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3. GRAVITATION AND FLOATATION, , 1. Introduction, Besides developing the three laws of motion, Sir lssac Newton also examined the motion of the heavenly, bodies - the planets and the moon. Newton recognized that a force of some kind must be acting on the, planets to keep them in nearly circular orbits, otherwise their paths would be straight lines. A falling, apple is attracted by the earth and the apple attracts the earth as well (Newton’s third law of motion)., Extending this idea, Newton proposed that each and every body in this universe attracts every other, body. This led to the discovery of the famous law of universal gravitation i.e. each object in this universe, attracts every other object. Note that gravitational force is attractive. Newton concluded that it was the, gravitational force that acted between the sun and each of the planets to keep them in their orbits. In this, chapter, we shall discuss the role of gravitational force of the earth on the objects, on or near the surface, of the earth., , 2. Gravitation or Gravitational Force, It was Newton, who said that every object in this universe attracts every other object with a certain, force. The force with which two objects attract each other is called the force of gravitation. The force of, gravitation acts even if the two objects are not connected by any means. If, however, the masses of the, objects are small, the force of gravitation between them is small and cannot be detected easily., The force of attraction between any two particles in the universe is called gravitation or gravitational, force., , PLANCESS CONCEPTS, Newton declared that the laws of nature are the same for earthly and celestial bodies. The, force operating between the earth and an apple and that operating between the earth and the, moon, must be governed by the same laws. This statement may look very obvious today but, in the era before Newton, there was a general belief in the western countries that the earthly, bodies are governed by certain rules and the heavenly bodies are governed by different rules., So, the Newton’s declaration was indeed revolutionary., Vipul Singh, AIR 1, NSO, , www.plancess.com

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Gravitation and Floatation, , 3.2, , 2.2 Newton’s Law of Gravitation, The magnitudes and the direction of the gravitational force between two particles are given by the, universal law of gravitation, which was formulated by Newton., 2.2.1 Universal Law of Gravitation, , The gravitational force of attraction between two particles is directly proportional to the product of the, masses of the particles and is inversely proportional to the square of the distance between the particles., The direction of the force is along the line joining the two particles., 2.2.2 Mathematical Derivation, , Let A and B be two particles of mass m1 and m2 respectively. Let the distance AB = r. By the law of, gravitation, the particle A attracts the particle B with a force F such that,, F, , ∝, , A, , m1m2 (for a given pair of particles), , And F ∝, , 1, r, , 2, , (for given separation between the particles), , m1, , F, , B, m2, , r, Figure 3.1: Forec between two masses, , mm, 2 or, So F ∝ m1m, F = G 12 2, 2, r, , F, , F, , r, , Here, G is a constant known as the universal constant of gravitation., 2.2.3 Universal Gravitational Constant, , (i) Introduction:, Force of gravitation between two bodies of mass m1 and m2 kept with distance r between their, centres, is given by: F = Gm12m2, r, , where constant of proportionality G is called universal gravitational constant (U.G.C.)., (ii) Definition:, In relation, , F=, , Gm1m2, r2, , If m1 = m2 = 1, r = 1, then F = G. Hence, universal gravitational constant may be defined as the, force of attraction between two bodies of unit mass each, when kept with their centres at unit, , distance apart., (iii) Units of G: F = Gm12m2, r, , www.plancess.com

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Foundation for Physics, , 3.3, , 2, We have, G = Fr, , m1m2, , In S.I. , , Nm2, =, G = Nm2kg−2, kgkg, , dynecm2, G = dynecm2 g−2, In C.G.S. =, g.g., , (iv) Values of G:, In S.I. , , G = 6.67 × 10-11 Nm2 kg2, , In C.G.S., , G = 6.67 × 10-8 dyne cm2 g-2, , 2.2.4 Important Characteristics of Gravitational Force, , (i) G,  ravitational forces between two bodies form an action and reaction pair i.e., the forces are equal, in magnitude but opposite in direction., (ii) Gravitational force is a central force i.e., it acts along the line joining the centres of the two, interacting bodies., (iii) Gravitational force between two bodies is independent of the nature of the intervening medium., (iv) Gravitational force between two bodies does not depend upon the presence of other bodies., (v) G,  ravitational force is negligible in case of light bodies but becomes appreciable in case of massive, bodies like stars and planets., (vi) Gravitational force is a long range force i.e., gravitational force between two bodies is effective, even if their distance of separation is very large. For example, gravitational force between the sun, and the earth is of the order of 1022 N, although distance between them is 1.5 × 108 km., (vii) Gravitational force is a conservative force., 2.2.5 Experimental Support for the Law of Gravitation, , (i) All the planets including the earth, rotate around the sun due to gravitational force between the, sun and the planet., (ii) Tides are formed in oceans due to gravitational force between the moon and the earth., (iii) It is the gravitational force between the planet and its satellite which makes the satellite to move, around the planet., (iv) The atmosphere of the earth is due to the gravitational force of the earth., (v) This is the force that binds us to earth., , www.plancess.com

