the focal length of his lens., , (d), , Which lens is used by selwyn?, , Ans. , , ∴ focal length = – 0.5 m, , Ans. , (a), , Presbyopia., , (b), , The focusing power of the eye lens decreases with, age. The muscles near the lens lose their ability to, change the focal length of the lens. The near point, of the lens shifts further than the eye., , (c), , P = + 3D, f =, , (a), , Hypermetropia., , (b), , (i), , Curvature of the cornea and the eye lens, decreases so that, the converging power of, the lens becomes less.., , 1, p, , 1, 3, = 0.33 m, =, , ∴ focal length = 0.33 m

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Master Key Science and Technology Part - 1 (Std. X), , 150, , (5), , (iv) In a ray diagram,, real image is, formed on opposite, side of lens., , Screen, Convex lens, , Candle, , (5), , 2F1, , F1, , 0, , F2, , (iv) In a ray diagram,, virtual image is, formed on same, side of lens., , B, , 2F2, , Fig 7.23, , A, , (a), , What changes in the position of the candle should, be made to obtain a virtual and enlarged image?, , (b), , What is nature of image formed if a concave lens, is used instead of a convex lens?, , (i), , Label A and B in the diagram., , (ii), , Explain the functions of A., , (c), , State any one point of difference between a real, and virtual image., , Ans. , , Ans. , (a), , To obtain a virtual and enlarged image, the candle, (object) should be placed between F1 and O., , (b), , If a concave lens is used, the image will be virtual, and erect., , (c), , Real image and Virtual image, , (i), , (ii), , Real Image, When the refracted, rays actually meet,, the real image is, formed., , (i), , Real image can be, (ii), taken on the screen., , (iii) Real image is, always inverted., , Fig 7.24, , (i), , A – Cornea, B – ciliary muscles., , (ii), , Maximum refraction of light occurs on the cornea., , (5), , Name of the two lenses A and B, , Virtual Image, When the refracted, rays appear to, meet, virtual image, is formed., Virtual image can, not be taken on the, screen., , A, , B, Fig 7.25, , Ans. , (i), , Lens A – positive meniscus., , (ii), , Lens B – negative meniscus., , (iii) Virtual image, is always erect, (upright)., , vvv

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Lenses, , 151, , ASSIGNMENT - 7, Time : 1 Hr., , , Marks : 20, , Q.1. (A) Fill in the blanks:, , (3), , (1), , Myopia can be corrected by using spectacle of, , (2), , The tendency of pupil to adjust the opening for light is called, , (3), , In simple microscope, , , , (1), , lens., , lens is used., , Q.1. (B) Choose and write the correct option:, , (2), , Farsightedness can be corrected by using, (a) cylindrical, , lens., , (b) concave, , (c) diverging, , (2) The perception of dim light is concerned with, (a) Rod, , (1), , ., , (b) Cone, , (d) converging, , cells., (c) Amoeboid, , (d) Squamous, , Q.2. Answer the following: (Any 2), , (4), , An object is placed at a distance of 15 cm from a convex lens. If the focal length of the lens is 60 cm,, find the image distance., , (2), , Give scientific reasons: “We cannot clearly see an object kept at a distance less than 25 cm from eye.”, , (3), , Distinguish between farsightedness and nearsightedness., , , , Q.3. Answer the following: (Any 2), , (6), , (1), , Draw a neat and labelled diagram of human eye., , (2), , A 5 cm high object is placed at a distance of 25 cm from a converging lens of focal length of 10 cm., Determine the position, size and type of the image., , (3), , Draw a neat and labelled ray diagram for convex lens., , , , (i) When object is beyond 2F1, , , , Q.4. Answer the following: (Any 1), , (ii) When object is at infinity, (5), , (1), , Explain the working of an astronomical telescope using refraction of light., , (2), , Write a short note on (i) Hypermetropia., , (ii) Myopia, , vvv

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8, , Metallurgy, CONCEPT MAP, Elements, Types, , Metals, rt, , e, op, , Pr, , Physical, , ies, , occur as, , oc, , cu, , ri, , n, , Ores, , Chemical, , Extraction of metal, , Processed for extraction, , Metalloids, , Non - metals, , e.g. Si, Ge, As, , e.g. C, S, P, , Natural state e.g., Gold, Platinum, , Concentration of ores, , Metals of high, reactivity, , Metals of medium, reactivity, , Metals of low, reactivity, , are present as, , Electrolysis of, molten ore, Pure metal, , are reduced, to oxides, , Carbonate ore, Calcination process, , e.g. Na, Al, Mg, , Sulphide ore, , occur as, , Sulphide ore, , Roasting method, converted to, , Heating, , Oxide of metal, , Metal, , Reduction of metal, oxide to metal, , Refining, , Purification of metal, , e.g. Cu, Hg, , e.g. Zn, Fe, , (4) The metals that produce a sound on striking on, hard surface are said to be, ., , MASTER KEY QUESTION SET - 8, Q.1. (A) 1. Fill in the blanks and rewrite the complete, statement :, , (5) The ability of metals to be drawn into wires is, called, ., , (1) , Elements that possess both the properties of, metals and non metals are called, ., , (6) The alkali metals from group 1 such as Lithium,, Sodium, and Potassium are very, ., , (2) , and, metals exist in, liquid state at room temperature., (3) The metallic lustre goes on, exposure to atmospheric oxygen., , (7) , Metal tungsten has the highest melting point, ., , due to, , (8) Non - metal bromine exists in, (152), , (9), , ., , is the hardest natural substance.

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Metallurgy, , (10), , 153, , Metals react with oxygen to form, , (21) Aqua Regia (22) Bauxite (23) Bayer’s process, (24) Calcination (25) Roasting (26) gold, silver, copper, (27) aluminium oxide (28) gangue (29) amalgam, (30) copper carbonate., , ., , (11) The metals; aluminium, iron and zinc do not react, with, but they react with, to form their oxides., (12), , Metals react with Nitric acid to form, ., , (13) , The arrangement of metals in the increasing, or decreasing order of reactivity is called the, ., (14) , liquid., (15), , is a highly corrosive and fuming, , Non - metals are also called, , ., , (16) , Non - metals combine with oxygen to form, ., , Q1., , (A) 2. Select the odd man out:, , (1), , Copper, Gallium, Gold, Silver., , Ans. Gallium: It is a metal in the liquid state and all, others are in solid state., (2), , Silver, Gold, Platinum, Calcium, , Ans. Calcium: Found in combined state,others are in, free state., (3) , Water, Ammonia, Sodium chloride, Hydrogen, chloride., , (17) The compounds formed from two units, namely, cation and anion are called, ., , Ans. Sodium chloride: It is an ionic compound others, are covalent compounds., , (18), , (4), , Ionic compounds are, , in nature., , K2O, Na2O, CaO, Al2O3., , (19) , The Science and Technology regarding the, extraction of metals from ores and their, purification for the use is called, ., , Ans. Al2O3: Others are basic oxides Al2O3 is an, amphoteric oxide., , (20), , Ans. Brittleness: It is the property of non - metals,, others are properties of metals., , is an ore of tin., , (21) A freshly prepared mixture of concentrated HCl, and concentrated HNO3 in the ratio of 3:1 is, known as, ., (22), , Aluminium is extracted from its ore, ., , (23) The conversion of bauxite ore into alumina is, done by a process called, ., (24) In, process carbonate ores are, strongly heated., (25) In, heated., , process sulphide ores are strongly, , (26) , Metals such as, ,, do not react with water at all., , and, , (27) Anodizing is a process of forming a thick layer of, , (5), , (6), , Malleability, Ductility, Brittleness, Lustre., , Graphite, Iodine, Silver, Phosphorous., , Ans. Silver: It is a metal, others are non metals., (7) , Sodium oxide, Zinc oxide, Potassium oxide,, magnesium oxide., Ans. Zinc oxide: It is an amphoteric oxide others are, basic oxides., (8), , Steel, Iron, Copper, Tungsten., , Ans. Steel: It is an alloy, others are metals., (9), , Galvanizing, Tinning, Anodizing, Roasting., , Ans. Roasting: It is a process of metallurgy, Others are, methods to prevent corrosion., (10), , Brass, Bronze, Steel, Antimony., , (28) The impurities present in an ore are called as, ., , Ans. Antimony: It is a metalloid and others are alloys., , (29) If one of the metal is mercury, then the alloy is, known as, , (1) , Metal : Reducing agent : : Non - metal :, , (30) Copper reacts with moist carbon dioxide in the, air and slowly loses its shine to gain a green coat, of, ., , Ans. Oxidising agent - Metals are electron donors, therefore metals are reducing agents. Non metals, accept electrons and so nonmetals are oxidising, agents., , Ans. (1) Metalliods (2) Mercury, Gallium (3) decreasing., (4) Sonorous (5) ductility (6) soft (7) 3422°C., (8) liquid state (9) Diamond (10) metal oxides, (11) water, steam (12) Nitrate salts (13) reactivity, series of metals (14) Aqua Regia (15) electronegative, elements (16) acidic oxides (17) ionic compounds, (18) crystalline (19) Metallurgy (20) Cassiterite, , Q.1, , (A) 3. Find out the correlation:, , (2), Brass : Copper and zinc : : Bronze :, Ans. Copper and Tin - Bronze is an alloy of copper and, Tin., (3), , Aluminium : Bauxite : : Mercury :, , Ans. Cinnabar - The ore of mercury is cinnavar

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154, , (4), , Master Key Science and Technology Part-1 (Std. X), , Cryolite : AlF3.3NaF : : Fluorspar :, , Ans. CaF2 - Flurspar is calcium flouride., (5), , Alumina : Al2O3 : : Sodium aluminate :, , Ans. NaAlO2 - Sodium aluminate is an intermediate in, the extraction of A1., (6) Coating of tin over metal : Tinning : : Coating of, zinc over iron :, , Q.1. (A) 5. State whether the following statements, are true or false, correct the false statements., (1) In the alloy, if one of the metals is mercury then,, it is known as Amalgam., (2) In the process of ‘Kalhaee’ a metal is coated with, Zinc to prevent its corrosion., (3) Nitric acid is a strong oxidizing agent., , Ans. Galvanization - The process of applying a coating, of zinc on Iron is called galvanisation., , (4) Non - metals react with acids to give a salt and, hydrogen gas., , Q.1, , (A) 4. Match the columns:, , *(1), , Substance, , (5) , The elements or compounds which occur, naturally is earth’s crust are known as minerals., , Property, , (1) Potassium bromide (a), , Combustible, , (2) Gold, , (b), , Soluble in water, , (3) Sulphur, , (c), , (4) Neon, , (d), , No chemical, reaction, High ductility, , Ans. (1 – b), (2 – d), (3 – a), (4 – c), *(2), , Column A, , Column B, , (6) Ionic compounds have low melting points., (7), , Alloys are resistant to corrosion., , (8) Non - metals are good conductors of heat and, electricity., (9) , Metals gain electrons and become negatively, charged ions., (10) Metal iron is more reactive than metal copper., (11) Non - metals combine with oxygen to form basic, oxides., , (1), , Bauxite, , (a), , Mercury, , (2), , Cassiterite, , (b), , Aluminium, , (3), , Cinnabar, , (c), , Tin, , (12) , The minerals from which the metal can be, separated economically are called ores., , Column B, , (13) , Stannic oxide (SnO2) is magnetic and Ferrous, tungstate (FeWO4) is a non - magnetic ingredient., , Ans. (1 – b), (2 – c), (3 – a), (3), , Column A, (1), (2), (3), (4), , (5), , React vigorously, with cold water., React with hot, water, React with steam, , (a), (b), , Silver and, Copper, Iron and Zinc, , (c), , Calcium, , React less, vigorously with, cold water, Do not react with, water, , (d), , Sodium and, Potassium, , (e), , Magnesium, , Ans. (1 – d), (2 – e), (3 – b), (4 – c), (5 – a), (4), , Column A, (1), (2), , (3), , (4), (5), , Wilfley table, method, Hydraulic, separation, method, Magnetic, separation, method, Froth Floatation, method, Leaching, , Column B, (a), , SnO2 and FeWO4, , (b), , CuFeS2, , (c), , Ball mill, , (d), , Ore soaked in a, solution, Law of gravitation, , (e), , Ans. (1 – c), (2 – e), (3 – a), (4 – b), (5 – d), , (14) Ore contains some impurities, these are called, gangue., (15) The process of separating gangue from the ores is, called concentration of ores., (16) , The reactivity of metals increase down to, reactivity series., (17), , Electronic configuration of Aluminium is 2,8,2., , (18) Cryolite (Na3AlF6) and Fluorspar ( CaF2) are used, to increase the melting point of alumina., (19) Electrolysis is the method used to obtain pure, metals from impure metals., (20) , The process of converting sulphide ores into, oxides by heating strongly in excess of air is, called calcination., (21), , Iodine is a lustrous metal., , Ans. , (1) True (2) False. In the process of ‘Kalhaee’ a metal, is coated with Tin to prevent its corrosion (3) True, (4) False. Metals react with acids to give a salt and, hydrogen gas (5) True (6) False. Ionic compounds have, high melting points (7) True (8) False. Non - metals are, bad conductors of heat and electricity (9) False. Metals, lose electrons and become positively charged ions, (10) True (11) False. Non - metals combine with oxygen

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Metallurgy, , 155, , to form acidic oxide (12) True (13) False. Stannic oxide, (SnO2) is a non - magnetic and Ferrous tungstate, (FeWO4) is a magnetic ingredient (14) True (15) True, (16) False. The reactivity of metals decreases down the, reactivity series (17) False. Electronic configuration, of Aluminium (2,8,3) (18) False. Cryolite (Na3AlF6), and Fluorspar (CaF2) are used to lower the melting, point of Alumina (19) True (20) False. The process, of converting sulphide ores into oxides by heating, strongly in air is called roasting (21) False. Iodine is a, lustrous non-metal., , Q.1. (B). Choose and write the correct options., , Q.1, , (A) 6. Name the following :, , *(1), , Alloy of sodium with mercury., , (4), , Ans. Sodium amalgam., , (1), , *(3) The oxide that forms salt and water by reacting, with both acid and base., Ans. Aluminium oxide, Zinc oxide (All Amphoteric, oxide), *(4), , The device used for grinding an ore., , Ans. Ball mill, *(5), *(6), , (b) Antimony, , (c) Germanium , , (d) Aluminium, , has the highest melting point., (a) Tungsten , , (b) Copper, , (c) Iron , , (d) Zinc, , (3), , is the most reactive metal., (a) Potassium , , (b) Magnesium, , (c) Calcium , , (d) Sodium, , is the formula of cuprite., (a) Cu2O, , (5), , (6), , (b) Cu2S, , (a) Copper , , (b) Silver, , (c) Calcium , , (d) Tin, , Metal oxides are generally, , (7) , electricity., , (8), , is a non metal which conducts, , (a) Diamond , , (b) Iodine, , (c) carbon , , (d) Graphite, , is an oxide which is amphoteric., (a) Copper oxide, (b) Magnesium dioxide, , *(7), , (c) Zinc oxide , , An alloy of copper and zinc, , (d) Calcium oxide, , Ans. Brass, , (9) The reactivity of metals with dil HCl in decreasing, order is, ., (a) Mg >Zn > Al > Fe , , Ans. Iron, , (b) Mg > Al > Zn > Fe, , The most lustrous substance., , (c) Fe >Zn > Al > Mg , , Ans. Diamond, *(10) An ore of Mercury., Ans. Cinnabar (HgS), , (d) Fe > Al > Zn > Mg, (10), , *(11) The process used for heating of carbonate ores., Ans. Calcination, *(12) , The metal which is the least reactive in the, reactivity series., Ans. Gold, *(13) The most reactive metal., Ans. Potassium, *(14) Alloy of copper and tin., Ans. Bronze, , (b) Basic, , (d) Both acidic and basic, , Ans. Aqua regia, , *(9), , in nature., , (c) Neither acidic nor basic, , The reagent that dissolves noble metals., , *(8) A metal which does not react with cold water but, reacts with steam., , (c) CuCO3 (d) CuCl2, , Cassiterite is an ore of, , (a) Acidic , , The non - metal having electrical conductivity., , Ans. Graphite, , (a) Silicon , (2), , *(2) , Molecular formula of the common ore of, aluminium., Ans. Al2O3.nH2O, , is not a metalloid., , (11), , Cinnabar is an ore of, , ., , (a) Aluminium , , (b) Sodium, , (c) Iron , , (d) Mercury, , The main constituent of bauxite is, (a) Al2O3 , , (b) Al2(SO4)3, , (c) CaSO4 , , (d) Na3AlF6, , ., , (12) Which method is used for the purification of, more reactive metals?, (a) Chemical reduction (b) Roasting, (c) Calcination, (d) Electrolytic reduction

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156, , Master Key Science and Technology Part-1 (Std. X), , (13) Substance used to decrease the melting point of, alumina in Hall’s process, ., (a) CuSO4 , , (b) Cryolite, , (c) Gypsum , , (d) Limonite, , (14) , Galvanisation is a method of protecting iron, from rusting by coating it with a thin layer of, ., , *(4), , Ans. , The Science and Technology regarding the, extraction of metals from ores and their, purification for the use is called metallurgy., (5), , Ans. The process of conversion of sulphide ores into, oxide by strongly heating in air is called roasting., , (b) Tin, , (6), , (c) Silver , , (d) Zinc, , Ans. The process of conversion of carbonate ores into, oxides by strongly heating in limited supply of, air is called calcination., , (a) Copper oxide, , (b) Iron oxide, , (c) Copper carbonate, , (d) None of the above, , (16) , react with dil.HNO3 to evolve, hydrogen gas., (a) Iron and Copper, (b) Manganese and Magnesium, (d) Aluminium and Magnesium, (17) , Silver articles become black on prolonged, exposure to air. This is due to the formation of, ., (a) Ag3NO3 , , (b) Ag2O, , (c) Ag2S , , (d) Ag2S and Ag3NO3, , (18) , In Tinning a layer of molten, deposited on metals., (a) Zinc , , (b) Iron, , (c) Tin , , (d) Copper, , (7), , is, , (8), , (10), , (11), , *(3), , Gangue, , Ans. Ores contain metal compounds with some of the, impurities like soil, sand, rocky substances etc., These impurities are called gangue., , Anodization, , Tinning, , Ans. The process of giving a protective coating of tin, over other metals is called tinning., , (14), , Ans. , The minerals from which the metals can be, separated economically are called ores., , Electroplating, , Ans. , The process where metals like copper and, aluminium are coated with a thin and strong, layer of their oxides by means of electrolysis is, called anodizing., , Q.2.1. Define the following :, , Ores, , Galvanizing, , Ans. The process where a less reactive metal is coated, on a more reactive metal using electrolysis is, called electroplating., , (12), , Ans. The compounds of metals that occur in nature, along with the impurities are called minerals., , Reduction, , Ans. Galvanizing is a process of giving a thin coating, of zinc on iron or steel to protect them from, corrosion., , Ans. , (1) (d) Aluminium (2) (a) Tungsten (3) (a) Potassium, (4) (a) Cu2O (5) (d) Tin (6) (b) basic (7) (d) Graphite, (8) (c) Zinc oxide (9) (b) Mg > Al > Zn > Fe, (10) (d) Mercury (11) (a) Al2O3 (12) (d) Electrolytic, reduction (13) (b) Cryolite (14) (d) Zinc (15) (c), Copper carbonate (16) (b) Manganese and magnesium, (17) (c) Ag2S (18) (c) Tin., Minerals, , Calcination, , Ans. The conversion of metal oxide into metal is called, reduction., , (9), , (c) Zinc and Manganese, , *(2), , Roasting, , (a) Aluminium , , (15) Copper reacts with moist carbon - dioxide in air, and slowly loses its shine to gain a green coat of, ., , *(1), , Metallurgy, , Alloy, , Ans. The homogeneous mixture formed by mixing a, metal with other metals or non - metals in certain, proportion is called an alloy., (13), , Ionic compounds, , Ans. The compounds formed from two units namely, cation and an anion are called Ionic compounds., Hydraulic Separation method., , Ans. It is the process in which finely grounded, (powdered) ore is washed with a stream of water., The lighter gangue particles are washed away, in the stream of water whereas heavier mineral, particles settle down at the bottom. It is based on, the difference in the density of ore and impurities., (15), , Froth Floatation process, , Ans. This process is used to concentrate some ores, especially sulphide ores. In this process, the finely, powdered ore is mixed with water and a suitable