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Gravitation and Floatation, , 3.4, , PLANCESS CONCEPTS, A spherically symmetric body can be replaced by a point particle of equal mass placed at its, centre for the purpose of calculating gravitational force due to this body on other bodies, placed outside its radius or on its surface., Anand K, AIR 1, NSO 2011, , 2.3 Newton’s Third Law of Motion and Gravitation, Newton’s third law of motion says that: If an object exerts a force on another object, then the second, object exerts an equal and opposite force on the first object. The Newton’s third law of motion also holds, good for the force on the earth in the opposite direction. Thus, even a falling object attracts the earth, towards itself. When an object, say a stone, is dropped from a height, it gets accelerated and falls towards, the earth and we say that the stone comes down due to the gravitational force of attraction exerted by, the earth. Now, the stone also exerts an equal and opposite force on the earth, then why don’t we see the, earth rising up towards the stone., From Newton’s second law of motion, we know that:, Force = Mass × Acceleration So, Acceleration =, , Force, Mass, , Or a =, , F, M, , It is clear from this formula that the acceleration produced in a body is inversely proportional to the mass, of the body. Now, the mass of a stone is very small, due to which the gravitational force produces a large, acceleration in it. Due to large acceleration of stone, we can see the stone falling towards the earth. The, mass of earth, however, is very-very large. Due to the very large mass of the earth, the same gravitational, force produces very-very small acceleration in the earth. Actually, the acceleration produced in the earth, is so small that it cannot be observed. And hence we do not see the earth rising up towards the stone., Illustration 1: Two persons having mass 50kg each, are standing such that the centre of gravity are 1m, , apart. Calculate the force of gravitation and also calculate the force of gravity on each., Solution: Given: m1 = m2 = 50kg., r = 1m. , G = 6.67 × 10-11 N. m2/kg2, Force of gravitation F = Gm12m2 ;, r, , =, F, , 6.67 × 10−11 × 50 × 50, = 1.667 × 10−7 N., 2, (1), , www.plancess.com

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Foundation for Physics, , 3.5, , Force of gravity due to earth,, Fₑ = GMm, 2, r, , Here r = R, radius of the earth and m1 = M = mass of earth, m2 = m = mass of object, =, Fe, , 6.67 × 10−11 × 6 × 1024 × 50, = 489 N , 2, 6.4 × 106, , (, , ... (ii), , ), , Fₑ is much greater than F so the persons will not move towards each other but each of them, moves towards the earth., , 2.4 Estimation of Gravitational Force between Different Objects, 2.4.1 Between the Sun and the Earth, , Mass of earth, m1 = 6 × 1024 kg, Mass of the sun, m2 = 2 × 1030 kg, Distance between the sun and the earth, r = 1.5 × 1011m, Gravitation force between the sun and the earth,, F=, , F=, , Gm1m2, r2, , 6.67 × 10−11 Nm2kg−2 × 6 × 1024 kg × 2 × 1030 kg, , (1.5 × 10 m), 11, , 2, , F = 3.6 × 1022 N, The gravitational force between the sun and the earth is very large (i.e. 3.6 × 1022 N). This force keeps, the earth bound to the sun., 2.4.2 Between the Moon and the Earth, , Mass of the earth, m1 = 6 × 1024 kg, Mass of the moon, m2 = 7.4 × 1022 kg, Distance between the earth and the moon, r = 3.8 × 108 m, , ∴ Gravitational force between the earth and the moon,, F = Gm12m2, r, , www.plancess.com

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Gravitation and Floatation, , 3.6, , F=, , 6.67 × 10−11 Nm2kg−2 × 6 × 1024 kg × 7.4 × 1022 kg, , (3.8 × 10 m), 8, , 2, , F = 2.05 × 1020 N, This large gravitational force keeps the moon to move around the earth. This large gravitational force is, also responsible for the ocean tides., Illustration 2: Two bodies P and Q having mass m and 2m respectively are kept at a distance apart., , Where a small particle should be placed, so that the net gravitational force on it due to the bodies P and, Q is zero?, Solution: It is clear that the particle must be placed on the line PQ, suppose it is at a distance  from P., Let its mass be m’., The force on m’ due to P,, F1 =, , G (m) m', , (), , 2, , A, , towards P., , and that due to Q is F2 =, , m, G ( 2m) m', , (d −  ), , 2, , d - = ± 2; d = ( 1, , ± 2), , ;  =, , , , towards Q., , The net force will be zero if F1 = F2 Thus,, d, , Gmm', 2, , (1 + 2 ), , As distance  cannot be negative So, , =, , or, , =, , B, , m', , G ( 2m) m', , (d −  ), , 2, , F1, , F2, d- , , 2m, , of (d −  )2 =, 2 2 (d −  )2 =, 2, , d, , (1 − 2 ), , d, , (1 + 2 ), , www.plancess.com

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Foundation for Physics, , 3.7, , PLANCESS CONCEPTS, Our solar system is part of a spiral galaxy, which contains roughly 1011 stars as well as gas,, dust, and other matter. The entire assemblage is held together by the mutual gravitational, attraction of all the matter in the galaxy., Vaibhav Gupta, AIR 2, NSO, , 2.5 Force of Gravitation of the Earth (Gravity), Gravitation and Gravity, Attraction between two bodies having mass of same order, is called gravitation and the force is called, gravitational force. Forces involved are very small and the attracting bodies do not move towards each, other., Attraction between a planet (earth) or its satellite and a body, having masses of widely different order is, called gravity and the force is called force of gravity. Forces involved are large and body moves towards, the planet., Thus, gravity becomes a special case of gravitation in which small bodies move towards huge planets., Then, force of gravity F = GMm, 2, r, , 3. Bodies Falling Near the Surface of the Earth, 3.1 Galileo’s Observations on Falling Bodies, The speed of falling body increases as it comes down. This means that the body accelerates, when it falls, freely. Suppose we drop a coin and a feather from the same height simultaneously. Which of them will, reach the ground first? The answer is obvious, the coin will reach earlier than lighter feather or we can say, that the heavier objects comes down more faster than lighter ones but such a generalization is not correct., If we take two solid balls of different masses, say, one of 1 kg and the other of 2 kg, and drop them from, the same height, we will find that they reach the ground almost simultaneously., It is said that Galileo dropped two stones of different masses from the Leaning Tower of Pisa (in Italy), and found that they reached the ground simultaneously. Galileo argued that the air resist on object, traveling through it. If the material is dense and its surface area is small, the resistance due to air is quite, small compared to the force of gravity. Thus, one can neglect the effect of air resistance while studying, falling stones, metallic blocks, coins etc. But the effect of air resistance is very important for small pieces, , www.plancess.com

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Gravitation and Floatation, , 3.8, , of paper, feather, leaves etc. each of which has a large surface area and low density. When a coin and a, feather fall through air, air offers greater resistance to the motion of the feather and less resistance to the, motion of the coin. According to Galileo’s argument, if air is totally removed, the coin and the feather, will fall simultaneously., Newton was born in the year Galileo died. Galileo did not have access to the equations for gravitational, attraction and the acceleration resulting from a force. Still, he correctly predicted something from his, observations that was contrary to everyday experience., Galileo’s prediction was tested by the British scientist Robert Boyle. He kept a coin and a feather in a, long glass tube and evacuated the air from inside the tube by using a vacuum pump. When the tube was, inverted, the coin and the feather fell together., , 3.2 Acceleration due to Gravity, If we drop a ball from a height, its speed increases as time passes. If we throw a ball upwards, its speed, decreases till it reaches the highest point. If we throw the ball at an angle to the vertical, its direction of, motion changes. In all these cases, the velocity of the ball changes, i.e., the ball is accelerated. Whenever, an object moves near the surface of the earth with no other object pushing or pulling it, it is accelerated., This acceleration is caused due to the force of gravity and is called the acceleration due to gravity. Consider, an object of mass m moving freely near the earth’s surface. Neglecting air resistance, the only force on it,, is due to gravity. The force has magnitude:, F=, , GMem, R 2e, , ........(i), , where Me = mass of the earth, m = mass of the object, and Re = radius of the earth., As the earth’s radius Re (6400 km) is large as compared to distance of the object from the earth’s surface,, we use Re in equation (i) to denote the distance of the object from the centre of the earth. As the force, a, given by equation (i), is the resultant force on the object, its acceleration is =, , F GMe, =, m, R 2e, , Note that this acceleration does not depend on the mass of the object. Thus, we have the following, conclusion:, If gravity is the only acting force (meaning that air resistance is neglected), all objects move with the same, acceleration near the earth’s surface. This acceleration is called the acceleration due to gravity,, 2, , −11 Nm, 24,  6.67 × 10,  × 6 × 10 kg, 2, GMe, kg, , whose magnitude, ‘g’ is given by g = 2 ; g =, 9.8ms −2, =, 2, Re, 6, 6.4 × 10 m, , (, , (, , ), , ), , www.plancess.com

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Foundation for Physics, , 3.9, , The direction of this acceleration is towards the centre of the earth, i.e., in the vertically downward, direction. The acceleration has the same value, both in magnitude (9.8 m/s2) and direction (towards, centre of earth), whether the particle falls, moves up or moves at some angle with the vertical. In all these, cases, we say that the particle moves freely under gravity., , 3.3 Value of ‘g’ on the Surface of the Moon, g=, , GM, R2, , where M is the mass of a heavenly body like earth and R is its radius. As all heavenly bodies (like, , planets, the sun and the moon) are of different masses and different radii, so the value of g is different on, different heavenly bodies., We know,, , gmoon =, , GMm, 2, Rm, , ...(i), , Mm (mass of the moon) = 7.4 × 1022 kg, Rm (radius of the moon) = 1.75 × 106 m, G = 6.673 × 10-11 Nm2 kg-2, Then, from equation (i), gmoon =, , , Now, , , 6.673 × 10−11 Nm2 kg−2 × 7.47 × 1022 kg, , (1.75 × 10 m), 6, , 2, , ;, , gmoon = 1.63ms−2, , gmoon 1.663ms−2 1, =, =, gearth, 6, 9.8ms −2, , or, , gmoon =, , 1, g, 6 earth, , 1, , Thus, acceleration due to gravity on the surface of moon is, times the acceleration due to gravity on, 6, the surface of the earth., , PLANCESS CONCEPTS, A uniform shell of matter exerts no net gravitational force on a particle located inside it., Vaibhav Gupta, AIR 2, NSO, , www.plancess.com