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Metallurgy, , oil, like pine oil. It is based on the principle that, the mineral particles due to their hydrophobic, property are wetted with oil, whereas gangue, particles get wetted with water due to the, hydrophilic property., (16), , Magnetic Separation, , Ans. It is a process in which magnetic ore is separated, from non - magnetic impurities with the help of, electromagnetic machines., (17), , Patination of Copper, , Ans. CO2 in moist air reacts with the surface of copper, vessel. Copper loses its lustre due to the formation, of greenish layer of copper carbonate CuCO3 on, its surface. This is called Patination of copper., , 157, , (6), , When steam is passed over aluminium., , Ans. , When steam is passed over aluminium,, aluminium oxide and hydrogen gas are formed., 2Al(s) + 3H2O(g) → Al2O3(s) + 3H2(g), Aluminium Steam, Aluminium Hydrogen, oxide, (7) Extraction of copper from its sulphide ore., Ans. 2Cu2S + 3O2 → 2Cu2O + 2SO2↑, , Copper, Oxygen, Copper, Sulphur, sulphide, oxide, dioxide, 2Cu2O + Cu2S → 6Cu + SO2↑, Copper, Copper, Copper, Sulphur, oxide, sulphide, dioxide, , Q.2.2. , Write chemical equation for the following, events:, , (8) When sodium oxide dissolves in water., Ans., Na O, + HO, → 2NaOH, , *(1), , Sodium oxide Water, Sodium hydroxide, (9) Copper reacts with concentrated nitric acid., Ans., , Aluminium came in contact with air., , Ans. , Aluminium combines with oxygen to form, aluminium oxide., 4Al, + 3O2 →, 2 Al2O3, Aluminum Oxygen, Aluminium, Oxide, *(2) Iron filings are dropped in aqueous solution of, copper sulphate. , Ans. , When iron filings are dropped in aqueous solution, of copper sulphate, iron displaces copper from, the solution. The blue colour of copper sulphate, solution fades out., Fe + CuSO4 → FeSO4 +, Cu, Iron, Copper, Iron, Copper, Sulphate, Sulphate, *(3) A reaction was brought about between ferric, oxide and aluminium., Ans. , When Ferric oxide reacts with aluminium,, aluminium oxide and iron are formed. This, reaction is known as “Thermit Reaction”., Fe2O3 + 2Al → 2Fe, + Al2O3 + heat, Ferric Aluminium, Iron, Aluminium, Oxide, Oxide, (4), , Electrolysis of alumina is done., , Ans. Electrolysis of molten mixture of alumina is done, in a steel tank. Electrode reaction, Anode reaction 2O2- → O2 + 4e- (Oxidation), , Cathode reaction Al3+ + 3e- → Al (Reduction), *(5) Zinc oxide is dissolved in dilute hydrochloric, acid., Ans. When zinc oxide dissolves in dilute hydrochloric, acid, zinc chloride and water is formed., ZnO + 2HCl → ZnCl2 + H2O, Zinc, Hydrochloric, Zinc, Water, Acid, Chloride, Oxide, , 2, , (s), , 2, , (l), , (aq), , Cu(s) + 4HNO3(aq) → Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l), Copper conc., Copper, Nitrogen, Water, Nitric, nitrate, dioxide, acid, (10) Copper reacts with dilute nitric acid, Ans. Cu(s) + 8HNO3(aq) → Cu(NO3)2(aq) + 2NO(g) + 4H2O(l), Copper Dilute, Copper, Nitric Water, , nitri acid, nitrate, Oxide, (11) Aluminium oxide reacts with sodium hydroxide., Ans., Al2O3 . 2H2O(s) + 2NaOH(aq) → 2NaAlO2(aq) + 3H2O(l), Aluminium, Sodium, Sodium, Water, oxide, Hydroxide, aluminate, Q.2.3 Answer the following in one or two sentences., (1) Name the categories into which metals can be, classified based on their reactivity., Ans. Based on reactivity, metals can be classified into, three groups:, (i), metals of high reactivity, (ii) metals of moderate reactivity, (iii) metals of low reactivity., (2) , Which metals are available in free state in, nature?, Ans. Less reactive metals like gold, silver and copper, are available in free state in nature., (3) In what form are ores / minerals generally found, in nature?, Ans. Generally, ores / minerals are found in the form, of oxides, sulphides, carbonates and nitrates., (4) Which impurities of bauxite ore are removed by, Bayer’s process?, Ans. Iron oxide (Fe2O3), silica (SiO2) and titanium oxide

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158, , Master Key Science and Technology Part-1 (Std. X), , (TiO2) are the impurities of bauxite ore which are, removed by the Bayer’s process., (5), , On what does the process of metallurgy depend?, , Ans. , The process of metallurgy depends on i) the, nature of the metal to be extracted ii) the type of, impurity present in the ore., (6), , What is concentration of ore?, , Ans. The process of separating gangue from the ore is, called concentration of ore., (7) , Write the cathode reaction in electrolytic, reduction of alumina., Ans. , Cathode reaction in electrolytic reduction of, aluminium., Al3+ + 3e-, , Al(l)., , (8) , Name the two properties on which Froth, floatation method is based., Ans. The Froth floatation method is based on the two, opposite properties, hydrophilic (getting wetted, with water) and hydrophobic (getting wetted, with oil) of the gangue particles and ore particles, respectively., (9), , What is the principle of froth floatation process?, , Ans. The Froth floatation process is based on the, principle of difference in wetting properties of, the ore and the gangue particles with oil and, water respectively., (10) What is the principle of magnetic separation, method?, Ans. Magnetic separation method uses the principle of, difference in the magnetic properties of the ore, and the gangue., (11) , Write an equation for the action of heat on, aluminium hydroxide., Ans. , When aluminium hydroxide is heated upto, 1000oC, it forms alumina and water., 2Al(OH)3 heat Al2O3 + 3H2O., 1000°C, (12) What is the purpose of roasting in metallurgy?, Ans. , In metallurgy, the purpose of roasting is the, conversion of Sulphide ore into oxide by strongly, heating it in air., (13) What acts as cathode in electrolytic reduction of, Al2O3?, Ans. Carbon (graphite) lining inside the Steel tank acts, as a cathode in electrolytic reduction of Al2O3., (14), , What works as a electrolyte in Hall’s cell?, , Ans. The mixture of molten Al2O3, Na3AlF6 and CaF2, works as an electrolyte in Hall’s cell., , (15), , What is reactivity series of metals?, , Ans. The arrangement of metals in an increasing or, decreasing order of reactivity is called reactivity, series of metals., (16), , What is the nature of the oxides of metals?, , Ans. Oxides of metals are basic in nature., (17) , In the extraction of aluminium, name the, process of concentration of Bauxite., Ans. The process used for concentration of Bauxite is, leaching process using either Bayer’s method or, Hall’s method. In both these methods, finally the, concentrated alumina is obtained by calcination., (18), , State the chemical composition of Bauxite., , Ans. In bauxite, Aluminium oxide (Al2O3.nH2O) is 30%, to 70% while the remaining portion is gangue, made of silica,sand,iron oxide etc., (19), , What do you mean by amphoteric oxides?, , Ans. Those oxides of metals which show both acidic, as well as basic behaviour in aqueous solution, are called amphoteric oxides. e.g. Zinc oxide,, Aluminium oxide are amphoteric oxides., (20) Why is hydrogen gas not liberated when metals, (except Mn and Mg) are treated with HNO3?, Ans. HNO3, being a strong oxidizing agent, oxidises, H2 to water (H2O), Hence, hydrogen gas is not, liberated when metals are treated with HNO3., , *(21) W,  hat is the electronic definition of Oxidation, and Reduction? (Can you recall; Textbook Page, no. 102), Ans: Definition in terms of electron transfer:, , , Oxidation means loss of electrons., , , , Reduction means gain of electrons., , *(22) , What are the moderately reactive metals?, (Can you tell; Textbook Page no. 104), Ans: The metals present in the middle of reactivity, series are called as, moderately reactive metals, e.g. Zinc, Iron, Lead, Tin, Copper., *(23) In which form do the moderately reactive metals, occur in nature? (Can you tell; Textbook Page no. 104), Ans: These metals are usually present as Sulphides or, Carbonates in nature., *(24) , What is meant by corrosion? (Can you recall;, Textbook Page no. 106), Ans: The degradation of materials due to reaction with, the environment is called corrosion., *(25) Which measures would you suggest to stop the, corrosion of metallic articles or not to allow the, corrosion to start ? (Can you recall; Textbook Page, no. 106), Ans: , Various methods are used to protect metals

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Metallurgy, , 159, , from corrosion. To stop the corrosion of metallic, articles; they should be kept isolated from a direct, contact with air., , of iron by applying a layer of paint, oil, grease, or varnish on their surface. Thus, we can prevent, rusting of iron windows and doors of our house., , *(26) , What is done so to prevent rusting of iron, windows and iron doors of your house?, (Can you tell; Textbook Page no. 106), , *(27) Which method do we use when we want to, study many things together and at the same, time? (Think about it; Textbook Page no. 93), , Ans:  We can lower the rate of the process of rusting, , Ans: W,  e use classification method if we want to study, many things together at the same time., , Q.2.4. Distinguish between:, (1) Metals and Non - metals. (based on physical properties)., Ans., , Metals, , Non - metals, , (i), , Metals are solid at room temperature with the, exception of mercury and gallium which are in, liquid state at room temperature., , (i), , Non - metals are either in solid or gaseous, state with the exception of bromine which is in, liquid state., , (ii), , Metals have lustre in their pure state and can, be polished to give a highly reflective surface., , (ii), , Non - metals usually do not have lustre except, iodine and diamond., , (iii), , Metals are malleable, i.e they can be, hammered into thin sheets., , (iii), , Non - metals are not malleable., , (iv), , Metals are ductile i.e they can be drawn into, wires., , (iv), , Non - metals are not ductile., , (v), , Metals are good conductors of heat and, electricity., , (v), , Non - metals do not conduct heat and electricity, except graphite as it is good conductor of electricity., , (vi), , Metals have high density., , (vi), , Non - metals have low densities., , (vii) Metals are sonorous., , (vii) Non - metals are not sonorous., , (viii) Metals are hard. Exceptions: Li, Na and K are, soft metals and can be cut easily with a knife., , (viii) Non - metals are not hard. Exception:, Diamond, , (ix), , (ix), , Metals have high melting and boiling points., Exception: Na, K, Hg and Ga are having low, melting and boiling points., , Non - metals have low melting and boiling, points., , (2) Metals and Non - metals. (based on chemical properties), Ans., , Metals, , Non - metals, , (i), , Atoms of metals usually have 1, 2 or 3, electrons in their outermost orbit., , (i), , Atoms of non - metals usually have 4, 5, 6 or 7, electrons in their outermost orbit., , (ii), , Metals lose electrons from their outermost, orbits to form positive ions., , (ii), , Non - Metals gain electrons to form negative, ions., , (iii), , Metals are electropositive, , (iii), , Non - metals are electronegative., , (iv), , Mostly metals form basic oxides., , (iv), , Non - metals form acidic oxides., , (v), , Metals are reducing agents., , (v), , Non - Metals are oxidizing agents., , (vi), , Metals displaces hydrogen from dilute acids, and form the corresponding salts., , (vi), , Non - metals generally do not react with dilute, mineral acids.

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160, , (3), , Master Key Science and Technology Part-1 (Std. X), , Calcination and Roasting, , Ans., , Calcination, , Roasting, , (i), , The process by which carbonate ores are first, converted into oxides is called calcination., , (i), , The process by which sulphide ores are, converted into oxides is called roasting., , (ii), , In this process, the ore is strongly heated in a, limited supply of air., , (ii), , In this process, ore is strongly heated in air., , (iii), , During calcination, CO2 is evolved., , (iii), , During roasting, SO2 is evolved., , (iv), , Example :, , (iv), , Example :, , ZnCO3 → ZnO + CO2↑, (4), , 2ZnS + 3O2 → 2ZnO + 2SO2, , Atoms and Ions, , Ans., , Atoms, , Ions, , (i), , An atom is electrically neutral., , (i), , Ion is an atom or group of atoms which is, either positively charged or negatively charged,, formed due to loss or gain of electrons., , (ii), , An atom has all the properties of the parent, element., , (ii), , An ion does not have properties of the parent, element., , (iii), , An atom may or may not have an independent (iii), existence., , An ion does not have an independent existence., , (iv), , In an atom, the outermost orbit may be unstable. (iv), Eg. : Na (Sodium), (2, 8, 1), , In an ion, the outermost orbit is stable. Eg.:, Na+ (sodium ion) (2, 8), K+ (potassium ion), (2, 8, 8)., , K (Potassium) (2, 8, 8, 1), (5), , Cations and Anions, , Ans., , Cations, , Anions, , (i), , Cations are positively charged ions., , (i), , Anions are negatively charged ions., , (ii), , Generally, metals produce cations. Eg. : Na+, (sodium ion), K+ (potassium ion)., , (ii), , Generally non-metals produce anions. Eg. : Cl, –, (chloride ion), Br (bromide ion)., , (iii), , During electrolysis, cations are discharged at, the cathode., , (iii), , During electrolysis, anions are discharged at, the anode., , Q.2.5. Give Scientific Reasons:, *(1) Lemon or tamarind is used for cleaning copper, vessels turned greenish., Ans., (i) Copper on exposure to moist air combines with, carbon dioxide and a green coloured copper, carbonate salt is formed, tarnishing the copper., (ii) Lime juice and tamarind contain citric acid and, tartaric acid respectively., (iii) These acids react with copper carbonate forming, water - soluble salts, which get washed off easily., (iv) Due to this, the tarnished copper becomes clean, and regains its lustre., , –, , (v) Hence, lemon or tamarind is used for cleaning, copper vessels turned greenish., *(2) , Generally the ionic compounds have high, melting points., Ans., (i) Ionic compounds are formed when an atom of a, combining metal element transfers its electrons, to the atom of another non - metallic element., (ii), , During this process, the atom of the combining, metal after losing electrons, forms a positively, charged ion called as cation and the atom of the, combining non-metal after gaining electrons, forms a negatively charged ion called as anion.

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Metallurgy, , (iii) Thus ionic bonds are formed between the cations, and anions during compound formation., (iv) Due to the presence of cations and anions, there, is a strong electrostatic attraction between the, molecules., (v) Hence to break these inter - molecular bonds a lot, of energy is needed., (vi) , Hence ionic compounds have high melting, points., , 161, , (ii), , The cathode and anodes are made of graphite., , (iii) , During the electrolysis process, aluminium is, deposited on the cathode on passing electric, current., (iv), , Oxygen is liberated at the anode., , (v) Some of this oxygen reacts with the carbon in the, graphite to form carbon - dioxide gas then slowly, burning away the anodes., , *(3) Sodium is always kept in kerosene., , (vi) Thus, the anodes need to be replaced from time, to time during the electrolysis of alumina., , Ans., , (6) Gold and silver are used to make jewellery., , (i) Sodium is a highly reactive metal., , Ans., , (ii) It reacts with the oxygen and moisture present in, the air to form sodium hydroxide and hydrogen, gas., , (i) Gold and silver are precious metals with attractive, lustre., , (iii) Hydrogen which is released, catches fire in, presence of oxygen due to liberation of heat, during this reaction., (iv) Sodium does not react with kerosene and hence,, it sinks in it., (v) Thus, to prevent sodium from coming in contact, with air, it is immersed in kerosene., , (ii) They have low reactivity. They do not react with, atmospheric oxygen, moisture, carbon dioxide., (iii), , They do not react with dilute acids., , (iv), , These metals are malleable and ductile., , (v) Due to these properties gold and silver are used, to make jewellery., (7) Aluminium foils are used to wrap food items., , Ans. Froth floatation is a process to purify sulphide, ores, in presence of pine oil., , Ans. Aluminium foils are used to wrap food items, because aluminium is highly malleable. It can, be beaten into thin sheets. Moreover, it does not, react with food., , , , (8) Aluminium oxide is called an amphoteric oxide., , *(4), , , , Pine oil is used in Froth floatation., , The froth floatation method is based on the, two opposite properties, hydrophilic and, hydrophobic, of the particles. Here the particles, of the metal sulphides, due to their hydrophobic, property, get wetted mainly with pine oil, while, due to the hydrophilic property the gangue, particles get wetted with water., Pressurized air is blown through the mixture, and simultaneously mixture is agitated using an, agitator, due to agitation particles of sulphide, ore form a foam with pine oil. This foam rises to, the surface of water and floats and can be easily, separated., , *(5) Anodes need to be replaced from time to time, during the electrolysis of alumina., Ans., (i) In extracting aluminium from aluminium oxide,, the molten the electrolyte is a liquid mixture, of cryolite (Na3AlF6) Fluorspar (CaF2) and, aluminium oxide. (alumina) (Al2O3)., , Ans., (i) An oxide which shows both acidic as well as, basic properties is called an amphoteric oxide., (ii), , Metal oxides are usually basic in nature., , (iii) , But some metal oxides like aluminium oxide,, zinc oxide, etc. react both with acids as well as, bases to produce salt and water., (iv) Aluminium oxide shows both acidic as well as, basic properties., (v) Hence, aluminium oxide is called an amphoteric, oxide., (9) , Carbonate and Sulphide ores are usually, converted into oxides during the process of, extraction., Ans. , (i) , Reduction of metal oxides is easier and, economical than the reduction of carbonate and, sulphide ores.

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162, , Master Key Science and Technology Part-1 (Std. X), , (ii) So, carbonate and sulphide ores are first converted, to metal oxide and then further reduced to form, metals., , Mg, +, 2HCl, →, MgCl2, +, H2↑, Magnesium Hydrochloric Magnesium Hydrogen, acid, Chloride, gas, , (10) Hydrogen gas is not evolved when metals like, copper, zinc, iron, etc. react with dilute nitric, acid., , *(3) What are the physical properties of metals and, Non - metals? (Can you tell; Textbook Page no. 93), , Ans. , , (i) , Metals are in solid state except Mercury and, Gallium which are in liquid state., , (i) Nitric acid is a strong oxidizing agent. It oxidizes, hydrogen (H2) to water., (ii) Hence, it oxidizes completely to form any of the, nitrogen oxide and water., (iii) Hence, these metals do not produce H2 gas when, they react with concentrated and dilute HNO3., E.g. Cu(s) +, , , 4HNO3(aq) → Cu(NO 3)2 + 2NO2(g), , Copper Conc., Copper, nitric acid nitrate, , + 2 H2O, , Nitrogen Water , dioxide, , Ans. Physical properties of metals:, , (ii) , Metals are hard, have lustre and high melting, and boiling points. They are ductile, malleable,, good conductors of heat and electricity., , Physical properties of non - metals:, (i) Some non - metals are in solid state while some, are in gaseous state., (ii) They do not posses lustre. Non - metals are not, hard, except diamond., , Q.3.1. Explain the following:, , (iii) They have low melting and boiling points., , *(1) When a copper coin is dipped in silver nitrate, solution, a glitter appears on the coin after some, time. Why does this happen? Write the chemical, equation., , (v) Non - metals are bad conductors of heat and, electricity, except graphite., , Ans. Copper being more reactive than silver, displaces, silver from silver nitrate to form element silver, and copper nitrate., Cu(s) + 2AgNO3(aq) → Cu(NO 3)2(aq) + 2Ag(s), Copper Silver, Copper, Silver, , nitrate, nitrate, , The silver displaced gets coated on to the coin,, hence coin gets the silver shine., *(2), ,  he electronic configuration of metal ‘A’ is 2,8,1., T, and that of metal ‘B’ is 2,8,2. Which of the two, metals is more reactive? Write their reaction, with dilute hydrochloric acid., , (iv), , Not malleable and not ductile., , *(4) , Write the electrode reaction for electrolysis of, molten Magnesium chloride and Calcium chloride., (Use your brain power; Textbook Page no. 103), Ans., (i), , For Magnesium chloride, , , , Mg2+ +2e– → Mg(magnesium metal at the cathode), , , , 2Cl– – 2e– → Cl2 (chlorine gas at the anode)., , , Magnesium ions gain electrons (reduction) to, form magnesium atoms. Chloride ions lose, electrons (oxidation) to form chlorine atoms., , , The overall reaction is, , , , MgCl2 (l) → Mg (s) + Cl2 (g) ↑, , Ans. Metal ‘A’ Electronic configuration - (2,8,1),, , (ii), , For Calcium chloride, , Metal ‘B’ Electronic configuration - (2,8,2),, , , , Ca2 + 2e– → Ca (calcium metal at the cathode), , Metal ‘A’ loses its one electron and form positive, ion. Metal ‘B’ also loses 2 electrons and forms, positive ion; but metal ‘A’ is more reactive than, metal ‘B’,as metal ‘B’ takes more energy to, remove two electrons where as metal ‘A’ requires, less energy to lose one electron., , , , 2Cl– – 2e– → Cl2 (chlorine gas at the anode), , Metal ‘A’ is sodium and metal ‘B’ is magnesium., Hence, sodium is more reactive than magnesium., 2 Na + 2HCl → 2NaCl + H2↑, Sodium Hydrochloric Sodium Hydrogen, acid, chloride, gas, , Calcium ions gain electrons (reduction) to form, calcium atoms. Chloride ions lose electrons, (oxidation) to form chlorine atoms., , , Overall reaction is, , , , CaCl2 → Ca(s) +Cl2↑, , *(5) Why do silver articles turn blackish while copper, vessels turn greenish on keeping in air for a long, time? (Think about it; Textbook Page no. 106), Ans. , (i) On exposure to air, Silver articles turn blackish