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Gravitation and Floatation, , 3.10, , 4. Mass of Earth and Mean Density of Earth, 4.1 Mass of the Earth, The mass of the earth can be calculated by using Newton’s law of gravitation. Consider a body of mass m, lying on the surface of the earth, then force of gravity acting on the body is given by, F=, , GMm, R2, , .....(i), , where, M = mass of the earth, , R = radius of the earth, Also,, , F = mg, , ....(ii), , From (i) and (ii), we have mg = GMm, 2, , or, , R, , Now, , M=, , gR 2, G, , g = 9.8 ms-2, R = 6400 km = 6.4 × 105 m, G = 6.67 × 10-11 Nm2 kg-2, , (, , ), , 2, , 9.8 × 6.4 × 106, =, M, = 5.98 × 1024 kg, −11, 6.67 × 10, , Thus, the order of the mass of earth is 1025 kg, , 4.2 Mean Density of Earth, We know, g = GM, 2, R, , Let ρ be the means density of the earth. Since earth is assumed to be a homogeneous sphere of radius R,, therefore, mass of the earth is given by, M = Volume × density =, , 4 3, πR ρ, 3, , Substituting this value in equation (i), we get, g=, , G, R, , 2, , ×, , 4 3, 4, πR ρ=, πGRρ ;, 3, 3, , Since, g = 9.8 ms-2,, , ∴ρ =, , 3g, 4πGR, , G = 6.67 × 10-11 N m2 kg-2, R = 6400 km = 6.4 × 106 m, , www.plancess.com

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Foundation for Physics, , ∴, , ρ=, , 3 × 9.8, 4 × 3.142 × 6.67 × 10−11 × 6.4 × 106, , or, , 3.11, , ρ = 5478.4 kgm‒3, , Density of earth 5478.4kgm−3, =, ~ 5.5, Density of water, 1000kgm−3 −, , Thus, density of earth is about 5.5 times the density of water., , 5. Equations of Motion For Freely Falling Object, Since the freely falling bodies fall with uniformly accelerated motion, the three equations of motion, derived earlier for bodies under uniform acceleration can be applied to the motion of freely falling, bodies. For freely falling bodies, the acceleration due to gravity is ‘g’ so we replace the acceleration ‘a’ of, the equations by ‘g’ and since the vertical distance of the freely falling bodies is known as height ’h’, we, replace the distance ‘s’ in our equations by the height ‘h’. This gives us the following modified equations, for the motion of freely falling bodies., General equations, of motion, , Equations of motion for, freely falling bodies, , (i) v = u + at, , changes to, , v = u + gt, , (ii) s = ut + at2, , changes to, , h = ut +, , (iii) v2 = u2 + 2as, , changes to, , v2 = u2 + 2gh, , 1, 2, , 1 2, gt, 2, , We shall use these modified equations to solve numerical problems. Before we do that, we should, remember the following important points for the motion of freely falling bodies., (i) When a body is dropped freely from a height, its initial velocity ‘u’ is zero., (ii) When a body is thrown vertically upwards, its final velocity ‘v’ becomes zero., (iii) The time taken by a body to rise to the highest point is equal to the time it takes to fall from the, same height., (iv) The distance traveled by a freely falling body is directly proportional to the square of time of fall., Sign Conventions, , (i) g is taken as positive when it is acting in the same direction as that of motion and g is taken as, negative when it is opposing the motion., (ii) Distance measured upward from the point of projection is taken as positive, while distance, measured downward from the point of projection is taken as negative., , www.plancess.com

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3.12, , Gravitation and Floatation, , (iii) Velocity measured away from the surface of earth (i.e. in upward direction) is taken as positive,, while velocity measured towards the surface of the earth is taken as negative., Illustration 3: If the force of gravitation between the earth and a body of mass M on its surface be, , 9 × 107 N, what would be the value of M? Mass of the earth = 6 × 1024 kg and radius of the earth = 6.4, ×106 m., Solution: Given that F = 9 ×107 N, Mass of the earth (m1) = 6 ×1024 kg, Mass of the body (m2) = m, Radius of the earth (r) = 6.4 ×106 m, G = 6.67 ×10–11 Nm2 kg–2, Using formula, F = Gm12m2 we get, r, , ⇒ 9 ×107 =, , 6.67 × 10−11 × M × 6 × 1024, (6.4 × 106 )2, , ∴Mass M = 9.2 ×106 kg., Illustration 4: Find the force of earth’s gravitation on a stone of mass 2 kg held near the surface of the, , earth. Also calculate the acceleration this force would produce in the stone and the earth. Given that,, mass of the earth = 6 ×1024 kg, radius of the earth = 6.4 ×106 m and G = 6.7 × 10–11 Nm2 kg–2., Solution: Force of earth’s gravitation on the stone,, F=, G×, , Mm, R2, , =, 6.7 × 10−11 ×, , 2 × 6 × 1024, (6.4 × 106 )2, , = 19.6 N, , Acceleration produced in the stone by this force = F/m =, produced in the earth by the same force = F/m =, , 19.6, = 9.8m / s 2 and acceleration, 2kg, , 19.6N, 24, , 6 × 10 kg, , = 3.3 ×10–24 m/s2, , Illustration 5: Consider a body in space which has a mass twice that of the earth and a radius thrice that, of the earth. What will be the weight of a book on this body in space, if its weight on the earth is 900 N?, Solution: Let m be the mass of the book, its weight on the earth is given by, , www.plancess.com