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Metallurgy, , 163, , after sometime because of the layer of silver, sulphide (Ag2S) formed by the reaction of silver, with hydrogen sulphide in the air., (ii) , Carbon dioxide in moist air reacts with the, surface of copper vessel and these copper vessels, turn greenish because of the formation of copper, carbonate (CuCO3) on its surface., *(6), , Why do pure gold and platinum always glitter?, (Think about it; Textbook Page no. 106), , hydrobromic acid loses hydrogen. In this reaction, Cl2 is the oxidizing agent., , *(10) Have you seen the following things?, , , Old iron bars of buildings, copper vessels not, cleaned for long time, silver ornaments or idols, exposed to air for long time, old abandoned, vehicles fit to be thrown away. (Can you recall;, Textbook Page no. 106), Ans. , , Ans. , (i) Gold and Platinum occur in free state., (ii), , They are noble metals, hence they are least reactive., , (iii) They are not affected by air, water and other, natural factors., (iv), , Hence, pure gold and platinum always glitter., , *(7), , Can we permanently prevent the rusting of an, iron article by applying a layer of paint on its, surface? (Use your brain power; Textbook Page no., 107), , Blackened silver, vessel, , The copper cladded statue, of Liberty made 300 years, ago has turned green., , Ans. , , Rusted shackles, Fig 8.1: Effects of Corrosion, , (i) , No, we cannot protect the articles permanently, from rusting by painting on them., (ii) The method of painting is suitable for short time., (iii) If there is a scratch on the paint on the surface, of the article and if a small surface of the metal, comes in contact with air, the process of rusting, starts below the layer of the paint., *(8), , Ans., , Divide the metals Cu, Zn, Ca, Mg, Fe, Na, Li, into three groups, namely reactive metals,, moderately reactive metals and less reactive, metals., Reactive, metals, , Moderately, reactive, , Na, Li, Ca, , Mg, Zn, Fe, , Less reactive, Cu, , *(9) , In the reaction between chlorine and HBr a, transformation of HBr into Br2 takes place. Can, this transformation be called oxidation? Which, is the oxidant that brings about this oxidation., (Use your brain power; Textbook Page no. 98), Ans. , When chlorine reacts with hydrobromic acid,, bromine is formed., , , Cl2(g) + 2HBr(aq) → 2HCl(aq)+ Br2., , Yes, this transformation is called as oxidation, as, , (i) After keeping for a long time in air or in open, when these above things come in contact with air, and water, it reacts with it and a layer is formed, above its, surface. They get corroded., (ii) Iron bars - They get rusted as coating of Iron, oxide (Fe2O3 . H2O) is formed., , Copper vessels - They react with CO2 in moist, air and a coating of copper carbonate (CuCO3) is, formed., Silver Ornaments - They react with H2S in air and, a layer of silver sulphide (Ag2S) is formed., , Old abandoned vehicles - They are rusted and, coating of Iron oxide (Fe2O3 . H2O) is formed., (11), , An ore on treatment with dilute hydrochloric, acid produces brisk effervescence. What type of, ore is this?, , , What steps will be required to obtain metal, from the enriched ore?, Ans. The ore is carbonate ore., , , Steps required to obtain the metal:, , (i) Calcination: Carbonate ore is heated in limited, supply of air and oxide is obtained., ∆ ZnO + CO ↑, , ZnCO, 3(s), , (s), , 2, , (ii) Reduction with Carbon: Metal oxide is heated, with carbon., , , ZnO(s) + C → Zn + CO ↑

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164, , Master Key Science and Technology Part-1 (Std. X), , (12), ,  metal ‘X’ acquires a green colour coating on, A, outer surface on exposure to air., , *Q.3.2. Draw a neat and labelled diagram., , (i), , I dentify the metal ‘X’ and name the process, responsible for this change., , Ans., , (1), , Magnetic separation method, , (ii) Name and write chemical formula of the green, coating formed on the metal., , Powdered ore, , (iii) , List two important methods to prevent the, process., Ans., (i), , The metal is copper. The process is corrosion., , (ii), , Copper carbonate (CuCO3), , (iii), , (a) It should be coated with tin., , , , (b) It should be mixed with other metals to form, alloys., , (13) How is the method of extraction of metals high, up in the reactivity series different from that for, metals in the middle? Why the same process, cannot be applied for them? Explain by giving, equation, the extraction of sodium., , Nonmagnetic, roller, , Magnetic, roller, , Nonmagnetic, ingredient, , Conveyor belt, , Collector, , Magnetic ingredient, Collector, Fig 8.2: Magnetic separations, (2), , Froth floatation method., , Ans., , Ans., (i) , Metals high up in the series are obtained by, electrolytic reduction because these metals are, strong reducing agents and therefore cannot be, obtained by chemical reduction., (ii) Metals in the middle of series are less reactive, and can be obtained by chemical reduction. The, same process can not be obtained by chemical, reduction. Extraction of sodium is done by, electrolysis of molten Sodium chloride., , , 2NaCl electrolysis 2Na(s) +Cl2↑, (Molten), , *(14) What are the various alloys used in daily life?, Where are those used? (Collect Information;, Textbook Page 108), , Air supply, Froth, , Bubbles, Water and, Pine oil, Agitator, , (3), , Electrolytic reduction of alumina, , Ans., Anode (Graphite), , Steel tank, , (iv) Bronze is used for making medals, sculptures,, statues cooking utensils and coins., *(15) What are the properties that the alloy used for, minting coins should have? (Collect Information;, Textbook Page 108), Ans. , Alloys used for minting coins should be ductile, and malleable., , Cathode, (Graphite lining), Mixture of, alumina,, cryolite and, fluorspar, , (ii) Brass is an alloy of copper and zinc and Bronze is, an alloy of copper and tin., (iii) , Brass is used for decorative articles and for, making musical instruments., , Gangue, , Fig 8.3: Froth floatation method, , Ans., (i) Brass and Bronze are the two alloys which we use, in our daily life., , Concentrated, sulphide ore, , Molten, aluminium, , Outlet for molten, aluminium, , Fig 8.4: Extraction of aluminium

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Metallurgy, , (4), , 165, , Hydraulic separation method., , (iii) The filtrate is warmed and neutralised by passing, carbon dioxide gas through it., , Ans., , (iv) , This results in the precipitation of aluminium, hydroxide., , powdered ore, , gangue, ore, suspension, water, , concentrated ore, Fig 8.5: Hydraulic separation, , Q.3.3. Complete the following statement using every, given options., , , Al2O3.2H2O(s) + Na2CO3(aq) → 2NaAlO2 (aq) + CO2 (g) + 2H2O(l), , (vi), , 2NaAlO2(aq) + 3H2O(l) + CO2(g) → 2Al(OH)3↓ + Na2CO3, , (vii) , The precipitate of Al(OH)3 obtained in Hall’s, process is filtered, washed, dried and then, calcinated by heating at 1000°C to obtain alumina., (viii) 2Al(OH)3→Al2O3 + 3H2O., (4) Heating the aluminium ore with concentrated, caustic soda., Ans. , Aluminium oxide being amphoteric in nature, it, reacts with the concentrated solution of sodium, hydroxide (caustic soda) at 140°C to 150°C under, high pressure for 2 to 8 hours in a digester to, form water soluble sodium aluminate., , During the extraction of aluminium, , (1) Ingredients and gangue in bauxite., Ans. , Aluminium is extracted from its ore Bauxite, (Al2O3.nH2O)., Bauxite contins 30% to 70% of Al2O3 and remaining, part is gangue. It is made up of sand,silica,iron, oxide etc. Silica (SiO2), ferric oxide (Fe2O3) and, titanium oxide (TiO2) are the impurities present, in bauxite., (2), , (v), , Use of leaching during the concentration of ore., , Ans. , (i) , Leaching is a method in which ore is soaked in a, certain solution for a long time., , , , Q.3.4. (1) Study the diagram and answer the questions:, , ↑, e–, , (iv) During the concentration of ore, separation of, impurities is done by leaching process using, either Bayer’s method or Hall’s method., (3) Chemical reaction of transformation of bauxite, into alumina by Hall’s process., Ans. , (i) , In the Hall’s process, the ore is powdered and, then leached by heating with aqueous sodium, carbonate to form water soluble sodium, aluminate., (ii) Then the insoluble impurities are filtered out., , Cell, , e–, ↓, , Cathode, Aluminium, spoon, , Anode, silver, plate, , (ii) The ore dissolves in that solution due to a specific, chemical reaction., (iii) The gangue does not react and therefore does not, dissolve in that solution., , Al2O3.2H2O(s) + 2NaOH(aq) → 2NaAlO2 (aq) + 3H2O(l), Aluminium, Sodium, Water, Sodium, oxide, hydroxide, aluminate, , Fig 8.6: Electroplating, (a), , What does the diagram represent?, , Ans. , The above diagram represents the process of, Electroplating of Silver., (b), , What can you say about the reactivity of metals, at anode and cathode., , Ans. , The reactivity of metal at anode is less than the, reactivity of metal at cathode., (c), , Write the reactions occuring at cathode and, anode., , Ans. , The electrode reaction are as follows:, Anode reaction - Ag → Ag+ + e–, Cathode reaction - Ag+ + e– → Ag

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166, , Master Key Science and Technology Part-1 (Std. X), , (2), , Refer the following diagram which shows, reaction of a metal with water and answer the, following questions., , (b) Name the technique used to coat the aluminium, article., , (a), , Name two metals which do not react in the, above experiment., , (c) Name the anode and electrolyte used in this, technique., , Ans. The two metals which do not react in the above, experiment are Gold (Au) and Platinum (Pt.), (b), , What are the properties of the gas produced in, the above experiment?, , Ans. The properties of hydrogen gas are as follows:, (i), , It is colourless, odourless and tasteless gas., , (ii), , It is lighter than air., , (iii), , It is insoluble in water., , (iv), , It is combustible gas but does not support, combustion., , (v), , Hydrogen burns with a pop sound in oxygen with, a pale blue flame forming water., , (c), , Write a balanced chemical equation for the, reaction of any one metal in the above process., , Ans. When Aluminium metal reacts with steam it forms, Aluminium oxide and hydrogen gas., , , 2Al(s), , + 3H2O(g) → Al2O3(s), , + 3H2(g), , Aluminium Water Aluminium Hydrogen, oxide, Q.4.1. Read the passage and answer the following, question:, , , (1), , Anodising is a process of forming a thick, oxide layer of aluminium. Aluminium, develops a thin oxide layer when exposed, to air. This aluminium oxide coat makes it, resistant to further corrosion. The resistance, can be improved by making the oxide layer, thicker. In this technique aluminium article, is used as an anode. Electrolyte used is dilute, sulphuric acid. The anode reaction results, in formation of a black coloured thin film of, aluminium oxide on the surface of anode. By, putting appropriate dyes in the electrolytic, bath coloured surface with decorative finish, can be achieved. Kitchen articles like anodised, pressure cooker, anodised pans and also, frames of sliding windows are the applications, of anodising techniques., , (a) How can we make an aluminium article resistant, to corrosion?, Ans. If an aluminium article is coated with aluminium, oxide, it becomes resistant to the corrosion., , Ans. The technique used is ‘Anodising’., , Ans. The aluminium article to be coated is used as, anode, while dilute sulphuric acid is used as an, electrolyte in this technique., (d) How can we use this technique to obtain articles, for decorative purpose?, Ans. By adding appropriate dyes in the electrolytic, bath, we can obtain decorative finish for the, articles which can be used for decorative purpose., (e) Name any two articles commonly used in day to, day life obtained by using this technique., Ans. , Anodised pressure cookers, pans and sliding, windows are some articles used in day to day, life., Q.4.2. Answer in brief., (1), , State any four properties of Ionic compounds., , Ans., (i) Ionic compounds are solids and are hard due to, strong force of attraction between positive and, negative ions., (ii) They are generally brittle and break into pieces, when pressure is applied., (iii) Ionic compounds have high melting and boiling, points, as a considerable amount of energy is, required to break the strong inter molecular, forces of attraction., (iv) They are generally soluble in water and insoluble, in solvents such as kerosene, petrol etc., (v) Ionic compounds do not conduct electricity in, solid state, due to their rigid structure. They, conduct electricity in molten state or in solution, in solvent like water., (2), , Describe Bayer’s Process., , Ans., (i) During extraction of aluminium from bauxite, the, bauxite ore is concentrated by Bayer’s Process., (ii) During concentration, bauxite is converted into, alumina., (iii) Crude bauxite contains impurities like iron oxide, and silica which are removed during Bayer’s, process., (iv) The ore is first ground in a ball mill. Then it is, leached by heating it with concentrated caustic

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Metallurgy, , soda solution under high pressure for 2 to 8 hours, at 140 °C to 150 °C in a tank called digester., (v) , Aluminium oxide (Al2O3) being amphoteric, dissolves in aqueous NaOH to form water soluble, Sodium aluminate (NaAlO2)., , 167, , (5), , How can corrosion of metals be prevented?, , Ans. , Corrosion of metals can be prevented by coating, the metal with something which does not react, with moisture and oxygen., , (vi) , The impurities like Fe2O3 do not dissolve in, aqueous NaOH and are removed by filteration., , (i) To fix a layer of any substance on the metal surface, so that the contact of the metal with moisture and, oxygen in the air is prevented and no reaction, would occur between them., , (vii) , Silica from the gangue dissolves in aqueous, NaOH forming water soluble sodium silicate, , (ii) Metals can be coated with paints , oils, grease or, varnish., , (viii) Sodium aluminate is diluted and cooled to 50°C, and hydrolyzed to give aluminium hydroxide as, precipitate., , (iii) Another method of preventing corrosion is done, by coating a corrosive metal with a non - corrosive, metal., , , , , , , , Al2O3.2H2O(s) + 2NaOH(aq) → 2NaAlO2 (aq) + 3H2O(l), , NaAlO2 + 2H2O → NaOH + Al(OH)3, , The precipitate of Al(OH)3 obtained is filtered,, washed,dried and then calcinated by heating at, 1000oc to obtain alumina., , , 2Al(OH)3 → Al2O3+3H2O., , (3), , Describe Hall’s Process., , Ans. Refer to answer Q.3.3 (3), (4) Describe the process of electrolytic reduction of, Alumina with the help of a diagram., (For diagram, refer fig 8.4), Ans. , (i) During the process of electrolytic reduction of, aluminium, a molten mixture of pure alumina, (m.p. is > 2000 °C) is electrolysed in a steel tank., (ii) This tank is lined inside with carbon graphite, which acts as cathode and a set of graphite rods, dipped in the molten electrolyte acts as anode., (iii) Cryolite (Na3AlF6) and Fluorspar (CaF2) are also, added to the mixture to lower the melting point, to about 1000 °C. Current is passed through, the electrolyte and aluminium is formed at the, cathode., (iv) The molten aluminium sinks to the bottom of, the tank as it is heavier from there it is removed, periodically., , This can be done by following methods:, , (a) Galvanizing : It is done by coating the metal, with a thin coating of Zinc on iron or steel., (b) Tinning is a process where a layer of molten, tin is coated over another metal e.g. copper,, brass, etc., (c) Electroplating :, In this method, a less reactive metal is coated on, a more reactive metal by electrolysis. e.g. silver, plated spoon., , , (d) Alloying :, , A homogeneous mixture of a metal with other, metals or non-metals in certain proportion is, called an alloy. This prevents corrosion., (e) Anodizing :, It is done to prevent corrosion of aluminium., Aluminium is coated with a strong film of its, oxide electrically., , +Power –, , Anode, , source, , Power, supply, , (v) Oxygen is liberated at the anode., (vi) The reaction occuring at the cathode and anode, are:, , , Cathode reaction: Al3+ + 3e– → Al(l) (Reduction), , , , Anode reaction: 2O2– + 4e– → O2 + 4e– (Oxidation), , (vii) The oxygen liberated reacts with the carbon anode, to form carbon dioxide gas., (viii) As a result of this oxidation, carbon anode have to, be replaced at intervals., , Cathode, Fig 8.7: Anodization, (6) , Describe the separation methods which are, based on gravitation., Ans. , Separation method based on gravitation - The, heavy particles of ores can be easily separated, from the light particles of gangue by the

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168, , (i), , Master Key Science and Technology Part-1 (Std. X), , gravitational method. The processes to carry out, this separation are as follows:, , (c) One of the rollers is nonmagnetic while the other, is electromagnetic., , Wilfley table method:, , (d) The conveyor belt moving around the rollers is, (non magnetic) made up of leather or brass., , (a) , In this method of separation, the Wilfley table is, made by fixing narrow and thin wooden riffles, on inclined surface., (b) The table is kept vibrating continuously., (c) Powdered ore obtained from lumps of the ore, using ball mill is poured on the table and a stream, of water is also released from the upper side., (d) As a result, the lighter gangue particles are carried, away along with the flowing water, while the, heavier particles in which proportion of minerals, is more and proportion of gangue is less, are, blocked by wooden riffles and get collected on, the slits between them., (ii) Hydraulic separation method:, (a) The hydraulic separation method is based on the, working of a mill., (b) There is a tapering vessel similar to that used in a, grinding mill., (c) It opens in a tank-like container that is tapering, on the lower side., (d) The tank has an outlet for water on the upper, side and a water inlet on the lower side., (e) Finely ground ore is released in the tank., (f) A forceful jet of water is introduced in the tank, from the lower side., (g) Gangue particles are lighter and therefore they, flow out along with the water jet from the outlet, on the upper side of the tank and get collected, separately., (h) At the same time the heavy particles of the ore, are collected at the bottom from the lower side of, the tank., (i) , In short, this method is based on the law of, gravitation, wherein particles of the same size are, separated by their weight with the help of water., , (For diagram, refer fig 8.5), (7) , Describe some general, concentration of ore., , methods, , for, , the, , (i) Magnetic Separation method:, requires, , an, , (f) , Two collector vessels are placed below the, magnetic roller., (g) The particles of the nonmagnetic part in the ore, are not attracted towards the magnetic roller., (h) Therefore, they are carried further along the belt, and fall in the collector vessel placed away from, the magnetic roller., (i) At the same time, the particles of the magnetic, ingredients of the ore stick to the magnetic roller, and therefore fall in the collector vessel near the, magnetic roller., (j) , In this way the magnetic and nonmagnetic, ingredients in the ore can be separated depending, on their magnetic nature., , , For example, cassiterite is a tin ore., , (k) It contains mainly the nonmagnetic ingredient, stannic oxide (SnO2) and the magnetic ingredient, ferrous tungstate (FeWO4)., (l) , These are separated by the electromagnetic, method., , , (For diagram, refer fig 8.2), , (ii), , Froth floatation method:, , (a) The froth floatation method is based on the, two opposite properties, hydrophilic and, hydrophobic, of the particles., (b) Here the particles of the metal sulphides, due to, their hydrophobic property, get wetted mainly, with oil, while due to the hydrophilic property, the gangue particles get wetted with water., (c) , By using these properties certain ores are, concentrated by froth floatation method., (d) In this method, the finely ground ore is put into a, big tank containing ample amount of water., (e) Certain vegetable oil such as pine oil, eucalyptus, oil, is added in the water for the formation of, froth., , Ans., (a) , This method, machine., , (e) The powdered ore is poured on the conveyor belt, near the nonmagnetic roller., , electromagnetic, , (b) The main parts of this machine are two types, of iron rollers and the conveyor belt moving, continuously around them., , (f) Pressurised air is blown through the water., (g) There is an agitator rotating around its axis in the, centre of the floatation tank. The agitator is used, as per the requirement. Bubbles are formed due, to the blown air.