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Foundation for Physics, , Weight W= F= G, , Mem, R 2e, , 3.13, , , , ...... (i), , and the weight of the book on the body in space is given by, W'= F'= G, , Mm, R2, , .......(ii), , Dividing equation (i) and (ii), 2, Then W' = M × R e = M ×  R e , W M, R2 M  R , e, , e, , , , 2, , , , Here, W = 900 N and M = 2Me and R = 3Re, 2, , W' 2M  R e , 1, =× , 2×,  =, 900 Me  3R , 9, , 1, 9, , ; W' = 2 × × 900 N = 200 N., , Illustration 6: What is acceleration due to gravity at the surface of mars if its diameter is 6760 km and, , mass one-tenth that of earth. The diameter of earth is 12742 km and acceleration due to gravity on earth, is 9.8 m/s2., Solution: We know that g =, , GM, R2, , , , , ; gM =  MM   RE , gE, , 2, ,  ME   RM , , 2, MM (D2E ) 9.8 × 0.1 × (12742), = 2, ⇒ gM = g E ×, = 9.8 × 0.35 = 3.48 m/s2, ME (D2M ), (6760), , 6. Variation in Acceleration due to Gravity, GM, , e, 9.8 m / s2, We know that the value of acceleration due to gravity at the earth surface =, is g =, 2, , Re, , where Me and Re are the mass and radius of earth respectively., , 6.1 Variation of ‘g’ With Altitude, On moving in vertical direction away from the earth’s surface, the value of g decreases because the, distance from the earth’s centre increases., , www.plancess.com

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3.14, , Gravitation and Floatation, , Suppose a body is taken to a height ‘h’ above the earth’s surface where the value of acceleration due, GMe, GMe , gravity is, gh: gh =, =, , 2, 2, (R e + h), , Re, , , , 2,  (R e + h) , R 2e, , or, gh = g  R e , , 2, ,  Re + h , , Since Re + h > Re ; ∴ gh < g, Illustration 7: How far away from the earth does acceleration due to gravity become one percent of its, , value at the earth’s surface? Assume that the earth’s surface is a sphere of radius R = 6.4 × 105 m., Solution: Let g’ be the acceleration due to gravity at height h from the surface of earth. Given g' =, , g', R2, =, If R is the radius of earth, then, or,, g (R + h)2, , 1, R2, =, 100 (R + h)2, , 1, g, 100, , ⇒ h = 9R = 9 × 6.4 × 106 m = 5.76 ×107 m., Illustration: 8 At what height above the surface of the earth the value of ‘g’ becomes 25% of its value on, , the surface of the earth? Given, radius of the earth = 6400 km., 2, ,  R , Solution: The value of g’ at height h above the surface of the earth is given by g' = g , R + h , , 2,  R , 25, 25, g=, g, Here, g’ = 25 % of g =, g ;∴,  or, 2R = R + h or, h = R = 6400 km., 100, 100, R +h, , 6.2 Variation of ‘g’ with Depth, If g is the value of acceleration due to gravity at the surface of the earth and g’ its value at a depth h below,  h, the surface of the earth, then=, g' g  1 −  which clearly shows that the value of g decreases as we go,  R, inside the surface of the earth. Hence the value of g is maximum at the surface of the earth and becomes, zero at the centre of the earth., Illustration 9: What is the distance from the centre of earth where the weight of the body is one fourth, , that of the weight measured on the surface of earth? (Assume R is the radius of the earth.), Solution: (i) The weight of the body at the center of the earth is equal to zero because,, ,  d,  R, gcentre = g  1 −  = g  1 −  = 0,  R,  R, , www.plancess.com

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Foundation for Physics, , (ii), , 3.15, , g1  d   R − d , = 1− =, where, d represents depth w.r.t, earth’s surface. Hence R – d = x = distance, g  R   R , from centre of earth., , ∴ g1 =g ×, , x, x, 1 x, ⇒ = ⇒ x = R / 4, ⇒ mg1 =mg ×, 4 R, R, R, , Illustration 10: Assuming the earth to be a sphere of uniform mass density, how much would a body, , weigh half way down to the centre of the earth if it weighed 250 N on the surface?, Solution: The weight, mg = 250 N, at the surface of the earth., ∴ m=, , 250 250, =, kg, g, 9.8, ,  h, g' g  1 − , The value of acceleration due to gravity at a depth h is given by=,  R,  1, 2, Here, g = 9.8 m/s2, h = R/2; g'= 9.8  1 − =, = 4.9m, 4.9 m/ s/ s2, ,  2, , Hence new weight of the body = mg' =, , 250, × 4.9 = 125 N, 9.8, , 6.3 Variation of ‘g’ with Latitude, The value of acceleration due to gravity changes with latitude due to following reasons:, 6.3.1 Shape of Earth, , The earth is not a perfect sphere. It is flattened at the poles and bulged at the equator. The equator radius, Re is greater than the polar radius by Rp, We know that, g = GM, 2, , Rp, , R, , Since, G and M are constants, ∴ g ∝, , 1, , Re, , R2, , Hence, the value of g increases as we move from the equator to the, pole., , Figure 3.2: Elleptical shape of, the earth, , www.plancess.com