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Metallurgy, , (h) Due to agitation, a foam is formed from oil, water, and air bubbles together. Due to the agitating., This foam rises to the surface of water and floats., That is why this method is called froth floatation, process., , 169, , Ans. For labelled diagram Refer to fig 8.2, The powdered ore is poured on the conveyor, belt near the nonmagnetic roller. Two collector, vessels are placed below the magnetic roller., The particles of the nonmagnetic part in the ore, are not attracted towards the magnetic roller., Therefore, they are carried further along the belt, and fall in the collector vessel placed away from, the magnetic roller. At the same time the particles, of the magnetic ingredients of the ore stick to the, magnetic roller and therefore fall in the collector, vessel near the magnetic roller., , (i) Particles of certain sulphide ore float with the, foam on water as they preferentially get wetted, by the oil. For example, this method is used for, the concentration of zinc blend (ZnS) and copper, pyrite (CuFeS2), , , (For diagram, refer fig 8.3), , (iii) Leaching:, (a) The first step in the extraction of the metals like, aluminium, gold silver from their ores is the, method of leaching., , (2), , (b) , In this method the ore is soaked in a certain, solution for long time., , Ans. For labelled diagram Refer to fig 8.2, , (c) The ore dissolves in that solution due to a specific, chemical reaction., , Ans. The above diagram indicates the process of Magnetic, separation method of concentration of ores., , (d) The gangue does not react and therefore does not, dissolve in that solution., , (b), , (e) So it can be separated. For example, concentration, of bauxite, the aluminium ore is done by leaching, method., (f) , Here, bauxite is soaked in aqueous NaOH or, aqueous Na2CO3 which dissolves the main, ingredient alumina in it., , (a), , Redraw the given diagram and label the parts., Complete the paragraph with the words given, in the bracket:, , What does the above diagram indicates?, , Name an ore which is concentrated by this, method and also write the two ingredients with, molecular formulae, present in this ore., , Ans. The ore which is concentrated by this method is, cassiterite., , , Cassiterite is a ore. It contains mainly the, nonmagnetic ingredient stannic oxide (SnO2), and the Magnetic ingredient ferrous tungstate, (FeWo4)., , (3), , Redraw the given diagram and label the parts., Complete the paragraph with the words given, in the bracket:, , Q.4.3. Solve the following questions:, (1), , Redraw the given diagram and label it correctly., Answer the questions given below. (For diagram, refer Q.4.3 (1)), , (magnetic roller, particles of the magnetic, ingredients, away, near, conveyor, belt,, nonmagnetic part), , (gangue particles, hydrophobic, oil, water, metal, sulphides, hydrophillic), , Ans. , , Ans. For labelled diagram Refer to fig 8.3, , , The froth floatation method is based on the, two opposite properties, hydrophilic and, hydrophobic, of the particles. Here the particles

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170, , Master Key Science and Technology Part-1 (Std. X), , of the metal sulphides, due to their hydrophobic, property, get wetted mainly with oil, while due, to the hydrophilic property the gangue particles, get wetted with water. By using these properties, certain ores are concentrated by froth floatation, method., (4), , Redraw the given diagram and label it correctly., Answer the questions given below: (For diagram, refer Q.4.3 (3)), , In hydraulic separation method, the finely, ground ore is released in the tank. A forceful jet, of water is introduced in the tank from the lower, side. Gangue particles are lighter and therefore, they flow out along with the water jet from the, outlet on the upper side of the tank and get, collected separately. At the same time the heavy, particles of the ore are collected at the bottom, from the lower side of the tank., (6), , Ans. For labelled diagram Refer to fig 8.3, (a), , What does the above diagram indicates?, , Ans. The above diagram indicates the process of Froth, floatation method of concentration of ores., (b), , Name two ores along with their molecular, formula which are concentrated by this method., , Ans. The two ores concentrated by this method are, (i) Zinc blende - Zns (ii) Copper pyrite - CuFeS2., , Redraw the given diagram and label it correctly., Answer the questions given below: (For diagram, refer Q.4.3 (5)), , Ans. For labelled diagram Refer to fig 8.5, (a), , What does the above diagram indicates?, , Ans. The above diagram indicates the process of, Hydraulic separation method of concentration of, ores., (b), , Explain the law on which this method is based., , Ans. The use of pine oil in this process is as follows:, , Ans. This method is based on the law of gravitation,, where in particles of the same size are separated, by their weight with the help of water., , (i), , Pine oil is hydrophobic in nature and also it, attracts the particles of sulphide ore., , (7), , (ii), , Thus the particles of sulphide ore get wetted by, pine oil., , (iii), , Due to air supply and agitation, pine oil forms a, froth which floats on the surface of water. Thus, the particles of sulphide ore are separated from, water., , (iv), , The gangue particles get wetted by water and are, removed from the bottom., , (5), , Redraw the given diagram and label the parts., Complete the paragraph with the words given, in the bracket:, , (c), , What is the use of pine oil in this process?, , Redraw the given diagram and label the parts., Completer the paragraph with the words given, in the bracket:, (heavier, ball mill, lighter, slits, inclined, stream, of water, vibrating), , (heavy particles, Gangue particles, bottom, lower,, lighter, upper), , , Ans. For labelled diagram Refer to fig 8.8, , Ans. For labelled diagram Refer to fig 8.5, , In the method of separation, Wilfley table is, made by fixing narrow and thin wooden riffles, on inclined surface. The table is kept vibrating, continuously. Powdered ore obtained from lumps, of the ore using ball mill is poured on the table,, a stream of water is also released from the upper, side. As a result, the lighter gangue particles, are carried away along with the flowing water,, while the heavier particles in which proportion

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Metallurgy, , 171, , of minerals is more and proportion of gangue is, less, are blocked by the wooden riffles and get, collected on the slits between them., (8), , Redraw the given diagram and label it correctily., Answer the questions given below: (For diagram, refer Q.4.3 (7)), , Ans. For labelled diagram Refer to fig 8.8, (a), , What does the above diagram indicates?, , Ans. The above diagram indicates the process of, Wifley table method of concentration of ores., (b), , What is the result of gangue particles and ore, particles in this process., , Ans. The result of this process is that the lighter, gangue particles are carried away along with, the flowing water, while the heavier particles, of ore are blocked by the wooden riffles and get, collected on the slits between them., (9), , Answer the questions based on the given, diagram., Label the diagram and answer the given, questions., , Ans. For labelled diagram Refer to fig 8.4, (a), , What is the importance of cryolite and fluorspar?, , Ans. Cryolite (Na3AIF6) and fluorspar (CaF2) are added, in the mixture to lower its melting point upto, 1000 0C., (b) Write the anode and cathode reactions., Ans. Anode reaction : 2O2– → O2 + 4e– (Oxidation), Cathode reaction : Al3 + 3e– → Al(l) (Reduction), (10), , Redraw the given diagram and label it correctly., Answer the questions given below: (For diagram, refer Q.4.3 (9)), , Ans. For labelled diagram Refer to fig 8.4, (a), What does the above diagram indicate?, Ans. The above diagram indicates the process of, Electrolytic reduction of alumina in Extraction of, aluminium., (b), , In the above process, name the two compounds, along with their molecular formula which are, added in the mixture to lower its melting point., , Ans. The two compounds cryolite (Na3AIF6) and, Fluorspar (CaF2) are added in the mixture to, lower its melting point up to 1000 0C, (c), , Give the cathode and anode reactions?, , Ans. The electrode reactions are as shown below., –, , –, , Anode reaction : 2O2 → O2 + 4e (Oxidation), +, , –, , Cathode reaction : Al3 + 3e → Al(l) (Reduction), , vvv

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172, , Master Key Science and Technology Part-1 (Std. X), , ASSIGNMENT - 8, Time : 1 Hr., , , Q.1. (A) Answer the following questions:, , (1), , , (3), , Fill in the blank:, (i) The metals that produce a sound on striking on hard surface are said to be, , (2), , (i) K2O, Na2O, CaO, Al2O3, , (3), , Find out the correlation, Brass : Copper and Zinc :: Bronze :, , , , Q.1. (B) Choose and write the correct options:, (1), , , , (b) Iron, , (c) Tungsten, , (d), , Zinc, , (b) Sodium, , (c) Iron, , (d), , Mercury, , Cinnabar is an ore of, , (a) Aluminium, , , Q.2. Answer the following: (Any 2), , (4), , (1), , Distinguish between metals and non-metals., , (2), , Write the balanced chemical equation for the reaction., , , (3), , , (2), , has the highest melting point., , (a) Copper, (2), , ., , Find the odd man out:, , , , , Marks : 20, , When copper reacts with concentracted nitric acid., How can corrosion of metals be prevented?, Q.3. Answer the following: (Any 2), , (6), , (1), , Describe Bayer's Process., , (2), , The electronic configuration of metal 'A' is (2, 8, 1), and that of metal 'B' is (2, 8, 2). Which of these two, metals is more reactive? Write their reaction with dilute hydrochloric acid., , (3), , Explain magnetic separation method., , , (1), , Q.4. Answer the following: (Any 1), , (5), , Give scientific reasons:, , , , (i) Lemon or tamarind is used for cleaning copper vessels turned greenish., , , , (ii) Gold and silver are used to make jewellery., vvv, Explain electrolytic reduction of Alumina with the help of a diagram., , (2), , vvv

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9, , Carbon Compounds, CONCEPT MAP, Living things, , are, , Substances in Nature, are, Organic, (all containing, carbon ), , Non living, are, are, can be, converted to, , are, , CO,, CO2,, Na2CO3, NaHCO3, Air, , Hydrocarbons, , Straight chain and, branched chain, are classified, as, , Saturated, , –, –, , e.g. Ethane (C2H6), H, , H, , H, , C, , C H, , H, , H, , Unsaturated, , Undergo, substitution, reaction, , (Alkenes), (– C C –), , (Alkynes), (– C C –), , e.g. Ethene, (C2H4), H H, , e.g. Ethyne, (C2H2), (H– C C –H), , C, , C, , H, , H, , not containing, carbon, O2, Fe,, sand, water., , Closed chain or Cyclic or, Ring compounds, are classified, as, , Shading (Alkanes), (– C –), , Inorganic, (may or may not, contain carbon), containing, carbon, , Alicyclic hydrocarbon, e.g. Cyclohexane, (C6H12), H, , Aromatic hydrocarbon, e.g. Benzene, (C6H6), H, , H, , C, H, C, H C, H, C, C, H, C, H, H, , H, H, H, H, , C, , H, , C, , C, , H, , C, , C H, , Undergo addition, reaction, , MASTER KEY QUESTION SET - 9, Q.1 , (A) 1. Fill in the blanks and rewrite the, completed sentence, (1), , he essential element in all the organic, T, compounds is, ., , (2), ,  he organic compounds having double or triple, T, bonds in them are termed as, ., , (3), ,  he chemical bonds in carbon compounds do not, T, produce, ., , H, , C, H, , (4), ,  he chemical bond formed by sharing of two, T, valence electrons between the two atoms is called, ., , (5), , , are known as parent organic, compounds., , (6), ,  aturated hydrocarbons have, S, only., , (7), , The general formula for alkanes is, , (8), ,  ydrocarbons necessarily contain, H, and, ., , bonds, ., , (173), , Third Proof : 09.02.2019, , D:\Dipa\2018\Books\Masker Key\X\Science\Xth_Science_Chapter 9_ Xth_Science_Chapter_9_Revised

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Master Key Science and Technology Part-1 (Std. X), , 174, , (9), , Covalent compounds are generally soluble in, , solvents., , (10), , he molecular masses of carbon compounds, T, range upto, ., , (11), ,  he property of carbon atom to form strong, T, covalent bond with itself and results in formation, of big molecules, is called, ., , (12), , Being, one carbon atom can form, bonds with four other atoms., , (13), , ydrocarbons having identical molecular, H, formula but different structures are called, ., , (14), , The molecular formula of cyclohexane is, , ., , (15), , , , (16), , The functional group , , (17), (18), , Q.1 (A) 2. Match the columns:, (1), , (2) Sodium ethoxide, (3) Acetic acid, (4) Cyclohexane, , thanol gets oxidized by alkaline Potassium, E, permanganate to form, .,  he reaction in which the place of one type of, T, atom/ group in a reactant is taken by another, atom/ group of atoms is called, ., , (20), , Ethanoic acid is commonly called, , (21), , Ethanol is also called, , *(2), , (22), ,  uring winter in cold countries ethanoic acid, D, freezes at room temperature itself and looks like, ice. Therefore, it is named, ., , (3) C2H5OH, (4) C3H6, (3), , oiling point of ethanol is, B, freezing point is, ., , (26), ,  olecular formula for the homologous series of, M, alkynes is, ., , (27), ,  he IUPAC name of the salt is Sodium ethanoate, T, while its common name is, ., , and, , Ans. , (1) Carbon (2) Unsaturated hydrocarbon (3) ions, (4) Covalent bond (5) Hydrocarbons (6) single, (7) CnH2n+2 (8) Hydrogen and Carbon (9) organic, (10) 1012 (11) Catenation (12) tetravalent (13) isomers, , Third Proof : 09.02.2019, , (c) C6H6, , (d) CH3CH2ONa, , Column B, (a) Unsaturated hydrocarbon, (b) Molecular formula of an, alcohol, (c) Saturated hydrocarbon, (d) Triple bond, , Column A, , Column B, , Name of the, Compound, , Structural Formula, , (1) Ethyl alcohol, , (a) CH - CH - CH - CH, 3, 2, 2, 3, , (2) Acetone, , (b) CH3 - CH2 - CH2 - OH, ––, , (25), , (b) C6H12, , O, (3) Propionic acid (c) CH – C – CH, 3, 3, , he giant carbon molecules formed from, T, hundreds and thousands of atoms are called, .,  macro molecule formed by regular repetition, A, of a small unit is called, ., , (a) CH3COOH, , Ans. (1 – c), (2 – d), (3 – b), (4 – a), , ., , (24), , Column A, (1) C2H6, (2) C2H2, , ., or, , Column B, , Ans. (1 – c), (2 – d), (3 – a), (4 – b), , O, , is called, ., C, Difference between successive members of, homologous series is, ., , Column A, (1) Benzene, , is a cyclic unsaturated hydrocarbon., , (19), , (23), , (14) C6H12 (15) Benzene (16) Ketone (17) –CH2, (18) ethanoic acid (19) substitution reaction, (20) Acetic acid (21) alcohol or sprit (22) glacial acetic, acid (23) macromolecules (24) polymer (25) 78°C,, –114°C (26) CnH2n – 2 (27) sodium acetate, , (4) Ethyne, , =O, (d) CH3 - CH2 - C –, OH, , (5) Propanol, , (e) CH3-CH2OH, , (6) Butane, , (f) CH ≡ CH, , Ans. (1 – e), (2 – c), (3 – d), (4 – f), (5 – b), (6 – a), (4), , Column A, , Column B, , Name of Polymer, , Constituent monomer, , (1) Polyethylene, (2) Polystyrene, (3) Polyvinyl, chloride (PVC), (4) Polyacrylo nitrile, (5) Teflon, (6) Polypropylene, , (a) CF2= CF2, , (b) CH3 – CH=CH2, , (c) CH2 = CH – C = N, (d) Cl – CH = CH2, (e) CH2 = CH2, , (f) C6H5 – CH = CH2, , Ans. (1 – e), (2 – f), (3 – d), (4 – c), (5 – a), (6 – b), , D:\Dipa\2018\Books\Masker Key\X\Science\Xth_Science_Chapter 9_ Xth_Science_Chapter_9_Revised

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Master Key Science and Technology Part-1 (Std. X), , 176, , Q.1. (A) 6. Answer the following questions in one, sentences:, (1), ,  ame the five things which contain carbon, N, essentially., Ans. Food, clothes, medicines, books, wooden table etc., (2),  ow can carbon become stable?, H, Ans. Carbon becomes stable by sharing four electrons, with atoms of other elements or carbon itself., (3), ,  hich organic compounds are named as alkanol, W, in IUPAC system?, Ans. The alcohol with functional group –OH are, named as alkanol in IUPAC system., (4),  hat is glacial acetic acid?, W, Ans. Acetic acid or ethanoic acid when frozen below, 17oC (290 K) solidifies to an ice-like mass, which, is known as glacial acetic acid., (5),  hat is covalent bond?, W, Ans. The chemical bond formed by sharing of two, valence electrons between the two atoms is called, covalent bond., (6), What are isomers?, Ans. Compounds having different structural formula, but having the same molecular formula are called, isomers., (7),  hat is a functional group?, W, Ans. The atom or group of atoms present in the molecule, which determine characteristic properties of an, organic compound is called the functional group., *(8), ,  hat are the types of compounds? (Can you, W, recall; Textbook Page no. 110), Ans. Organic and Inorganic compounds., *(9), , Objects in everyday uses such as foodstuff,, fibers, paper, medicines, wood, fuels are made of, various compounds.,  , Which constituent elements are common in these, compounds? (Can you recall; Textbook Page no. 110), Ans. All these materials are made up of organic, compounds and the constituent element common, in their compounds is carbon., *(10) T,  o which group in the periodic table does the, element carbon belong? Write down the electronic, configuration of carbon and deduce the valency, of carbon. (Can you recall; Textbook Page no. 110), Ans. Carbon belongs to group 14 of the Periodic table., Its atomic number is 6. Electronic configuration is, (2, 4) and valency of carbon is 4., *(11) W,  hat is meant by a chemical bond? (Can you tell;, Textbook Page no. 110), , Third Proof : 09.02.2019, ,  chemical bond is an attraction between atoms,, Ans. A, ions or molecules that enables the formation of, chemical compounds., *(12) W,  hat is the number of chemical bonds that an, atom of an element forms called? (Can you tell;, Textbook Page no. 110), Ans. The number of chemical bonds that an atom of an, element forms is called its valency., *(13) W,  hat are the two important types of chemical, bond? (Can you tell; Textbook Page no. 110), Ans. The two important types of chemical bond are, Ionic or Electrovalent bond and Covalent bond., *(14) W,  ith which bond C atom in CO2 is bonded to each, of the O atoms? (Use your brain power; Textbook, Page no. 112), Ans. As carbon and oxygen share the valence electrons, and fulfill their valency, the bond formed is, covalent bond., *(15) M,  olecular formula of propane is C3H8. From the , Molecular formula draw its structural formula., (Use your brain power; Textbook Page no. 115), Ans. Structural formula : C3H8., H H H, H, , C, , C, , C, , H, , H H H, *(16) Propane (C3H8) is one of the combustible, component of LPG. Write down the reaction for, propane (C3H8). (Use your brain power; Textbook, page no 123), Ans. C3H8 + 5O2 → 3CO2 + 4H2O + heat + light, *(17) , Which is the component of biogas that makes it, useful as fuel. (Can you recall; Textbook Page no 123), Ans. Methane present in biogas makes it useful as fuel., *(18) W,  hich product is formed by the combustion, of elemental carbon? (Can you recall; Textbook, Page no 123), Ans. Carbon dioxide and water are formed by the, combustion of elemental carbon., *(19) I s the biogas combustion reaction endothermic or, exothermic (Can you recall; Textbook Page no 123), Ans. It is an exothermic reaction., *(20) W,  hich one of ethanoic acid and hydrochloric acid, is stronger? (Use your brain power; Textbook Page no, 129), Ans. Hydrochloric acid is stronger than ethanoic acid, (acetic acid)., *(21)  Which indicator paper out of blue litmus paper, and pH paper is useful to distinguish between, ethanoic acid and hydrochloric acid? (Use your, brain power; Textbook Page no 129), , D:\Dipa\2018\Books\Masker Key\X\Science\Xth_Science_Chapter 9_ Xth_Science_Chapter_9_Revised

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Carbon Compounds, , 177, , Ans. pH paper is useful to distinguish between the, ethanoic acid and hydrochloric acid., *(22) , When fat is heated with sodium hydroxide, solution, soap and glycerin are formed. Which, functional groups might be present in fat and, glycerin? (Use your brain; Textbook page no 131), Ans. Ester –COO group, Alcohol- OH, and (–COOH), carboxyl functional groups, are present in fat and, glycerin., *(23) What are the chemical names of the nutrients that, we get from the food stuff, namely cereals, pulses, and meat? (Can you tell; Textbook page no 131), Ans. Carbohydrates, fats and proteins, *(24) , What are the chemical substances that make, cloth, furniture, and elastic objects? (Can you tell;, Textbook page no 131), Ans. Manmade fibres, elastomers., (25) Hydrogen peroxide decomposes on its own by, the following reaction. (Use your brain power;, Textbook Page no. 113), H – O – O – H → 2H-–O – H +O2, From this, what will be your inference about the, strength of O – O, Covalent bond., Ans. The strength of O - O, Covalent bond is weaker., *(26) Tell from the above example whether oxygen has, catenation power or not? (Use your brain power;, Textbook page no. 113), Ans. No, oxygen shows catenation property only in, the formation of ozone., (27) What is the full from of PET?, Ans. PET - Poly ethylene terephthalate., (28), , (2), , (3), , (b) polystyrene, , (c) polyvinyl chloride, , (d) polyacrylo nitrile, , (4), , (5), , Q.1. (B) Choose and write the correct option., (1), The fundamental organic compounds are also, , known as, compounds., (a) parent , (b) father, (c) mother , , (d) daughter, , (c) both, , (d) none of the above, , The, hydrocarbon is also called, marsh gas., (a) ethane , (b) propane, (d) butane, , Ethane with the molecular formula C2H6 has, covalent bonds., (b) seven, , (6), , (c) eight, , (d) nine, ,  hat are the products obtained on complete, W, combustion of hydrocarbons?, (a) CO + H2O , (b) CO2 + H2, (c) CO2 + H2O , , (d) CO + H2, , The functional group of alcohol is, , O, , O, (a) –COOH (b) –OH, (7), , (8), , (9), , (c) H C (d), , ., , C, , thanol is used as an additive to increase, E, the efficiency of petrol such a fuel is called, (a) ethanol , , (b) cananol, , (c) gashol , , (d) methanol, ,  fter the formation of four Covalent bonds,, A, Carbon attains the electronic configuration of, ., (a) Helium , , (b) Neon, , (c) Argon , , (d) Krypton, , Diamond, , , and, ., , graphite, , have, , the, , same, , (a) chemical properties, (b) degree of hardness, (c) electrical conductivity, (d) physical forms, (10), , (c) polyvinyl chloride : It is used to produce, P.V.C. pipes, door mats, tubes and bags in, hospital kits., (d) polyacrylo nitrile : It is used in producing, winter clothing and blankets., , (b) chemical, , (a) six, , Ans. (a) polyethylene : It is used to make carry bags,, and sports wear, (b) polystyrene : It is used to make thermocol, articles., , (a) physical, , (c) methane , , Give uses of the following polymer’s., (a) polyethylene, , A functional group mainly determines the, , properties., , (11), , (12), ,  he reaction in which two molecules react to, T, form a single product is known as, reaction., (a) substitution , , (b) addition, , (c) hydrogenation, , (d) polymerisation, , IUPAC name of CH3-CH3 is, (a) ethene , , (b) ethane, , (c) ethyne , , (d) ethylene, , A saturated hydrocarbon will have suffix, (a) –ene, , (b) –yne, , (c) –ane, , (d) –one