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Gravitation and Floatation, , 3.16, , 6.3.2 Rotation of Earth, , The value of g is larger at the poles and smaller at the equator due to rotation of earth about its NorthSouth axis., Illustration 11: What is the height at which the acceleration due to gravity decreases by 36% of its value, , on the surface of the earth? (Radius of earth = 6400 km), Solution: g =, , GM, r, , 2, , at surface, g =, , GMe, R 2e, , , , ...(i), , GMe, At distance ‘r’ from centre g', =, ;=, r R e + h ...(ii), 2, r, , Given that, g’ = 0.64 (g), Solving (i) and (ii) Re = 4h ⇒ h =, , Re, = 1600km, 4, , PLANCESS CONCEPTS, A black hole is a place in space where the force of gravity is so strong that even light cannot, get out. The force of gravity is so strong because matter has been squeezed into a tiny space so, that the density is tremendously high. This can happen when a star is dying., Because no light can get out, black holes are invisible. Space telescopes with special tools can, help find black holes. The special tools can see how stars that are very close to black holes act, differently than other stars., Anand K, AIR 1, NSO 2011, , 7. Mass and Weight, 7.1 Mass, The mass of a body is the quantity of material (or matter) contained in it. It is a scalar quantity. The unit, of mass is kilogram. The mass of a body is constant and does not change from place to place., , 7.2 Weight, “The force with which the earth attracts an object is called the weight of the object”. Weight is a force,, and hence is expressed in newton. The relation between the mass of an object and its weight W is as, , www.plancess.com

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3.17, , Foundation for Physics, , GM m, ,  GM , , e, W =, m  2 e  = mg ⇒ W = mg, =, shown below.,  R , R 2e,  e , , Thus, the weight of an object of mass 1 kg will be W = (1kg) × (9.8 m/s2) = 9.8 N., The weight of a one kg object is sometimes called 1 kg weight. Thus, 1 kg weight = 9.8 N. When a, weighing machine gives weight as 35 kg, it means the actual weight is 343 N. So kg. wt. is a unit of force., , 7.3 Centre of Mass, A body consists of many particles. A point where the total mass can be supposed to be concentrated is, called centre of mass., If we assume earth to be a sphere of uniform density then its centre of mass lies at its centre. The force of, the earth on any body is therefore in the direction of line joining their centres., , 7.4 Centre of Gravity, g, , GMe, , where ‘r’ is distance from centre of earth. As we move at high altitude, from surface of earth, the value of ‘g’ changes. The centre of gravity of a, body, is therefore, a point at which force of gravity of the body can be assumed to, act. For bodies of regular shape and which have uniform density, the centre of, gravity lies at geometrical centre of body, if the field due to gravity is uniform., g=, , r2, , 7.5 Difference between Mass and Weight, , r, , Re, , Figure 3.3: Value of, ‘g’ at any ‘r’, , Table 3.1: Mass and weight comparison, , Mass, , Weight, , 1. Mass is quantity of matter possessed by a body, , 1. , Weight is the force with which a body is, attracted towards the centre of the earth., , 2. It is a scalar quantity., , 2. It is a vector quantity., , 3. Its S.I. units is kilogram (kg.), , 3. Its S.I. unit is Newton (N)., , 4. Mass of a body remains constant at all places., , 4. Weight of the body changes from place to, place., , 5. Mass of a body is never zero., , 5. Weight of a body becomes zero at the centre, of the earth., , 6. Mass of measured by a beam balance., , 6. Weight is measured by a spring balance., , www.plancess.com

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3.18, , Gravitation and Floatation, , 7.6 Weight of Object on Moon, A body of mass m has weight, W = mg, Calculation:, For earth,, , ge = 9.8 ms-2, , For moon,, , gm = 1.7 ms-2, , Hence,, For earth,, , We = mge, , For moon,, , Wm = mgm, , Ratio,, , Wm mgm gm 1.7 1, =, = =, ≈, We mge ge 9.8 6, , i.e. Weight on moon =, , 1, th weight on earth., 6, , 7.7 Variation in the Weight of a Body, Weight of the body is given by, W = mg, So the weight of a body depends upon (i) the mass of the body and (ii) value of acceleration due to, gravity (g) at a place., The mass of a body remains the same throughout the universe, but as the value of ‘g’ is different at, different places, the weight of a body is different at different places., (i) The value of ‘g’ is more at poles and less at the equator. Therefore, weight of a body is more at, the poles and less at the equator. In other words, a body weighs more at the poles and less at the, equator., (ii) The value of ‘g’ on the surfaces of different planet of the solar system is different, therefore, the, weight of a body is different on different planets., (iii) The value of ‘g’ decreases with height from the surface of the earth. Therefore, the weight of a, body also decreases with height from the surface of the earth. That is why, the weight of a man is, less on the peak of Mount Everest than the weight of the man at Delhi., (iv) The value of ‘g’ decreases with depth from the surface of the earth. Therefore, the weight of a, body decreases with depth from the surface of the earth., (v) The value of ‘g’ at the centre of the earth is zero hence weight (=mg) of the body is zero at the, centre of the earth., , www.plancess.com

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Foundation for Physics, , 3.19, , 8. Difference between ‘g’ and ‘G’, Table 3.2: Comparison of ‘g’ and ‘G’, , Acceleration due to gravity (g), , Universal gravitational constant (G), , 1. The acceleration produced in a body falling, freely under the action of gravitational pull, of the earth is known as acceleration due to, gravity., , 1. The gravitational force between two bodies, of unit masses separated by a unit distance is, known as universal gravitational constant., , 2. The value of ‘g’ is different at different points, on the earth., , 2. The value of ‘G’ is same at every point on the, earth., , 3. The value of ‘g’ decreases as we go higher from, the surface of the earth or as we go deep into, the earth., , 3. The value of ‘G’ does not change with height, and depth from the surface of the earth., , 4. The value of ‘g’ at the centre of the earth is, zero., , 4. The value of ‘G’ is not zero at the centre of the, earth or anywhere else., , 5. The value of ‘g’ is different on the surface of, different heavenly bodies like the sun, moon,, and the planets., , 5. , The value of ‘G’ is same throughout the, universe., , 6. The value of ‘g’ on the surface of the earth is, 9.8 ms-2., , 6. The value of G = 6.673 × 10-11 Nm2 kg-2, throughout the universe., , 9. Weightlessness, Introduction: When a man stands on weighing machine at rest, his weight compresses its spring, , downwards. Due to upward reaction, the pointer of the machine moves over the scale and the machine, records the weight of the man., But when the same machine starts falling down freely, there is no reaction and the pointer stays at zero,, thus recording a zero weight., The man falling freely under the action of gravity becomes weightless., Definition: Weightlessness may be defined as the state in which a body loses its weight due to free fall., , www.plancess.com

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3.20, , Gravitation and Floatation, , Demonstration, , Let a stone piece be suspended from a spring balance suspended by a hand finger. The balance shows the, actual weight of the stone., When the balance is released from the hand finger, the balance falls freely with the hanging stone piece., The balance shows a zero reading. This proves that the freely falling stone is weightless., , 0, 100, 200, , 200 gf, , 300, 400, 500, , 0, , 0 gf, , 100, 200, 300, 400, 500, , Spring, balance, Figure 3.4, (a): Theshowing, spring, balance, showsof, theaweight, weight, stoneof, the stone, (a), , Freely, rallying, balance, Figure, 3.4 (b):spring, Freely falling, spring, balance, withweight, the stone, shows, zero, showing a (b), zero reading, , Weightlessness of an Astronaut in a Satellite (Space Ship):, , A satellite is a free falling body orbiting round the earth. It tries to reach the earth but its path being parallel, to earth’s surface, it does not reach the earth. Hence the satellite and all the bodies inside it become weightless., It is due to this situation of weightlessness of astronauts that they are shown floating in spaceship in films, on television., , PLANCESS CONCEPTS, A person feels weightlessness on a satellite. This is because the gravitational force exerted by, the earth on the man is just equal to the necessary centripetal force there and the normal, reaction between the foot of the man and the floor of the satellite becomes zero., , Vipul Singh, AIR 1, NSTSE 2009, , www.plancess.com

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3.21, , Foundation for Physics, , 10. Motion of Planets, M1, , Planets move around the sun or satellites around planets under the action of, gravitational force of attraction. The centripetal force required is provided by, gravitational force of attraction., , V0, r, , GM1M2 M2V02, 2πr, =, ⇒ V=, 0, 2, r, T, r, , M2, , Figure 3.5: Orbits of, planets, , Illustration 12: A remote sensing satellite of the earth is revolving in a circular orbit at a height of 400 km, , above the surface of earth. What is the (a) orbital speed and (b) period of revolution of satellite? Radius, of earth = 6 × 106 m and acceleration due to gravity at the surface of earth is 10 m/s2., Solution: Here, R = 6 × 106 m, , g = 10 m/s2 h = 400 × 103 m = 0.4× 106 m, (a) Orbital speed v = R, , 10, g = 6 × 106, R +h, 6 × 106 + 0.4 × 106, , (b) Period of revolution T =, , 2 × 3.14, 2π (R + h)3, =, R, g, 6 × 106, , = 7.5 × 103 m/s, , (6 × 106 + 0.4 × 106 )3, = 5368.5 sec =1hr 30 min, 10, , Illustration 13: The satellite Aryabhatta is revolving round the earth at a height of 650 km and completes, , one revolution in 96 minutes. If the earth’s radius be 6400 km, find acceleration of Aryabhatta towards, the centre of earth., 2, Solution: Mv = centripetal force, , (R + h), , ∴ Acceleration=, , v2, (R + h), , ;, , 2, ,  2π , =, g'   (R + h) ; g’ = 8.3 m/s2,  T , C, , 10.1 Kepler’s Laws of Planetary Motion, , D, , The motion of planets are also governed by gravitational forces of, attraction. The inverse square law by Newton can be explained on, the basis of Kepler’s laws., , O, Sun, , B, , A, Figure 3.6: Path of planet, around sun, , www.plancess.com