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Master Key Science and Technology Part-1 (Std. X), , 178, , (13), , The valency of carbon is, (a) 2, , (b) 3, , (14), , (15), , (17), , (18), , (d) 6, , is a natural macromolecule., (a) Polythene , , (b) Monosaccharide’s, , (c) Polysaccharides, , (d) Disaccharides, , (b) CO, , (c) H2, , (d) CO2, ,  small unit that repeats regularly to form a, A, polymer., (a) Macromolecule, , (b) Polysaccharides, , (c) Monomer , , (d) Dinomer, , Monomer of polythene is, (a) CH ≡ CH , , (b) CH2=CH2, , (c), , (d) C2H5 – C2H5, , CH3-CH3 , , (b), , Ethanol, , (c) Ester , , (d), , Ethanoic acid, , (19), , is used in nonstick cookware., (a) PVC , , (b) Teflon, , (c) Polystyrene , , (d) Polypropylene, , Ans. , (1) (a) parent (2) (b) chemical (3) (c) methane, (4) (b) seven (5) (c) CO2+H2O (6) (b)–OH, (7) (c) gashol (8) (b) neon (9) (a) chemical properties, (10) (b) addition (11) (b) ethane (12) (c) -ane (13) (c) 4, (14) (c) polysaccharides (15) (d) CO2 (16) (c) monomer, (17) (b) CH2=CH2 (18) (c) Ester (19) (b) Teflon, Q.2.1. Define the following:, *(1) Structural isomerism, Ans. , The phenomenon in which compounds have, different structural formulae but have the same, molecular formula is called structural isomerism., e.g. Structural formula of compound C4H10, , H, , H, , H, , H, , H, , C, , C, , C, , C, , H, , H, , H, , H, , H, H, , H, , H, , H, , C, , C, , C, , H, , e.g. CH3OH. In this example, one hydrogen atom, in methane CH4 is replaced by –OH group. OH is, a functional group. So, OH is a heteroatom., Ans. The hetero atom or group of hetero atoms, present, in a compound due to which the compound, acquires specific chemical properties irrespective, of the length and nature of the carbon chain in, it, are called as functional group. e.g. alcohol, group (-OH). Any hydrocarbon that has – OH, substituent. e.g. methanol (CH3OH), H, H, , OH, , *(5) Alkane, Ans. , Alkanes are chemical compounds that consist, of carbon and hydrogen atoms, so they are also, called hydrocarbons, the chemical structure of, alkanes only consists of single covalent bonds., As they contain only single bonds, and because, of that each carbon atom is saturated with, hydrogen atoms. Alkanes are, therefore referred, to as saturated hydrocarbons., e.g.:Methane Ethane, Pentane, C2H6, , ↓, , H, , H, , C, H, , CH4, , H, , H C, , Ans. The atoms of the elements, substitute one or more, hydrogen atoms in the hydrocarbon chain and, thereby, the tetravalency of carbon is satisfied., The atom of the element which is substitute for, hydrogen is referred as hetero atom., , *(4) Functional group, , , are used for making fragrance and, flavouring agents, (a) Ethers , , H O H, *(3) Hetero atom in a carbon compound, , Gas evolved during fermentation, (a) O2, , (16), , (c) 4, , Each of the covalent bonds contains two electrons, – one from a hydrogen atom and one from the, oxygen atom. Both atoms share the electrons., , H, , H, , C, H, , H, H H, , ↓, , C5H12, , H, , C, , C, , H, , H, , H H, , ↓, , H, , H, , H, , H, , H, , C, , C, , C, , C, , C, , H, , H, , H, , H, , H, , H, , H, *(2) Covalent bond, , *(6) Unsaturated hydrocarbons, , Ans. The chemical bond formed by sharing of two, valence electrons between the two atoms is called, covalent bond., , Ans. Hydrocarbons in which carbon atoms are linked, to each other by double or triple bonds are known, as unsaturated hydrocarbons., , e.g. Covalent bond is formed between oxygen and, each of the hydrogen in a water molecule (H2O)., , Third Proof : 09.02.2019, , e.g. Ethene; (C2H4),, , Ethyne; (C2H2), , D:\Dipa\2018\Books\Masker Key\X\Science\Xth_Science_Chapter 9_ Xth_Science_Chapter_9_Revised

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Carbon Compounds, , H, , C, , ↓, , 179, , C, , H, , ↓, C=C, , H, , H, , H H, *(7) Homopolymer, Ans. When the polymers are formed by repetition of, single monomer., , *(8) Monomer, Ans. A macromolecule formed by regular repetition, of a small unit is called polymer. The small unit, that repeats regularly to form a polymer is called, monomer e.g. Vinyl chloride polymerizes into, polyvinyl chloride of PVC. Glucose polymerises, , , These are called homopolymer. e.g. Polyvinyl, Chloride (PVC), into starch, cellulose, laminarin and glucans., Synthesis of polyethylene is as follows:, H, , H, C=C, , H, , H, +, , H, , H, C=C, , H, , H, +, , H, , H, C=C, , H, , +..., , Polymerisation, , H, , ethylene monomer, *(9), , Reduction:, , Ans. Reduction is the type of reaction in which loss of, oxygen and gain of hydrogen occurs or in terms, of electron transfer, we can define reduction, reactions are the type of reactions where gain of, electrons takes place., e.g: CH3CHO → CH3CH2OH, , , Reduction by gain of hydrogen, , H, , H, , H, , H, , H, , H, , C, , C, , C, , C, , C, , C, , H, , H H H H, Polyethylene, , H, , n, , same functional group in the place of a particular, hydrogen atom in the chains having sequentially, increasing length is called Homologous Series., (14), , Addition reaction, , Ans. When a carbon compound combines with another, compound to form a product that contains all the, atoms in both the reactants is called as Addition, Reaction., , *(10) Oxidant:, , (15), , Ans. , Substances that can give oxygen to other, substances are called oxidants or oxidizing, agents. e.g. Potassium permanganate or, potassium dichromate are commonly used as, oxidizing agents., , Ans. The reaction in which the place of one type of, atom or group in a reactant is taken by another, atom or group of atoms is called Substitution, Reaction., For example CH4+Cl2, , (11) Catenation, Ans. , Carbon has a unique ability to form strong, covalent bonds with other carbon atoms, this, results in formation of big molecules. This, property of carbon is called Catenation power., , (16), , (12), , (17), , Structural Isomerism, , Ans. The phenomenon in which compounds having, different structural formulae but have the same, molecular formula is called Structural Isomerism., (13), , Homologous series, , Ans. The series of compounds formed by joining the, , Substitution reaction, , Sunlight, , CH3 - Cl + HCl, , Aliphatic compounds, , Ans. A group of hydrocarbons that contains carbon, atoms joined together in straight chains, branched, chains or simple rings., Aromatic Compounds, , Ans. , A cyclic compound containing at least one, benzene ring and characterized by the presence, of alternating double bonds within the ring. e.g., Benzene – C6H6

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Master Key Science and Technology Part-1 (Std. X), , 180, , Q.2.2 Differentiate between:, (1), Saturated and unsaturated Hydrocarbons., Ans., , Saturated Hydrocarbons, , Unsaturated Hydrocarbons, , (i), , They have C-C single bonds., , (i), , They have C-C double or triple bonds., , (ii), , These compounds are chemically less reactive., , (ii), , These compounds are more reactive., , (iii), , Substitution reaction is the characteristic, property of these hydrocarbons., The general formula is CnH2n+2, , (iii), (iv), , Addition reaction is the characteristic property, of these hydrocarbons., The general formula is CnH2n–2 or CnH2n, , (v), , e.g. ethene C2H4. H2C = CH2, , (iv), , H, (v), , e.g. Methane CH4. H, , C, , ethyne C2H2. HC ≡ CH, , H, , H, (2), , Alkanes and Alkenes, , Ans., , Alkanes, (i), , Hydrocarbons containing carbon to carbon, single bonds (C-C) only are called alkanes or, Paraffins., , (i), , The hydrocarbons which contain carbon to, carbon double bonds (C=C) are called Alkenes, or Olefins., , (ii), , Their general formula is CnH2n+2, , (ii), , They are chemically less reactive., , (iii), , Their general formula is CnH2n, , They undergo substitution reaction., H H, , (iv), , They undergo addition reaction., , (v), , e.g. ethene C2H4. C = C, , (iii), (iv), (v), (3), , Alkenes, , e.g. ethane – C2H6. H, , C, , C, , H, , H, , H, , H, , H, , H, , H, , Alkenes and Alkynes., , Ans., , Alkenes, (i), , The hydrocarbons which contain carbon to, carbon double bonds (C = C) are called alkenes., , (ii) Their general formula is CnH2n, H, , H, , H, , H, , (iii) e.g. ethene C2H4. C = C, , (4), , They are chemically more reactive., , Alkynes, (i), , The hydrocarbons which contain carbon to, carbon triple bonds (C C) are called alkynes., , (ii), , Their general formula is CnH2n-2, , (iii), , e.g. ethyne C2H2. HC ≡ CH., , Covalent compounds and Ionic compounds., , Ans., , Covalent compounds, (i), (ii), , These compounds are formed due to sharing of, electrons., They exist as solid, liquid and gases., , (iii) Covalent compounds have low melting and, boiling points., (iv) They are generally insoluble in water., (v), , They do not conduct electricity in molten or in, aqueous solution., , (vi) e.g. Methane (CH4), ethane (C2H6)., , Ionic compounds, (i), (ii), , These compounds are formed due to transfer of, electrons., They exist as solids., , (iii) Ionic compounds have high melting and boiling, points., (iv) They are mostly soluble in water., (v), , They conduct electricity in molten state and in, aqueous solution., , (vi) e.g. Sodium Chloride (NaCl)., Potassium Chloride (KCl)

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Carbon Compounds, , (5), , 181, , Ethanol and Ethanoic acid (Physical properties), , Ans., , Ethanol, , Ethanoic Acid, , (i), , It has a specific smell., , (i), , It has vinegar like smell., , (ii), , It has a burning taste., , (ii), , It is sour to taste., , (iii) It does not freeze in winters., , (iii) It freezes in winters., , (iv) Boiling point is 78°C., , (iv) Boiling point is 118°C., , (6) Ethanol and Ethanoic acid (Chemical Properties), Ans., , Ethanol, , Ethanoic acid, , (i), , It does not react with NaHCO3., , (i), , It evolves CO2 with NaHCO3., , (ii), , It burns with a blue flame., , (ii), , It does not burn with a blue flame., , (iii) It does not affect blue litmus., Q.2.3 Answer the following questions:, (1) , Why are carbon and its compounds used as, fuels for most application?, Ans. Carbon and its compounds undergo combustion, to produce heat, the amount of heat released can, be handled and used, so that they can be used as, fuels for most application., (2) How is ethene prepared from ethanol? Give the, reaction involved in it., Ans. When ethanol is heated at the temperature 170°C, with excess amount of concentrated sulphuric, acid, one molecule of water is removed from, its molecule to form ethene, an unsaturated, compound., CH3 – CH2 – OH, Ethanol, , 170°C, CH2 = CH2 + H2O, Conc H2SO4, Ethene Water, , (iii) It turns blue litmus red., *Q.2.4 Give Scientific reasons:, (1) , Covalent compounds have low melting and, boiling points., Ans. , (i), ,  hen substances melt or boil, bonds between the, W, molecules are broken due to heat supplied., , (ii), , In covalent compounds, the bonds between two, or more molecules are weak., , (iii), , ence, intermolecular bonds in covalent, H, compounds are broken easily due to which they, have low melting and boiling points., , (2) Carbon can form a large number of compounds., Ans. , (i), ,  tomic number of carbon is 6 and its electronic, A, configuration is (2,4)., , (ii), ,  n atom of carbon has four valence electrons., A, It can share these electrons with other carbon, atoms and also with hydrogen, oxygen, nitrogen,, sulphur etc., , (iii), ,  arbon also shows the property of catenation., C, By virtue of this property, it can form a variety of, chain and ring structures, with different shapes,, sizes and molecular mass., , (iv), , hus, carbon can form a large number of, T, compounds., , (3) State the physical properties of ethanoic acid., Ans. Physical properties of ethanoic acid are:, (i), , Ethanoic acid is a colourless liquid., , (ii), , Boiling point of ethanoic acid is 118°C., , (iii), , Melting point of ethanoic acid is 17°C., , (iv), , Ethanoic acid is commonly called as acetic acid and, the frozen ethanoic acid is called glacial acetic acid., , (4) State the number of bonds essentially present, between carbon and carbon in alkenes and, alkynes., Ans. In alkenes, at least one carbon is linked to another, carbon by a double bond and in alkynes at least, one carbon is linked to another carbon by a triple, bond., , (3) Graphite is a good conductor of electricity., Ans. , (i), , Graphite is an crystalline allotrope of carbon., , (ii), , I n its structure, each carbon is attached to three, other carbon atoms, forming a hexagonal planar, structure.

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Master Key Science and Technology Part-1 (Std. X), , 182, , (iii), ,  hese hexagonal rings form layer on one another., T, Thus, in graphite many free electrons move, throughout the entire layers., , (iv), , hese free electrons help in conduction of, T, electricity., , (v), , Hence, graphite is a good conductor of electricity., , Ans. , (a), Methane - molecular formula - CH4, H, H C H, H, (b), , Ethene - molecular formula - C2H4, H H, C C, H H, , (c), , Methanol - molecular formula - CH3OH, H, H C O H, H, , (d), , Water - molecular formula - H2O, , (4) Many hydrocarbons are used as fuels in day to, day life., Ans. , (i), , , H, ydrocarbons are compounds consisting, of carbon and hydrogen and they are easily, combustible, , (ii), ,  hen they burn in air, large amount of heat is, W, evolved with the formation of carbon dioxide., , (iii), ,  ue to evolution of heat on combustion, many, D, hydrocarbons are used as fuels in day to day life., , (5) , Conversion of ethanol to ethanoic acid an, oxidation reaction, Ans. , (i), ,  ddition of oxygen or removal of hydrogen or, A, loss of electrons is known as oxidation., , (ii), , thanol gets oxidised by alkaline potassium, E, permangnate to form ethanoic acid., The following equation will explain conversion, of ethanol acid., O, [O], CH3 – C – OH, CH3 – CH2 – OH, alkaline, (ethanol), KMnO4 (ethanoic acid), , (iii), , 2 Hydrogen atoms are replaced by one Oxygen, atom, therefore it is an oxidation reaction., (6) , Vegetable oils are healthy as compared to, vegetable ghee., , H O H, (2) , Draw all possible structural formula of, compounds from their molecular formula given, below., (A) C3 H8, (B) C4 H10 (C) C3H4, Ans. , (A) C3H8 → Propane, 2D - formula, H, , (ii), , (iii), , H, , H, , C, , C, , C, , H, , H, , H, ,  egetable ghee contains saturated fatty acids, V, which are harmful, they increase cholesterol and, increase lipids in blood., , H, H, H C, H, , H, C, C, , (C3H8), , H, H, , H, H C, H, , *Q.2.5. Answer the following questions:, (1) Draw an electron dot structure of the following, molecules. (without showing the circles)., (a) Methane , , (b) Ethene, , (c) Methanol , , (d) Water, , H, C, , H, , C, , H, H, , H H H, , H, , (C3H8), , H, C, , H, , H, C C H, H, H, (C3H8), , (B) C4 H10→ Butane, H, H, , H, , H, , H, , H, , C, , C, , C, , C, , H, , H, , H, , H, ,  ence, vegetable oils are healthy as compared to, H, vegetable ghee., , Third Proof : 09.02.2019, , H, , (C3H8), ,  egetable oils which contain unsaturated fatty, V, acids are good for health as they do not increase, cholesterol., , H, H, C, H, H C, H, H, C, H H, , (C3H8), , Ans. , (i), , H, , 3D - formula, , H, H, , C4 H10, , H, H, , C, , H, H, , C, , C, , C, , H, , H, , H, , H, , C4 H10, , (C) C3H4 → Propyne, H, H, , C, H, , H, C, , C, , H, , H, , C, , C, , C, , H, , H, , D:\Dipa\2018\Books\Masker Key\X\Science\Xth_Science_Chapter 9_ Xth_Science_Chapter_9_Revised

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Carbon Compounds, , 183, , *(3) , Write structural formulae for the following, IUPAC names., , (ix) Methanol. Molecular formula CH2O, Structural formula, , (i) Pent - 2 - one - Molecular formula- C5H10O, Structural formula, H H H O H, H, , C, , C, , C, , H, , H, , H, , C, , H, , C, H, , (ii) 2 - Chlorobutane : Molecular formula C4H9Cl, Structural formula, H H Cl H, H, , C, , C, , C, , C, , C, , H, , C, , (i), , Having 5 isomers., H H, H, , (v) Butanoic acid : Molecular formula- C4H8O2 or, CH3CH2CH2 COOH, Structural formula, H H H O, C, , C, , C, , H, , H, , H, , C, , O, , H, , C, , C, , C, , H, , H, , H, , (iii), H, , (viii) Butanone: Molecular formula C4H8O, Structural formula, H, H H, C, , C, , C, , C, , H, , O, , H, , H, , H, , H, , H, , C, , C, , C, , C, , H, C1, , C2, , C3, , C4, , C5, , H, H, , C, , H, H, , H, , H, , H, , H, , H, , H, , H, , H, , H, , C1, , C2, , C3, , C4, , C5, , H, , H, H, , C, , H, H, , H, , H, , H, 3- methyl pentane, H, , (iv), , H, , Br, , (vii) Ethanamine: Molecular formula C2H5NH2, Structural formula, H H, H, H C C N, H, H H, , H, , H H H H, n- Hexane, H H H H, , H, , (vi) 1- bromopropane : Molecular formula - C3H7Br, Structural formula, H H H, , H, , H, , H, , H, Iso- hexane or 2 methyl pentane, , OH, , H, , H, , C, , (ii), , (iv) Methanol : Molecular formula- CH3OH, Structural formula, H, , H, , C, , H, , C, , H, , Ans. Molecular formula is C6H14 Its Hexane., , H OH H, , H, , C, , *(4) Draw all possible structural formulae having, molecular formula C6H14. Give names to all the, isomers. Which difficulties were faced by you, while naming? (Use your brain power; Textbook, Page No. 121), , H, , C, , H H H H, (iii) Propan - 2 - ol : Molecular formula- C3H7OH, Structural formula, H H H, H, , H, , O, , H, , H, H, , C, , H, H, , C4, , C3, , C2, , C1, , H, , H, H, , C, , H, H, , H, , H, 2, 2 – dimethyl butane or neohexane, (v), , H, , H, , H, , H, , H, , C1, , C2, , C3, , C4, , H, H, , C, , C, , H, H, , H, , H H, , H, , H, , 2, 3 dimethyl butane, , The difficulty faced by us is naming the isomer., As the number of carbon atoms increases number

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Master Key Science and Technology Part-1 (Std. X), , 184, , of isomers will also increase and their common, name creates confusion., *(5) The molecular formula of sulphur is S8 in which, eight sulphur atoms are bonded to each other to, form a ring. Draw an electron – dot structure of, S8 without showing circles. (Use your brain power;, Textbook Page no. 112), Ans. , Electron - dot, structure, S, , S, , S, , S, , S, , S, , S, , S, , S, , S, , S, , Line, S structure, , S, , S, , S, *(6), , S, ,  raw electron dot structure of cyclohexane, D, (Use your brain power; Textbook Page no. 116), , Ans. Molecular formula is C6H12, H H, H, H, , C, , H, , H, , C, , C, , H, , C, , C, , H, , H, , H, , C, H H, , (7), ,  he proportion of carbon atoms in ethanol, T, (C2H5OH) and naphthalene (C10H8), , Ans. , Compound, , The number of, carbon atoms, , Ethanol, , 2, , Naphthalene, , 10, , (8), , *(1) , By how many CH2 (methylene) units do, the formulae and the first two members of, homologous series of alkane, methane (CH4) and, ethane (C2H6) differ? Similarly, by how many, CH2 Units do the neighbouring members, ethane (C2H6) and propane (C3H8) differ from, each other? (Use your brain power; Textbook Page, no. 119), Ans. Methane and ethane differ by one – CH2- unit., Also in ethane and propane differ by one – CH2unit., *(2) , How many methylene units are extra in the, formula of the fourth member than the third, members of the homologous series of alcohols?, (Use your brain power; Textbook Page no. 119), Ans. , There is only one extra unit of methylene, (–CH2–) in the formula of the fourth member, than the third member of the homologous series, of alcohol., *(3) , How many methylene units are less in the, formula of the second member than the third, member of two homologous series of alkenes?, (Use your brain power; Textbook Page no. 119), Ans. , One unit of methylene (–CH2–) is less in the, formula of the second member than the third, member of the homologous series of alkenes., *(4) Atomic number of Chlorine is 17. What is the, number of electrons in the valence shell of, Chlorine? (Use your brain power; Textbook Page no., 112), Ans. , Atomic Number of Chlorine is 17, Electronic, configuration in 2, 8, 7. Therefore, number of, electrons in valence shell is 7., *(5) Inspect the molecular formulae of the members, of Alkenes., ,  ow is the transformation of ethanol into, H, ethanoic acid an oxidation reaction?, Ans. , Ethanol gets oxidised by alkaline potassium, permanganate to form ethanoic acid. Potassium, permanganate is an oxidizing agent which is, the source of oxygen. Addition of oxygen is, oxidation., O, (O), CH2 – CH2 – OH, CH3 – C – OH, alkaline KMnO4, ethanoic acid, , Third Proof : 09.02.2019, , Q.2.6. Answer the following questions:, ,  o you find any relationship in the number, D, of carbon atoms and the number of hydrogen, atoms in the molecular formulae (Use your brain, power; Textbook page no. 120), Ans. , In the Alkene series, First member is ethene, where molecular formula is C2H4, Next is propene, C3H6, then 1- Butene – C4H8. We can see from, the molecular formula that if number of carbon, atoms increases by ‘1’, then number of hydrogen, atoms increases by ‘2’, , D:\Dipa\2018\Books\Masker Key\X\Science\Xth_Science_Chapter 9_ Xth_Science_Chapter_9_Revised

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Carbon Compounds, , 185, , *(6) If the number of carbon atoms in the molecular, formulae of alkenes is denoted by ‘n’ what will, be the number of hydrogen atom? (Use your, brain power; Textbook page no. 120), Ans. General formula of alkenes is CnH2n. When the, value of n = 2, then the number of hydrogen, atoms will be 2n = 2 × 2 = 4. Or if n = 3, then the, hydrogen atoms will be 2n = 2 × 3 = 6, *(7) What causes the existence of very large number, of carbon compounds?, Ans. , (i), , (ii), , (iii), , (iv), ,  arbon is the only element that can form so many, C, different compounds because each carbon atom, can form four chemical bonds with other atoms., Having the atomic number 6, every carbon atom, has a total of six electrons. The carbon atom’s four, valence electrons can be shared by other atom,, forming covalent bonds, gives them the ability, to form long carbon – to – carbon chain, they, can even join “head – to – tail” to make rings of, carbon atoms., Carbon atoms can share not only a single electron, with another atom to form a single bond, but it, can also share two or three electrons, forming, double or triple bond., This makes a huge number of different possible, molecules. Hence, carbon forms large number of, compounds, because of all these properties., , *(8) What is meant by vinegar and gashol? What are, their uses?, Ans., (i), Vinegar: Ethanoic acid is most commonly known, as acetic acid. Its dilute solution in water (5-8 %), is known as vinegar., (ii) Uses: Vinegar is used as a cooking ingredient, as, a preservative in pickles, food storage, vinegar is, also used for fermentation and also in medicinal, field., (iii) Gashol: Ethanol is a clean fuel. Therefore, it is, used as an additive to increase the efficiency of, petrol. Such a fuel is called gashol., (iv) Uses: It is used as a fuel., *(9) What is a catalyst? Write any one reaction which, is brought about by use of a catalyst., Ans. Catalyst: A substance which increases the rate, of reaction without causing any disturbance is, called a catalyst., Vegetable oils are converted into vanaspati ghee, by passing hydrogen in presence of nickel as, catalyst., Vegetable oil + H Ni vanaspati ghee, H, , H, , 2, , H, , C = C, , H2, Pt/Ni, , H, , H, , C, , C, , H, , H, , (10) State some of the physical properties of covalent, compounds., Ans. Some physical properties of covalent compounds, are i) They have low melting and boiling points., ii) They are generally insoluble in water but, soluble in other organic solvents iii) They are, poor conductors of heat and electricity., , (11) Explain by writing a reaction, what will happen when pieces of sodium metal are put in n- propyl alcohol., Ans. When pieces of sodium metal are put in n-propyl alcohol, hydrogen is liberated and sodium n-propoxide salt, is formed., 2Na +, Sodium, , 2CH3 – CH2 – CH2OH → 2CH3 – CH2 – CH2 – ONa, n-propyl alcohol, Sodium n-propoxide, , + H2 ↑, Hydrogen, , (12) Explain by writing a reaction, which product will be formed on heating n - butyl alcohol with concentrated, sulphuric acid., Ans. When n - butyl alcohol is heated at the temperature 170°C with excess amount of concentrated sulphuric acid,, one molecule of water is removed from its molecule to form butene, an unsaturated compound., CH3 – CH2 – CH2 – CH2OH, , 170°C, CH3 – CH2 – CH = CH2 + H2O, Conc H2SO4, Butene, , Here concentrated sulphuric acid acts as a dehydrating agent.

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Carbon Compounds, , 187, , Carboxylic Acids, Molecular formula, , Structural Formula, , Common Name, , IUPAC Name, , HCO OH, , O, H – C – OH, , Formic acid, , Methanoic acid, , CH3CO OH, , H O, H–C–C, H O–H, , Acetic acid, , Ethanoic acid, , CH3CH2 COOH, , H H O, H – C – C– C – O – H, H H, , Propionic acid, , Propanoic acid, , CH3CH2 CH2COOH, , H H H O, H – C – C– C – C – O – H, H H H, , Butyric acid, , Butanoic acid, , Same way students can write with Alkenes and, Alkynes., , *(4) Give the names n-pentane, and Neo-pentane to, the above structural formulae., , *(2) General formula and the homologous series of, alkanes is CnH2n+2. Write down the molecular, formula of the 8th and 12th member using this., (Use your brain power; Textbook page 120), , (Use the same logic as used in the names of the, isomeric butanes for their purpose). (Use your, brain power; Textbook page 121), , Ans. General formula of alkanes is CnH2n + 2, , (i), , Structure is n-pentane, , Molecular formula of 8th member n=8, , (ii), , Structure is i-pentane or isopentane, , C8 H2 × 8 + 2 = C8 H18, , (iii), , Structure is Neopentane., , There are two isomers of Butane(C4H10). n-Butane, and isobutane., , Molecular formula of 12th member, i.e n = 12., C12 H2 × 12 + 2 = C12 H26, *(3) , Draw three structural formulae having, molecular formula C5H12. (Use your brain power, Textbook page 121), Ans. Molecular formula = C5H12., i., , H, , H, , H, , H, , H, , H, , C, , C, , C, , C, , C, , H, , H, , H, , H, , H, , H, , H, , H, , H, , C, , C, , C, , C, , H, H, , C, , H, H, , H, , ii., H, , H, , C, , H, H, , C, , C, , C, , H, H, , H, C, H, , H, , H, , H, , C, , C, , C, , C, , H, , H, , H, , H, , H, , H, , H, , C, , C, , C, , C, , H, H, , H, H, , H Or CH -CH -CH -CH -CH, 3, 2, 2, 2, 3, , H, , Or, , CH3-CH-CH2-CH3, CH3, , H, , Or, , CH3-CH2-CH2-CH3, n - butane, , H, , Or, , CH3-CH-CH3, CH3, Isobutane, , H, *(5) Molecular formula of chlorine is Cl2. Draw an, electron dot and line structure of a chlorine, molecule. (Use your brain power; Textbook Page no., 112), Ans. Molecular formula Cl2, Electron-dot Structure → outermost shell of, chlorine contain 7 electrons., , H, H, H, , H, , H, , H, , H, iii., , Ans. In –the above question, , CH3, H, , Or, , CH3, , C, , Cl, , CH3, , +, , Cl, , Cl, , Cl, , CH3, , H, , , , Line Structure – Cl + Cl → Cl2 or Cl – Cl

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Master Key Science and Technology Part-1 (Std. X), , 188, , *(6) The molecular formula of water is H2O. Draw, electron-dot and line structure of this triatomic, molecule. (use dots for electron of oxygen, atom and cross for electrons of hydrogen atom), (Use your brain power; Textbook Page no. 112), , *(8) , The molecular formula of carbon-dioxide is, CO2 – Draw the electron dot structure (without, showing circle) and line structure of CO2, (Use your brain power; Textbook Page no. 112), , Ans. Molecular formula of water is H2O, , Atomic number of carbon = 6;, Electronic configuration = (2, 4), Atomic number of oxygen = 8;, Electronic configuration (2, 6), Electron – dot structure., , Electron dot structure:, H ×+, , O, , +× H, , H×, , Hydrogen Oxygen Hydrogen, atom, atom, atom, , O, , ×H, , Ans. Molecular formula and Carbon dioxide is CO2, , Water, molecule, , O, , Ans. Molecular formula of Ammonia is NH3., Atomic number of Nitrogen is 7., Electronic configuration : (2, 5), Its valency is 3, which will be satisfied by three, hydrogen atoms., , H ×+, , +× H, , N, , Line Structure: , H N H, , , H, , +, , O, , O, , C, , O, , CO2 - Compound, , *(9) , In the Chlorination, substitution reaction of, propane, two isomeric products containing, one chlorine atom are obtained. Draw their, structural formula and give their IUPAC names., (use your brain power; Text book page no. 126), Ans., 2CH3 CH2 CH3 + Cl2→ CH3CH2-CH2Cl + CH3-CH Cl-CH3, (45%), ((55%), Structural formula and two isomers, , Electron – dot structure, H ×+, , C, , Oxygen, Carbon, Oxygen, atom, atom, atom, Line Structure: O = C = O, , Line structure:, H O H, *(7) , The molecular formula of Ammonia is NH3., Draw electron dot structure and line structure, of ammonia molecule. (Use your brain power;, Textbook Page no. 112), , +, , H × N ×H, ×, H, Ammonia, molecule, , H, , H, , H, , H, , C, , C, , C, , Cl, , H, , H H H, 1- Chloropropane, , H, , H, , H, , C, , C, , C, , H Cl H, 2- Chloropropane, , *Q.3.2. Complete the given tables:, (1), , Complete the table by writing their IUPAC names (Complete the table; Textbook Page No. 123), Sr. No., , Common Name, , Structural Formula, , H, , IUPAC Name, , (1), , Ethylene, , CH2 = CH2, , ethene, , (2), , Acetylene, , HC CH, , ethyne, , (3), , Acetic acid, , CH3 - COOH, , ethanoic acid, , (4), , Methyl alcohol, , CH3 - OH, , methanol, , (5), , Ethyl alcohol, , CH3 - CH2 - OH, , ethanol, , (6), , Acetaldehyde, , CH3 - CHO, , ethanal, , (7), , Propionaldehyde, , CH3 - CO - CH3, , propanone or propan – 2 – one, , (8), , Ethyl methyl ketone, , CH3 - CO - CH2 - CH3, , butanone or butan – 2 – one, , (9), , Ethyl amine, , CH3 - CH2 - NH2, , ethanamine or 1 – Aminoethane, , (10), , n- propyl chloride, , CH3 - CH - CH2 - Cl, , 1 – chloropropane

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Master Key Science and Technology Part-1 (Std. X), , 190, , *Q.3.3. Answer the questions based on the reaction., (1), , Observe the given reaction and answer the, questions given below., , Test tube, , (i), , What reaction is shown in the above diagram?, Name the reactant and products., , (ii), , Name the chemical equation., , (iii), , What is the special characteristic of the group, that is formed in this reaction?, , Ans., (i), Water, , Beaker, , Reactant - Ethanoic acid and ethanol in presence, of concentrated sulphuric acid as catalyst., , Wire guage, , Products - Ethyl ethanoate (ester) and water., , Tripod stand, , Mixture of, ethanoic acid,, ethanol and, sulphuric acid, , (ii), , Ethanoic, , Ethanol, , Ethyl ethanate, , acid , , Water, , (ester), , The special characteristic of esters is that they, have sweet odour and hence used for making, fragrances and flavouring agents., , Answer the questions based on the reaction. Observe the given reaction and answer the questions given below., H, , H H, C=C, , H, , +, , H H, C=C, , H H, , +, , H, C=C, , H H, , (i), , What type of reaction is it? Define it., , (ii), , Name the reactant and the products., , (iii), , What are the uses of the product / products?, , + ..., H, , H, , H, , H, , H, , H, , C, , C, , C, , C, , C, , H, , H, , H, , H, , H, , H, C, H, , polymerization reaction, ethylene (monomer), molecules are converted into polyethylene, (polymer)., , Ans., (i), , Chemical equation., CH3COOH + C2H5OH → CH3COO – C2H5 + H2O, , Burner, , (iii), , (2), , Reaction - Esterification, , It is a Polymerization reaction. The reaction by, which monomer molecules are converted into a, polymer is called polymerization. In the above, , (ii), , he reactant is ethylene (monomer) and the, T, product is polyethylene (polymer)., , (iii), , he uses of the product polyethylene is for, T, making carry bags and sportswear., , Q.4.1. Answer the following questions:, *(1) Give any four functional groups containing oxygen as the heteroatom in it. Write name and structural formula, and one example each., Ans., Functional group, Hetero atom, , Name, , Structural formula, , Condensed, Structural formula, , –O–H, , – OH, , Oxygen, , (2) Aldehyde, , O, –C–H, , – CHO, , O, –C–, , – CO–, , Oxygen, , (4) Carboxylic acid, , O, –C–O–H, , – COOH, , CH3 – C – H, Propanone, O, CH3 – C – CH3, –, –, , (3) Ketone, , –, –, , Oxygen, , Ethyl alcohol, C2H5–OH, Acetaldehyde, O, =, , (1) Alcohol, , –, –, , Oxygen, , Example, , –, –, , Acetic Acid, CH3COOH

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Carbon Compounds, , *(2), , 191, ,  ive names of three functional groups containing three different heteroatoms, write names and structural, G, formulae and one example each., , Ans., Functional group, Hetero atom, , Name, , Halogen, , Halo (chloro,, bromo, iodo), , (Chlorine,, Bromine, Iodine), , –X (–Cl, –Br, –I), , –X (–Cl, –Br, –I), , Example, Ethyl Chloride or, chloro ethane, C2H5–Cl, , O, –C–, , Propanone, O, , – CO–, , –, –, , Ketone, , –, –, , Oxygen, , Condensed, Structural formula, , Structural formula, , CH3 – C – CH3, Methyl amine, , *(3), , –N–H, , Amines, , H, , Saturated hydrocarbons are classified into three, types. Write these names giving one example, each., , Ans. , Saturated hydrocarbons in which the carbon, atoms are linked to each other only by single bonds, (C – C) are known as saturated hydrocarbons., They are classified into:, Saturated Hydrocarbon, Branched chain, , Straight chain, , e.g. Butane, (C4H10), , e.g. Ethane, (C2H6), , H, , H, , H, , C, , C, , H, , H, , C2H6, , H, , H, , H, , H, , H, , C, , C, , C, , H, H, , C, , H, H, , H, C4H10, , e.g. Cyclohexane, (C6H12), H, , H, , Polymer, , H, , C, H, C, C, H, H, C, C, H, C, H, H, C6H12, , H, H, H, H, , *(4) Give names of three natural polymers. Write, the place of their occurrence and name of, monomers from which they are formed., Ans. A macromolecule formed at regular repetition, of a small unit is called polymer. The small unit, that repeats itself regularly to form a polymer is, called monomer. Most of the structure of living, things are composed of natural polymers., , H, , Name of the, monomer, , C, , N, , H, , H, , H, , Occurence, , Polysaccharide Glucose, Proteins, , DNA, Cyclic or ring, carbon atoms, , H, , – NH2, , –, , Nitrogen, , Starch and, Cellulose, Alpha amino Muscles, hair,, acids, enzyme, skin,, egg., Nucleotide, (base- deoxy, ribose, phosphate), , Chromosomes, of animals, , (5) With the help of an example explain what is, meant by homologous series., Ans. , (i), You have seen that chains of different length are, formed by joining the carbon atoms to each other., Moreover you have also seen that a functional, group can take place of hydrogen atom on these, chains., (ii) As a result, large number of compounds are formed, having same functional group but different length, of carbon chain. For example, there are many, compounds such as CH3-OH, CH3-CH2-OH,, CH3-CH2-CH2-OH, CH3-CH2-CH2-CH2-OH which, contain alcohol as the functional group., (iii) Though the length of the carbon chain in them is, different, their chemical properties are very much, similar due to the presence of the same functional, group in them., (iv) The series of compounds formed by joining the, same functional group in the place of a particular

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Master Key Science and Technology Part-1 (Std. X), , 192, , hydrogen atom on the chains having sequentially, increasing length is called homologous series., (v), , pillars of the living world. (ii) We get food, clothing, and shelter from polysaccharides, namely starch, and cellulose. (iii) Proteins constitute a large part, of the bodies of animals and also are responsible, for their movement and various physiological, processes. (iv) Nucleic acids control the heredity, at molecular level. Rubber is another type of, natural macromolecule., , There are different homologous series in, accordance with the functional group. For, example, homologous series of alcohols,, homologous series of carboxylic acids,, homologous series of aldehydes, etc., , (vi) All the members of the homologous series are, homologues of each other. Homologous series of, alkanes is:, CH4 Methane , , C2H6 Ethane, , C3H8 Propane , , C4H10 Butane, , Man-made macromolecule: (i) Today manmade, macromolecules are in use in every walk of, life. (ii) Manmade fibres which have strength, along the length similar to natural fibres cotton,, wool and silk (iii) Elastomers which have the, elastic property of rubber, plastics from which, innumerable types of articles, sheets, pipes are, made. (iv) All these are examples of manmade, macromolecules., , (6) , Explain with an example what is meant by, substitution and addition reactions., Ans., (i), ,  he reaction in which the place of one type of, T, atom or group of atoms in a reactant is taken, by another atom or group of atoms is called, Substitution reaction., sunlight, CH4 + Cl2, CH3 – Cl + HCl, CH3Cl + Cl2, , sunlight, , CH2Cl2 + Cl2, CHCl3 + Cl2, (ii), , H, , sunlight, sunlight, , CHCl3 + HCl, , H, , C, , C, , Ethene, , H + Br – Br, Bromine, , (i), ,  sterification is the reaction in which esters are, E, formed by reaction between carboxylic acid, and an alcohol in the presence of concentrated, sulphuric acid, , Alcohol + Carboxylic acid →, , CCl4 + HCl, , H, , What is esterification and give its uses?, , Ans., , CH2Cl2 + HCl, ,  hen a carbon compound combines with another, W, compound to form a product that contains all, the atoms in both the reactants, it is called an, addition reaction. For example - When ethene, is heated with bromine, bromine gets added to, ethene to form ethylene dibromide., H, , (8), , H, , H, , C, , C, , H, , Br Br, Ethylene dibromide, , (7) What are macromolecules? Explain Natural and, manmade macromolecule., Ans. , (i), , The number of the known carbon compounds is, as large as about 10 million and the range of their, molecular masses is as large as 101-1012-, , (ii), , The giant carbon molecules formed from, hundreds or thousands of atoms are called, macromolecules., , , Natural macromolecules: (i) The natural, macromolecules, namely, polysaccharides,, proteins and nucleic acids are the supporting, , Ester, , + water, , For e.g. Ethanoic acid reacts with ethanol in, the presence of an acid catalyst and ester, ethyl, ethanoate is formed., acid, C2H5OH + CH3COOH catalyst CH3COO – C2H5 + H2O, Ethanol, Ethanoic, Ester (ethyl, water, acid, ethanoate), (ii), ,  ses: They are used for making fragrances and, U, flavouring agents. e.g.: flavouring in ice-cream, and sweets., , (9), , What is saponification?, , Ans., (i), ,  aponification is the process of making soap, S, molecule from the chemical reaction in which, fatty esters are hydrolysed in basic conditions., , (ii), , his process converts fats and oils via basic, T, reaction medium to salts of fatty acids., , (iii), ,  hen an ester is reacted with the alkali sodium, W, hydroxide, the corresponding alcohol and, carboxylic acid (in the form of its sodium, salt) are obtained back. This reaction is called, saponification reaction as it is used for preparation, of soap from fats., , Ester + Sodium hydroxide → Sodium Carboxylate, , + Alcohol

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Carbon Compounds, , 193, , Q.4.2. Study the diagram and answer the following, questions:, (1), Stand, , Thistle funnel, Gas delivery tube, , Cork, , Small Test, tube, Big test, tube, , (i), , Which acid is present in the big test tube?, , (ii), , What is the observation and conclusion of this, experiment?, , (iii), , Explain neutralization reaction of this acid with, a base., , Ans. , (i), , Acetic acid (ethanoic acid) is present in the big test, tube., , (ii), , Observation : Acetic acid (ethanoic acid) reacts, wtih basic salt namely, sodium carbonate and a, gas with effervescence is given out. This gas, when, passed through lime water in the small test tube,, turns lime water milky., , Freshly prepared, lime water, , Acetic, acid, , Sodium carbonate, , Conclusion : Acetic acid (ethanoic acid) reacts with basic salt namely, Sodium carbonate, to form a salt named, sodium ethanoate, water and carbon dioxide gas., 2CH3COOH(aq) + Na2CO3(s) → 2CH3COONa(aq) + H2O(I) + CO2(g), Acetic acid, , , Sodium, carbonate, , Sodium, Water, Carbon, ethanoate dioxide, , This carbon dioxide gas reacts with lime water Ca (OH)2 in the small test tube and forms a white colour precipitate, of calcium carbonate due to which lime water turns milky., CO2(g) + Ca(OH)2(aq) → H2O + CaCO3 ↓, Carbon, Lime, Water, Calcium, dioxide, Water carbonate, (water), (iii) The neutralization reaction of this acid with a base to form salt and water., CH3COOH + NaOH → CH3COONa(aq) + H2O(I), (2), , Observe the apparatus and chemicals given below and answer the following questions:, Apparatus : Test tube, beakers, burner etc., Chemicals : Glacial ethanoic acid, ethanol, concentrated Sulphuric acid etc., , (a), , Which reaction will you study using above apparatus and chemicals? Draw a neat labelled diagram for the, experimental set up., , (b), , How will you identify that the reaction has occurred?, , (c), , Write chemical equation involved in this experiment., (For fig. see pg. 172 Q.3.3(1)), , Ans. , (i), , We will be able to study the esterification reaction. Substances having ester as the functional group are formed by, the reaction between a carboxylic acid and an alcohol., , (ii), , 1 ml of ethanol is mixed with 1 ml of glacial ethanoic acid in a test tube and a few drops of concentrated sulphuric, acid is added to this mixture. This tube is heated using hot water bath for five minutes. This mixture is powered, into 20-0 ml of water in a breaker. If we get a sweet odour the reaction is complete, as the esters have sweet odour., , (iii), , Ethanoic acid reacts with ethanol in the presence of an acid catalyst and ester, ethyl ethanoate is formed., Acid, CH3COOH + CH3CH2 – OH Catalyst CH3 – COO – CH2 – CH2+ H2O, Ethanoic Ethanol, (ethyl ethanoate), water, acid, , vvv

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Master Key Science and Technology Part-1 (Std. X), , 194, , ASSIGNMENT - 9, Time : 1 Hr., , , Marks : 20, , Q.1. (A) Fill in the blanks:, , (3), , (1), , Hydrocarbons having identical molecular formula but different structures are called, , (2), , Ethanoic acid is commonly known as, , ., , (3) A macromolecule formed by regular repetition of a small unit is called, , (1), , ., , ., , Q.1. (B) Choose and write the correct option:, , (2), , What are the products obtained on complete combustion of hydrocarbons?, (a) CO + H2O (b) CO2 + H2 (c) CO2 + H2O (d) CO + H2, , (2) Ethanol is used as an additive to increase the efficiency of petrol. Such a fuel is called, (a) Ethanol, (b) Cananol, (c) Gasohol, (d) Methanol, , , Q.2 Answer the following: (Any 2), , (4), , (1), , Why acetic acid is called glacial acetic acid?, , (2), , Give IUPAC names for the given compounds, (b), H H H, (a), H H OH H, H, , (3), , , C, , C, , C, , H, , H, , H, , H, , C, , C, , C, , C, , H, , H, , H, , H, , H, , Differentiate between natural and man-made macromolecules., Q.3 Answer the following: (Any 2), , (6), , (1), , Draw electron dot structure of (i) CO2 (ii) H2O, , (2), , Detergents are superior to soaps - Give scientific reasons., , (3), , Write short notes on (i) Structural Isomerism (ii) Covalent bonds., , , , Q.4 Answer the following: (Any 1), , (5), , (1), , Explain with example: (i) Substitution reaction (ii) Addition reaction, , (2), , Classify saturated compounds. Give 1 example of each., , vvv

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10 Space Missions, Points to Remember:, �, ,  he universe consists of vast emptiness called space with unknown boundaries. It comprises of millions of, T, galaxies which are cluster of Stars., , �, ,  stronomy is a branch of Science which deals with the study of all the Celestial or Heavenly bodies in this, A, Universe. Unlike other sciences, astronomy is an observational Science., , �, ,  he advancement of technology, specially space technology, has taken a huge step in the last part of, T, 20th century. Hence, space travel got a boost., , �, , Thousands of artificial satellites are revolving around the earth in specific orbits. This enables man to, know and understand more about the Solar system., , �, , Space mission is broadly classified into two types:, - Launching an artificial satellite in the orbit around the earth for research and various other useful, applications., -, , �, , Sending spacecrafts to outer space to observe objects closely and to understand them., , Russian Yuri Gagarin was the first to go into the space and revolve around the Earth., , In 1969, Neil Armstrong was the 1st man to land on the surface of the Moon., �, , In 1984, Rakesh Sharma, was the first Indian to orbit the earth., , �, ,  unita Williams and Kalpana Chawla of Indian origin also participated in space exploration through missions, S, organized by NASA., , �, , Due to space explorations, the world has becomes a universal or global village., , �, , Moon is the natural satellite of the earth. Jupiter has maximum, i.e. 69 natural satellites., , �, , Artificial satellite is a man-made object that revolves regularly around a planet or earth in a fixed orbit., , �, , The first artificial satellite ‘Sputnik’ was launched in space by Russia in 1957., , �, ,  here are different types of artificial satellites; they are : Weather satellites, Communication satellites,, T, Broadcast satellites, Navigational satellites, Military satellites and Earth Observation Satellites., , �, , All artificial satellites do not revolve around the Earth in the same orbit., , �, , The satellite takes 24 hours to orbit around the earth if its height from the surface of the earth is 35780 km., -, ,  he specific velocity given to a satellite in a tangential direction to the orbit so that it starts revolving, T, around earth is called as critical velocity (vc)., , -, ,  he orbit of a satellite at a height greater than or equal to 35780 km above earth’s surface is called High, T, earth orbit., , -, ,  he orbit of a satellite at a height between 2000 km and 35780 km above earth’s surface is called Medium, T, earth orbit., , -, ,  he orbit of a satellite at a height between 180 km and 2000 km above earth’s surface is called Low earth, T, orbit., , -, , Satellite launch vehicles are used to place satellites in their specific orbits., , -, ,  eosynchronous satellites revolve in an orbit parallel to the equator and have a time of revolution of 24, G, hours., , -, , Polar satellites revolve in elliptical orbits passing over polar regions and revolve in 2 to 24 hours., (195)

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Master Key Science and Technology Part-1 (Std. X), , 196, �, ,  he Moon is the closest celestial object around the Earth. Hence, the earlier explorations were towards Moon, T, only., , �, , The second closest celestial body is Mars. The Mars expeditions were very tough. ISRO launched a space, Mangalyaan Shuttle to Mars in Nov. 2013 and in Sept. 2014, it got placed in the orbit of Mars and started, functioning., , �, ,  ikram Sarabhai is known as the father of the Indian space programme. The first Indian artificial satellite, V, ‘Aryabhatta’ was launched in 1963 under his efforts., , �, ,  here are many other explorations carried on other planets which were successful. Some space shuttles, T, moved around those planets and some landed on them., , �, , I n the space due to explorations, some man-made objects revolve around the earth. In 2016, it was noticed that, nearly 2 crore objects having length of more than 1 cm are revolving around the earth. This is called space, waste or space garbage. The worldwide efforts are now underway to manage these space debris., (16), , he earth takes almost, T, rotate about itself., , Q.1. , (A) 1. Fill in the blanks and rewrite the, statement:, , (17), ,  he first satellite launched by India was named, T, as, ., , (1), ,  he man made devices that revolve around the, T, earth, like any other planet are called, ., , (18), , he earth takes almost, T, revolve around the Sun., , *(2), , I f height of orbit of a satellite from earth’s surface, is increased, the tangential velocity of the satellite, will, ., , (19), , The planet Jupiter has, , MASTER KEY QUESTION SET - 10, , *(3), ,  he initial velocity (during launching) of the, T, Mangalyaan must be greater than, of the earth., , (4), ,  he presence of, T, on the Moon was, predicted by the space shuttle, ., , (5), , India’s first successful inter-planetary mission, was, ., , (6), ,  he first man to travel in space through a, T, spacecraft was, ., , (7), , I n 1969, the first person who landed on the Moon, was, ., , (8), ,  he first Indian, T, travelled around the, earth in a Russian spacecraft in 1984., , (9), , Due to, global village., , (10), , In 1957, Russia launched a satellite named, , ., , (11), ,  he launching of a satellite is based on, T, third law of motion., , (12), , The nearest celestial object from the earth is, , ., , (13), , , T, he Father of, programme is, , , the world has become a, , Indian, , ., , space, , (14), , The nearest planet to earth is, , (15), , In India, ISRO is headquatered at, , exploration, ., ., , (20), , hours to, , days to, satellites., , are the Broadcast satellites of India., , Ans. , (1) Artificial satellite (2) decrease (3) escape velocity, (4) water, Chandrayaan 1 (5) Mangalyaan (6) Y., Gagarin (7) N. Armstrong (8) Rakesh Sharma, (9) Space exploration (expedition) (10) Sputnik, (11) Newton’s (12) Moon (13) Vikram Sarabhai, (14) Mars (15) Bengaluru (16) 24 (17) Aryabhatta, (18) 365 (19) 69 (20) INSAT., Q.1. (A) 2. Find the odd man out:, (1), ,  uri Gagarin, Neil Armstrong, Rakesh Sharma,, Y, Vikram Sarabhai, , Ans. Vikram Sarabhai - He was a scientist while others, were astronauts., (2), , Moon, Sputnik, INSAT, Mars, , Ans. Mars - It is a planet while others are satellites of, earth., (3), , INSAT, GSAT, IRS, PSLV, , Ans. PSLV - It is a satellite launch vehicle while others, are satellites., (4), , H.E.O, L.E.O, M.E.O, GSLV, , Ans. GSLV - It is a satellite launch vehicle while others, are earth orbits., Q.1. (A) 3. Complete the analogy:, (1), , USSR: Yuri Gagarin :: India :, , Ans. Rakesh Sharma: The first person to go in space, from USSR was Yuri Gagarin, whereas the 1st, Indian to go in space was Rakesh Sharma.

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Space Missions, , (2), , 197, , PSLV : Polar Satellite Launch Vehicle : : GSLV :, , (3), , Ans. Geosynchronous Satellite Launch Vehicle : PSLV, stands for Polar Satellite Launch Vehicle whereas, GSLV stands for Geosynchronous ‘Satellite, Launch Vehicle., (3), , 2 000 km to 35780 km : M.E.O : : 180 km to 2000, km :, , Ans. (LEO) : If the height of the satellite orbit above, earth’s surface is in between 2000 km and 35780, km, the orbits are called Medium Earth Orbits,, (MEO), Whereas if the height of the satellite, orbit above the earth’s surface is in between 180, km and 2000 km, the orbit are called Low Earth, Orbits, (LEO), (4), , (4), , Moon : Chandrayan - 1 : : Mars :, , Column B, , (1) INSAT, , (a) Polar Satellite, Launch Vehicle, , (2) GSAT, , (b) Indian National, Satellite, , (3) GSLV, , (c) Geo synchronous, Satellite, , (4) PSLV, , (d) Geo synchronous, Satellite Launch, vehicle, , Ans. (1 - b), (2 – c), (3 - d), (4 - a), (2), , Column A, , Column B, , (1) Weather satellite (a) Information of the, area on protection, point, (2) Communication (b) To decide accurate, satellite, latitude and, longitude, (3) Navigational, satellite, , (c) Communicate, various places, through waves, , (4) Military satellite (d) To predict weather, forecast, Ans. (1- d), (2 – c), (3 - b), (4 - a), , (1) Neil Armstrong, , (a) The first man in, space, , (2) Yuri Gagarin, , (b) The father of, Indian Space, Research, , (3) Rakesh Sharma, , (c), , (4) Vikram Sarabhai, , (d) The first Indian, in space, , The first man on, Moon, , Column A, , Column B, , (1) Gravitational, Constant, , (a) 6 × 1024 kg, , (2) Gravitational, acceleration, , (b) 6.4 × 106 m, , (3) Mass of the earth (c) 9.8 m/s2, (4) Radius of the, earth, , Q.1. (A) 4. Match the columns:, Column A, , Column B, , Ans. (1 – c), (2 – a), (3 - d), (4 – b), , Ans. Mangalyan : In 2008, ISRO successfully launched, Chandrayan I and placed it into an orbit around, the moon, while the spacecraft made by ISRO, in November 2013 was placed into orbit around, Mars., (1), , Column A, , (d) 6.67 × 10-11 Nm2/kg2, , Ans. (1 – d), (2 – c), (3 – a), (4 – b), Q.1. (A) 5. State whether the following statements, are ‘True’ or ‘False’. Correct the false statements:, *(1) If a spacecraft has to be sent away from the, influence of Earth`s gravitational field, its velocity, must be less than the escape velocity., (2), , INSAT is an educational satellite., , *(3) The escape velocity on the Moon is less than that, on the Earth., *(4) A satellite needs a specific velocity to revolve in, specific orbit., *(5) If the height of the orbit of a satellite increases, its, velocity must also increase., (6) All artificial satellites revolve in similar orbits, around earth., (7), , ISS and Hubble revolve in L.E.O., , Ans. (1) False. The escape velocity is responsible to launch, a rocket in space which can overcome the earth’s, gravitational force. Hence, its velocity should be more, than the escape velocity. (2) False. EDUSAT is the, satellite used in the field of education. (3) True. As the, gravitational force of attraction on the moon is 1/6th, of the Earth, hence, less escape velocity is required to, overcome the gravitational force. (4) True. (5) False., If height increases, the velocity decreases. (6) False., All artificial satellites do not revolve in similar orbits, around earth. (7) True

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Master Key Science and Technology Part-1 (Std. X), , 198, , Q.1. (A) 6. Name the following:, (1), , I ndian origin female astronauts who travelled by, NASA satellite., , (4) , Who is called the father of Indian Space, programme ?, , Ans. Sunita Williams and Kalpana Chawla, , Ans. Vikram Sarabhai is called the father of Indian, Space programme., , (2), , (5) Name two female astronauts of Indian Origin?, , atellite designed by the student of COEP in, S, Pune., , Ans. Swayam, (3), , In 2008, the ISRO launched space shuttle., , Ans. Chandrayan 1, (4), , First Satellite launched by India., , Ans. Aryabhatta, (5), , Value of gravitational constant., , Ans. 6.67 × 10-11 Nm2/kg2, (6), ,  rbit of a satellite between 180 km. to 2000 km, O, from Earth’s surface., , Ans. Low Earth Orbit, (7), , Energy on which satellites work., , Ans. Solar energy, (8), ,  atellite which appears stationary with respect to, S, Earth, revolving parallel to the equator., , Ans. Geosynchronous satellite, (9), ,  rbits in which satellites complete one revolution, O, in 2 to 24 hours., , Ans. Polar Orbits, (10), ,  he velocity required for launching a remote, T, sensing satellite., , Ans. Two female astronauts of Indian origin are Sunita, Williams and Kalpana Chawla., (6) , What are the applications of geo-stationary, satellite?, Ans. , Geo-stationary satellites are used for weather, report, telephone, television, Radio, broadcasting, and telecommunication., Q.1 (A) 8. Write the full forms of the following:, (1), , INSAT : Indian National Satellite, , (2), , GSAT : Geosynchronous Satellite, , (3), , IRNSS : , lndian Regional Navigation Satellite, System, , (4), , GSLV : , Geosynchronous, Vehicle, , (5), , PSLV : Polar Satellite Launch Vehicle, , (6), , ISRO : Indian Space Research Organisation, , (7), , NASA : , National Aeronautics, Administration, , Indian satellite working for monitoring and, management of natural resources and disaster, management., , Ans. IRS Satellite, Q.1. (A) 7. Answer the following in one sentence:, (1), , (1), , (2), , (2) Which planet in our solar system has maximum, satellites?, Ans. Jupiter has maximum satellites, i.e. 69., (3) What is the name of the first Indian satellite?, Ans. The first Indian Satellite is named as ‘Aryabhatta’., , and, , Space, ,  hich of the following is the communication, W, satellite of India?, (a) INSAT , , (b) EDUSAT, , (c) Astrosat , , (d) Resourusat–1, , ‘Launching of a rocket’ is based on Newton’s, law of motion., (a) first, , (3), , What is an artificial satellite?, , Ans. , A man made object that revolves around the, Earth or any other planet in a fixed orbit is called, artificial satellite., , Launch, , Q.1. (B) Choose and Re-write with the correct option:, , Ans. Escape velocity, (11), , Satellite, , (4), , , satellites., (a) Earth, , (d) fourth, , planet has maximum number of, (b) Jupiter, , (c) Mars, , (d) Saturn, ,  hich of the following is a satellite launch, W, vehicle?, (a) PSLV, , (5), , (b) second (c) third, , (b) IRS, , (c) INSAT (d) GSAT, , , is known as Pioneer of Indian space, Programme., (a) Neil Armstrong, (b) Yuri Gagarin, (c) Rakesh Sharma, , (d) Vikram Sarabhai

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Space Missions, , (6), , 199, , , is a High Earth Orbit (HEO), satellite?, (a) Navigational satellite, (b) Geosynchronous, (c) International Space Station, (d) SPUTNIK, , (7), , To find: ve(p) = ?, , Solution:, (a), ,  hich of the following is Low Earth Orbit (LEO), W, statellite?, (a) Navigational satellite, (c) International Space Station, , For earth,, ve =, , (b), , (b) Geostationary satellite, , o use a satellite for communication or, T, meteorology, what type of orbit will be best, suited?, (a) Circular orbit, (b) Geosynchronous orbit, (c) Elliptical orbit, , (9), , (10), , (d) Polar orbit, ,  wo satellites revolving in a LEO and geoT, synchronous orbit have speed x and y respectively., Which of the following relation is correct?, (a) x > y , , (b) x < y, , (c) x = y , , (d) None of the above, ,  hich is the best suited orbit for a remote sensing, W, satellite?, (a) Geosynchronous orbit, (b) Elliptical orbit, (c) Circular orbit, (d) Sun synchronous Polar orbit, , Ans. (1) (a) INSAT (2) (c) third (3) (b) Jupiter (4) (a) PSLV, (5) (d) Vikram Sarabhai (6) (b) Geosynchronous, orbit (7) (c) International Space Station, (8) (b) Geosynchronous orbit (9) (a) x > y (10) (a), Geosynchronous orbit, Q.2.1. Solve the following:, Escape velocity:, , Type: A, , 2GM, R, for earth: vesc = 11.2 km/s., , vesc =, , , *(1) If mass of a planet is eight times the mass of, the earth and its radius is twice the radius of the, earth, what will be the escape velocity for that, planet?, Ans. Given:, , Mp = 8 Me, , , , Rp = 2 Re, , 2GMe, , = 11.2 km/s, Re, For the planet given,, vc =, , (d) All of the above, (8), , 2GM, R, , Formula: ve =, , =, =, =, , 2GMp, Rp, , , , ...(ii), ∴, , 2G8Me, 2Re, , 8, 2, , ×, , ...(i), , (Mp = 8 Me , Rp = 2Re), , 2GMe, Re, , 4 × 2GMe, Re, , = 2 ×, , 2GMe, , , , Re, , ...[from (i)], , = 2 × 11.2, = 22.4 km/s, , ∴ The escape velocity for that planet will be, 22.4 km/s., Type: B, Orbital velocity:, GM, R+h, 2πr, v= T, Note: At height h from surface, 2π (R + h), v=, T, , v=, , , *(1) Suppose the orbit of a satellite is exactly 35780, km above the earth’s surface. Determine the, tangential velocity of the satellite., Ans. Given: h = 35780 km = 35780000m, , R = 6.4 × 106 m = 6400000m, , M = 6 × 1024 kg, , G = 6.67 × 10–11 Nm2/kg2, To find: v = ?, GM, Formula: v =, R+h, Solution: v =, , 6.67 × 10–11 × 6 ×1024, (35780000 + 6400000)

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Master Key Science and Technology Part-1 (Std. X), , 200, , =, , 40.02 × 1013, 42180000, , Formulae: v =, , =, , 40.02 × 1013, 42.18 × 106, , Solution: v =, , =, , 40 × 107, 42, , =, , 0.95 × 107, , =, , 9.5 × 106, , GM, R+h, let vn be the new velocity,, vn =, vn =, , = 3.08 × 103 m/s or 3.08 km/s, ∴ Tangential velocity will be 3.08 km/s., , vn =, , *(2) In the previous example, how much time the, satellite will take to complete one revolution, around the earth?, Ans. Given:, , h = 35780 km, , , , R = 6400 km, , *(3) How much time a satellite in an orbit at height, 35780 km above earth’s surface would take, if, the mass of the earth would have been four, times its original mass?, Ans. Method 1, Given: h = 35780 km, , v = 3.08 km/s, , M=4M, , R = 6400 km, To find: T = ?, , G × 4M, R+h, 4 × GM, R+h, GM, , , , R+h, , ...(i), , (R + h), Also, vn = 2π, Tn, 2π (R + h), Tn =, vn, , Formula: v =, , ∴ Satellite will take 23 Hrs. 54 mins to, complete one revolution., , R+h, , vn = 2v , , To find: T = ?, , 2πr, T, 2πr, Solution: v =, T, 2πr, T =, v, 2π, T =, (R + h), v, 3.14, T = 2 ×, × (6400 + 35780), 3.08, 3.1, = 2 ×, × 42180, 3.1, = 2 × 42.18 × 103, = 2 × 43 × 103, = 86 × 103 s or 86000 s, 86000, Also, T =, hrs = 23.89 hrs = 23 hrs 54 mins, 3600, , GMn, , vn = 2 ×, , v = 3.08 km/s, , , , GM , v = 2πr (R + h), T, R+h, , Tn =, , 2π (R + h), 2v, , ...from equation (i), , 3.14, × (35780 + 6400), 3.08, 3.1, Tn =, × 42180, 3.1, 42180, Tn = 42180 s or Tn =, = 11.71 hrs. ~ 12 hrs., 3600, Tn =, , ∴ Time take will be ~ 12 hrs., *(3) How much time a satellite in an orbit at height, 35780 km above earth`s surface would take, if, the mass of the earth would have been four, times its original mass?, Ans. Method 2, Given: New Mass of Earth (M) = 4 × (6 × 1024) kg, , = 24 × 1024 kg., Altitude of the satellite (h) = 35780 km, , = 35780 × 103 m., To find: Time required for revolution (T) = ?, Formulae: v =, , , T=, , Solution: v, ∴, , v =, , GM, R+h, 2 π ( R+h), v, =, , GM, R+h, , 6.67 × 10-11 × 24 ×1024, 6.4 ×106 + 35780 ×103

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Space Missions, , ∴, , 201, , Q.2.2. Define the following:, , v = 37.95 × 106, v = 6.160 km/s., , (1), , Ans. If the satellite is revolving in an orbit parallel to, the equator, the time of revolution of the Earth, around itself and that of the satellite to revolve, around the Earth is the same. Thus the satellite, will appear to be stationary with respect to the, Earth. This satellite is called Geo-stationary, satellite., , 2 π ( R+h), T =, v, =, T =, , , 2 × 3.14 × 6400 + 35780, 6.16, 2 × 3.14 × 42180, 6.16, , (2), , = 43001.68 sec, , T = 11.94 hr ~ 12 hours, ∴ The satellite will take approximately 12, hours to revolve around earth., *(4) , If the height of a satellite completing one, revolution around the earth in T seconds, is h1 meter, then what would be the height, of a satellite taking 2√2 T. seconds for one, revolution?, Ans. Given:, , Height of 1st satellite = h1, , , , Time of rev. of 1st satellite = T, , , , Time of rev. of 2nd satellite = 2√2 T, , 2π (R + h), T2, = k,, 3, T, r, Solution: Case (i) Time = T, r = R + h1, Formulae:, , Taking cube root on both sides,, 3, , , , (R + h ), 2, , 3, , =, , 3, , 8 (R + h1 ), , , , R + h2 = 2 (R + h1), , , , R + h2 = 2R + 2h1, , 3, , , , h2, , = 2R + 2h1 – R, , , , h2, , = R + 2h1, , ∴ Height of the 2nd satellite will be R + 2h1, , Low Earth Orbits, , Ans. , The height of the satellite orbit above the earth, surface is in between 180 km and 2000 km are, called Low Earth Orbits., Polar Orbit, , Critical Velocity (vc), , Ans. , The specific velocity given to a satellite in a, tangential direction to the orbit, so that it starts, revolving around earth is called as critical, velocity (vc)., , (ii), , (7), , From (i) and (ii), T2, 8T2, =, (by cross multiplying), (R + h1) 3, (R + h2) 3, , (R + h2)3 = 8 (R + h1)3, , (4), , (i), , r = R + h2, , 8T, 3 = k, (R + h2), , Medium Earth Orbits, , Ans. I f the height of the satellite’s orbit above the, Earth’s surface is in between 2000 km to 35780, km, the orbits are called Medium Earth Orbits., , (6), , Case (ii) Time = 2√2 T, 2, , (3), , Ans. , In order to observe and study polar regions,, geo-stationary satellites have limitations. Hence,, elliptical medium earth orbits passing over the, polar regions are used. They are called Polar, Orbits., , To find: Height of 2 satellite (h2) = ?, , , , High Earth Orbits, , Ans. If the height of the satellite’s orbit above the, Earth’s surface is greater than or equal to 35780, km, the orbit is called High Earth Orbits., , (5), , nd, , T2, 3 = k, (R + h1), , Geo-stationary satellite, , Satellite Launch vehicle, , Ans. , The space vehicles which are used to place, satellites in their specific orbits are called Satellite, Launch vehicles., Q.2.3. Write short notes:, (1), , Space Exploration, , Ans., (i), , I n the second half of the 20th century, there, was tremendous development in technology,, especially space technology. Hence, space shuttles, were developed and launched in the space., , (ii), ,  ore than thousands of artificial satellites now, M, are revolving around the Earth in various orbits., , (iii), ,  pace Exploration helps to observe and study, S, the various aspects of our solar system and the, universe outside the solar system.

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Master Key Science and Technology Part-1 (Std. X), , 202, , (iv), ,  he collective effect of space exploration can, T, be seen in the development of communication, technology, weather forecasting and many other, fields where artificial satellites are playing a very, important role in the world today., , (v), ,  pace explorations are basically of two types,, S, a) Satellite revolving around the Earth which, study and observe various things happening, on or around earth and sending information to, the earth. b) Sending satellite outside our solar, system to know more about the universe., , (2), , (iv), , I SRO has made an expedition towards Mars,, a success. In Nov, 2013 a space shuttle was, launched towards Mars and in Sept 2014, valuable, informations were gathered about Mars., , Q.2.4. Label the diagram:, (1), , Orbits of satellites, (b), , Artificial Satellite, , (c), , (a), , Ans., (i), ,  man-made object which revolves around the, A, Earth in a definite orbit is called artificial satellite, of the earth., , (ii), ,  putnik was the first artificial satellite launched, S, in 1957., , (iii), ,  oday, there are more than thousands of artificial, T, satellites revolving around the earth., , (iv), ,  hese satellites use solar energy for their, T, functioning., , (v), ,  atellites are used for various purposes, like, S, weather, communication, broadcasting etc., , (vi), , Aryabhatta was the first Indian satellite., , (3), Lunar expeditions, Ans., (i), Moon is the natural celestial body nearest to the, Earth., (ii), , It is our natural satellite. Therefore, initial, explorations were always caused on the moon., , (iii), , Soviet Union, USA, European countries, China,, Japan and now India carried out such expeditions, towards the moon., , (iv), , Russia had sent series of space shuttles called, Luno on the moon., , (v), , From 1962 to 1972, USA carried out expedition to, Moon., , (vi), , In 1969, Neil Armstrong was the first man who, landed on Moon., , (4), , Mars expeditions, , (d), Fig 10.1 : Orbits of Satellites, , Ans. , (a) LEO (b) MEO (c) MEO (d) HEO, (2), , Structure of PSLV made by ISRO, , (a), (b), (c), (d), , (e), , (f), , Ans., (i), , The planet nearest to the Earth is ‘Mars’., , (ii), ,  any countries had sent space shuttles towards, M, Mars for various purposes., , (iii), ,  ars expeditions are very tough and nearly 50%, M, of the expeditions have failed., , Ans. , (a) Place for satellite (b) Fourth stage using liquid, fuel (c) Third stage using solid fuel (d) Second, stage using liquid fuel (e) First stage using solid, fuel (f) Engine using solid fuel which provides, the initial thrust

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Space Missions, , 203, , Q.2.5. Answer the following:, , (6) How are satellites launched in an orbit?, , (1) , What are the applications of geo-stationary, satellite?, , Ans. The satellites are taken to the desired orbits using, a satellite launcher. The satellite is then given a, specific velocity, called critical velocity (vC) in a, tangential direction., , Ans. Geo-stationary satellites are used for weather, report, telephone, television, Radio, broadcasting, and telecommunication., (2) , What is space debris ? How this debris is, managed?, Ans., (i) , In addition to artificial satellites some other objects, also revolve around the earth. They include nonfunctional parts of satellites, detached parts of, launcher, pieces due to collision of satellites., , *(7), , (ii) , This debris can be harmful to the artificial, satellites. They can collide with the satellites or, space crafts and damage them. Soon it will be, very difficult to launch new space crafts., (iii), , The escape velocity for the Earth is 11.2 km/s., , *(4) Why are geo-stationary satellites not useful for, studies of polar regions?, Ans. , (i), Geo-stationary satellites have orbits parallel to, the equator and hence the observation of polar, regions is not carried out properly., (ii), , IRNSS, , Fix the location of, any place on the, earth’s surface in, terms of its very, precise latitude and, longitude, , Navigational, satellite, , INSAT, and, GSAT, , Weather study and, prediction, , Weather, satellite, , IRS, , Study of forest,, deserts, ocean, etc., and management of, natural resources, , Earth’s, observation, , How is it managed., , (3) , Explain escape velocity. Write the value of, escape velocity of earth?, Ans., (i), The velocity required to overcome the Earth’s, gravitational force and move the objects in space, is called escape velocity., (ii), , Complete the following table:, ,  or this purpose, elliptical medium earth orbits, F, passing over polar regions are used., , *(5) Which types of telescopes are orbiting around, the earth? Why it is necessary to put them in, space? (Can you tell; Textbook Page no. 135), Ans., (i), ,  he different types of telescopes orbiting around, T, the earth are optical telescopes (Hubble), X ray, telescope (Chandra), etc., , (ii), , These telescopes are placed in space because, (a) , Some of the light coming from heavenly, bodies is absorbed by earth’s atmosphere., (b) , The position of image is disturbed due, to changes in pressure and turbulence in, atmosphere., (c) , Optical telescopes cannot be used during, the day due to sunlight., (d) During night, city lights and cloudy weather, causes difficulty in observations., , *(8), , What is difference between space and sky? (Can, you recall; Textbook page no 135), , Ans., (i),  pace: The region between the celestial bodies, S, which have different gases and dust particles, is called space. There are many stars, planets,, galaxies present in the space., (ii), ,  ky: On a cloudless night, if we observe up above, S, the earth, from an open ground, we can see stars, twinkling on a dark background. What we see, beyond the earth’s atmosphere with our naked, eyes which appears like a ceiling is called the sky., , *(9) What are different components of Solar System?, Ans. , Our Solar system comprises of Sun, Moon,, Planets, dwarf planets, Comets, Meteors, etc., *(10) What is meant by Satellite?, Ans. , An astronomical object in space that orbits, or circles around the earth or planet is called, satellite., *(11) , How many natural satellites does the earth, have?, Ans. , The Moon is the only natural satellite of the Earth., *(12) Where does the signal in your cell phone come, from?, Ans. Mobile receives its signals from communication, (mobile) towers on the earth which are in the, vicinity.

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Master Key Science and Technology Part-1 (Std. X), , 204, , *(13) , Where from do mobile towers receive the, signals?, Ans. M,  obile towers receive the signals through various, communication satellites like INSAT, GSAT, series of satellites., , artificial satellites which overcome the earth’s, gravitational force and move into space., (ii), , *(14) , Where does the signal to your TV set come, from?, , Ans. T,  hey are obtained with the help of weather, satellites like INSAT and GSAT., (16), , The broadcast signals that originate from a, radio station are sent to an artificial satellite for, redistribution to other locations. Name the orbit, where such satellites should be placed and the, launch vehicle used., , Ans. T,  he satellite should be placed in High Earth Orbit, (HEO). The launch vehicle used is GSLV., (17), , What is the range from the Earth’s surface, where, an artificial satellite, to detect a precise latitude, and longitude of a place, should revolve?, , Ans. 2000 km. to 35780 km., (18), , An artificial satellite is at a height of 35780 km., from the Earth’s surface. What is the period of, revolution of this satellite?, , Ans. S,  ince the orbit is at a height of 35780 km. from the, Earth’s surface, its period of revolution is 24 hrs., (19), , Satellites need a specific velocity to revolve in, specific orbits. Justify whether true of false., , Ans. True., (i), , The specific velocity given to a satellite in, tangential direction i.e. the critical velocity is, given as vc =, , (ii), , GM, R+h, , Since the gravitational constant (G), mass of, Earth (M) and the radius of Earth (R) is constant,, the critical velocity depends on the height of the, satellite from the Earth’s surface. Thus satellites, need a specific velocity to revolve in specific, orbits., , *Q.3. Answer the following:, (1) , Calculate the minimum velocity required by, spacecraft to escape the earth’s gravitational, force., Ans., (i), Space launch vehicles are used for the launching, , 2GM, R, , vesc =, , Ans. T,  V sets receive signals from various broadcasting, satellites which revolve around the earth., *(15) You may have seen photographs showing the, position of monsoon clouds over the country, in, the newspaper. How are these images obtained?, ,  he velocity of an object should be more than, T, the escape velocity, only then the object can be, launched into space., , G - Gravitational constant = 6.67 × 10-11 Nm2/kg2, M - Mass of the planet (earth) = 6 × 1024 kg, ∴, , R - Radius of the earth = 6.4 × 106 m, 2GM, v=, R, v=, , 2 × 6.67 × 10-11 × 6 × 1024, 6.4 ×106, , v = 11.18 × 103 m/s = 11.18 km/s., Hence, the escape velocity for earth is 11.18 km/s., The satellite launch vehicle should have velocity, more than or at least 11.2 km/s., (2) Derive the formula for critical velocity (vc)., Ans. , If a satellite of mass ‘m’ is revolving around the, earth in an orbit of height ‘h’ with speed ‘vc’, then, mvc2, centripetal force F =, will act on it., r, Here, ‘r’ is the orbital radius of the satellite from, the centre of the earth., , This centripetal force is provided by the, gravitational force of the earth., Therefore, centripetal force = gravitational force, between the Earth and the satellite., mvc2, , GMm, (R + h)2, R+h, vc2 = GM, , R+h, Taking square root on, both sides,, , , , , vc, , =, , =, , GM, R+h, , satellite, , vc, h, R, , Earth, , r, , satellite, orbit, , Fig 10.2: Orbit of an artificial, satellite, , *(3) What is meant by the ‘orbit’ of a satellite? On, what basis and how are the orbits of artificial, satellites classified?, Ans., (i), The particular path that a satellite follows to, , revolve around the earth is known as its orbit., Different artificial satellites revolve in different, orbits around the earth.

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Space Missions, , (ii), , 205, ,  he orbits of artificial satellites are classified on, T, the basis of following factors -, , (v), , (a) Height of the orbit above the Earth’s surface., (b) , The shape of the orbit (i.e. circular or, elliptical), , (iii), , The structure of the launch vehicle is decided by, the weight of the satellite and the type of satellite, orbit., , (c) Whether it is parallel to the equator or makes, a particular angle with the equator., , Place for, satellite, ,  ccording to the altitude, the orbits are classified, A, as follows:, , Fourth stage, using liquid fuel, , (a) High Earth Orbits: Height 35780 km or, higher above the Earth’s surface., , Third stage, using solid fuel, , (b) Medium Earth Orbits: Height 2000 km to, 35780 km above the Earth’s surface., , Second stage, using liquid fuel, , (c) Low Earth Orbits: Height 180 km to 2000 km,, , above the Earth’s surface., , First stage using, solid fuel, , *(4) Why is it beneficial to use Satellite Launch, Vehicles made of more than one stage?, Ans., , Engine using, solid fuel which, provides the, initial thrust, , (i), ,  atellite Launch Vehicle uses a special type of, S, fuel in huge amounts., , (ii), ,  ence, the maximum weight of the launcher is, H, due to the fuel., , (iii), ,  ue to this, different stages of fuels are made so, D, that as the fuel gets utilized, the weight of the, satellite launch vehicle can gradually reduce., , (vi), , (iv), ,  s the fuel in the 1st step is burnt, that part get, A, automatically detached from the launcher. Due, to this, the satellite launch vehicle can acquire, more velocity., , (vii) To overcome this problem, launch vehicles with, more than one stages are used. Due to this, the, weight of the vehicle can be reduced step by step,, after its launching., , (v), ,  hus, ultimately escape velocity is achieved and, T, satellite can be launched successfully., , (viii) For example, consider a launch vehicle having, two stages., , Fig 10.3 : Structure of PSLV made by ISRO, , Q.4. Answer in detail:, , For launching the vehicle, the fuel and engine in, the first stage are used. This imparts a specific, velocity to the vehicle and takes it to a certain, height., , *(1) , What is meant by satellite launch vehicles?, Explain a satellite launch vehicle developed by, ISRO with the help of a schematic diagram., Ans., (i), , Satellite launch vehicles are used, to place the, satellites in their specific orbits. The functioning, of the satellite launch vehicle is based on the, Newton’s third law of motion., , (ii), , The launch vehicle uses specific type of fuel., , (iii), , The gas produced due to combustion of the, fuel expands due to its high temperature and is, expelled forcefully through the nozzles at the, rear of the launch vehicle., , (iv), , As a reaction of this, a thrust acts on the vehicle,, which drives the vehicle high into the space., , The fuel forms a major portion of the total weight, of the launch vehicle., , (ix), , Once the fuel in this first stage is exhausted, the, empty fuel tank and the engine are detached, from the main body of the vehicle and fall either, into a sea or on an unpopulated land., , (x), , As the fuel in the first stage is exhausted, the fuel, in the second stage is ignited., , (xi), , Since, the vehicle now contains only one stage,, the weight has now been reduced and the vehicle, can move with higher speed. Almost all vehicles, are made of either two or more stages., , (xii) As an example, Polar Satellite Launch Vehicle, (PSLV), , vvv

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Master Key Science and Technology Part-1 (Std. X), , 206, , ASSIGNMENT - 10, Time : 1 Hr., , (1), , (2), , (3), , Marks : 20, , Q.1. (A) Answer the following questions., , (3), , Fill in the blanks and rewrite the statement., The nearest planet to earth is, , ., , Find the odd one out., INSAT, GSAT, IRS, PSLV, Complete the analogy., , , , 2000 km to 35780 km : M.E.O : : 180 km to 2000 km :, , , , Q.1. (B) Choose and write the correct option:, , (2), , (1) Launching of rocket is based on Newton’s, (a) first, (2) , , (b) second, , (c) third, , (b) Yuri Gagarin, , (c) Rakesh Sharma, , (d) Vikram Sarabhai, , Q.2. Answer the following: (Any 2), , (4), , (1), , Write short note on: Artificial satellite., , (2), , What is the difference between space and sky., , (3), , Define (a) High Earth orbits (b) Low Earth orbits, , , , (d) fourth, , is known as Pioneer of Indian Space Programme., , (a) Neil Armstrong, , , law of motion., , Q.3. Answer the following: (Any 2), , (6), , (1), , How are satellites classified based on their functions?, , (2), , Complete the following table., IRNSS, , Weather study and prediction, , Earth’s observation, , , (3), , , How much time a satellite in an orbit at height 35780 km above Earth’s Surface would take, if, the mass of earth would have been four times its original mass?, Q.4. Answer in detail: (Any 1), , (5), , (1), , Derive the formula for escape velocity., , (2), , What are Satellite Launch Vehicles? Explain a Satellite Launch Vehicle developed by ISRO with the, help of schematic diagram., , vvv

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MODEL ACTIVITY SHEET, , 207, , MODEL ACTIVITY SHEET, Time : 2 Hrs., , (1), , Marks : 40, , Q.1. (A) Solve the following questions., (5), If the height of the orbit of a satellite from the earth surface is increased, the tangential velocity of the, satellite will, ., , (2), , The conversion of Ferrous Sulphate to Ferric Sulphate is, , (3), , Write the proper answer in the box: H =, , (4), , (5), , (1), , × R × t., , What is the odd one out from:, Fuse wire, bad conductor, rubber gloves, generator., During transformation of liquid phase to solid phase; the latent heat is, Q.1. (B) Choose the correct alternative and rewrite the sentence:, The force of gravitation between two bodies in the Universe does not depend on, (a) the distance between them, (b) the product of their masses, , (2), , (b) H2, N2, O2, , (d) Na, Br, Ar, ., , (c) glowing effect of current, (d) lighting effect of current, , Which of the following has the highest heat capacity., (b) Air, , (c) Soil, , (d) None of the above, , The part of the lens through which the ray of light passes without suffering deviation is called, (a) Focus, , , (1), , (c) Li, Na, K, , When electric current is passed through a bulb, the bulb gives light because of, , (a) Water, (5), , (5), ., , (c) the sum of their masses, (d) the gravitational constant, , (a) electric effect of current, (b) heating effect of current, (4), , ., , Which of the following is a Dobereinei’s triad?, (a) Ne, Ca, Na, , (3), , reaction., , (b) Optical centre, , ., , (c) Centre of curvature (d) Pole, , Q.2. Solve the following questions: (Any 5), Complete the following table., , (10), , IRNSS, Weather study and predict, (2), , Write the IUPAC names of the following structural formulae., (a) CH3 – CH2 – COOH, , (b) CH3 – CO – CH2 – CH3, , (3), (4), , Explain the terms: (a) Ores (b) Gangue, Distinguish between: Farsightedness and Nearsightedness., , (5), , The absolute refractive index of water is 1.36. What is the velocity of light in water? (velocity of light in, vaccum is 3 × 108 m/s)., What is meant by latent heat? How will the state of matter transform if latent heat is given off., , (6), (7), , (1), , , (2), , Give scientific reasons: Tungsten metal is used to make a solenoid type coil in an electric bulb., Q.3. Solve the following questions: (Any 5), The gravitation force between two objects is F. How will this force change when:, (a) Distance between them is reduced to half., (b) The mass of each object is quadrupled., , (15), , What is the cause of non - metallic character of elements? What is the expected trend in the variation, of non - metallic character of element from left to right in a period?

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Master Key Science and Technology Part-1 (Std. X), , 208, , (3), Some examples of redox reaction are as follows:, , (a) Identify the reductants and oxidants from them:, 2H2S + SO2 → 3S↓ + 2H2O, MnO2 + 4HCl → MnCl2 + 2H2O + Cl2↑, , (b) If oxidation means losing electrons, what is meant by reduction?, , (c) Write the reaction of formation of Fe2+ by reduction Fe3+ by making use of the symbol (e–)., (4), An electron enters a magnetic field at right angle to the, , field direction., , (a) State the rule to find the direction of force acting, Magnetic field, on the electron., , (5), , (b) What will be the direction of force acting on the, electron in the above case?, , Electron, , Observe the following graph. Considering the, change in volume of water as its temperature is raised from 0°C, discuss the difference in the, behaviour of water and other substances. What is this behaviour of water called?, , 1 kg Volume of water, , 0.30, 25, 0.20, 15, 0.10, 05, 0.00, 0, , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9 10, , Temperature °C, , (6), , Prove the statement: A rainbow is the combined effect of the refraction, dispersion and total internal, reflection of light., , (7), , What is meant by Satellite launch vehicles? Explain a Satellite launch vehicle developed by ISRO with, the help of a schematic diagram., , , (1), , , , , , (2), , , , , , , Q.4. Solve the following questions: (Any 1), (5), Most of the children of a school in a village bring parantha and pickle in their tiffin in school. These, children do not share their food with anyone. Some children bring chapati, fruits, salad, curd,, vegetable. Those children share their food with each other., (a) Which acid is present in curd and vinegar used in pickles?, (b) What class of compounds should be present in our diet?, (c) Which group of children bring healthy food? Give two benefits of this food., (d) Which group of children bring unhealthy food? Give two harmful effects., (e) Which group of children have better value system and why?, An element is placed in 2nd group and 3rd period of the periodic table, burns in presence of oxygen to, form a basic oxide., (a), (b), (c), (d), (e), , Identify the element., Write the electronic configuration., Write the balanced equation when it burns in the presence of air., Write the balanced equation when this oxide is dissolved in water., Draw the electron dot structure for the formation of this oxide., , vvv