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Gravitation and Floatation, , 3.22, , The three laws of Kepler are as follows:, 1. The orbit of a planet is an ellipse with the sun at one of foci., 2. The line joining the planet and the sun sweeps equal areas in equal intervals of time. (Areal velocity, is constant)., 3. The cube of mean distance of a planet from sun is proportional to square of its orbital time period, T. i.e. T 2 ∝ r 3, Suppose orbital velocity is v and radius is r. Then force acting on orbiting planet is given by, F∝, , v2, r, , v=, , 2,  1   r3 , 2πr, , so v 2 ∝ r 2 ; v 2 ∝    2 , T, T,  r   T , , r3, T, , 2, , = constant by Kepler’s 3rd law, , Hence, or v 2 ∝, , 2, 1, and F ∝ v ∝ 12, r, r, r, , Illustration 14: Compare the period of rotation of planet Mars about the Sun with that of the earth about, , Sun. The mean distance of Mars from the sun is 1.52 times the distance of earth from sun., Solution: According to Kepler’s third law, T 2 ∝ R 3 or,, , 2, , 3, ,  T1 ,  R1 ,   =, ,  T2 ,  R2 , , = (1.52)3, , T1, = (1.52)3/2 = 1.8741, T2, Illustration 15: A Saturn year is 29.5 times the earth year. How far is the Saturn from the Sun if the earth, , is 1.50 × 108 km away from the sun?, Solution: According to Kepler’s law T2 = R3, , ∴ To find the distance of Saturn from sun we take ratio of time period of both planets, 2, , 3, , i.e.  T1  =  R1 ,  T2 ,  R2 , , Given, T1 = 29.5T2; R2 = 1.50 × 108 km = 1.50 × 1011 m, Hence, (29.5)2 =, , R13, , (1.50 × 1011 )3, , or, R13 = (29.5)2 × (1.50 ×1011)3 = 2.937 ×1036, , www.plancess.com

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Foundation for Physics, , ∴ R1, =, , 3, , 3.23, , 2.937 × 1036 = 1.43 × 1012 m, , Illustration 16: What will be the ratio of the distance moved by a freely falling body from rest in 4th and, , 5th second of its journey?, Solution: The distance travelled in nth second, ,  2n − 1 , Sn= u + , g,  2 ,  2× 4 −1 , 0+, Hence h4 =,  × 9.8 ,  2 , , . . . (i), ,  2×5 − 1 , 0+, and h5 =,  × 9.8 ,  2 , , . . . (ii), , From (i) and (ii), , h4 7, =, h5 9, , PLANCESS CONCEPTS, If an artificial satellite revolves around the earth in an equatorial plane with a time period of, 24 h in the same sense as that of earth, then it will appear stationary to the observer on earth., Such a satellite is known as geostationary satellite or Parking satellite. The orbital radius of, geostationary satellite is 42,400 km and its height above the surface of earth is 36,000 km., The orbital velocity of geostationary satellite is 3.08 km/s., Neeraj Toshniwal, AIR 23 , NSO, , 11. Fluids and Pressure, Matter is basically divided into three categories, solids, liquids and gases. In solids, the inter-molecular, forces are strong, so that the shape and size of solids do not change easily. In liquids, this force is, comparatively less and so the shape is easily changed. Even though the shape of a liquid can be easily, changed, the volume of a given mass of liquid is not so easy to change. But in case of gases, since the, intermolecular forces are very small, both the shape and density can be changed easily. Liquids and gases, together are called fluids, i.e., that which can flow., , www.plancess.com

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3.24, , Gravitation and Floatation, , 11.1 Pressure, When a force acts perpendicular on a surface and is uniformly distributed over an area A of the surface,, F, A, , then the pressure on the points over the area is defined as P= =, , Force, Area, , Hence, pressure is the force acting perpendicular on a unit area of the object., Unit of Pressure, , SI unit of pressure is Newton/metre2 (N/m2) which is also called Pascal (Pa)., 1 Pascal = 1 N/m2., Another unit of pressure is atmosphere (atm). 1 atm = 1.013 × 105 Nm-2 (or Pa), 1 atm or one atmosphere is the pressure exerted by our atmosphere on earth surface due to the weight, of atmosphere., Pressure is a Scalar Quantity: Scalar quantities are those which do not have any direction. The physical, quantities which have both magnitude and direction are called vector quantities. Pressure is a scalar, quantity because at one level inside the liquid, the pressure is exerted equally in all direction, which, shows that a direction is not associated with hydrostatics pressure or pressure due to a static fluid., , PLANCESS CONCEPTS, A fluid contained in a vessel exerts pressure at all points and in all directions. A solid exerts, pressure only on the surface on which it is placed i.e., at its bottom, but a fluid exerts pressure, on the bottom as well as on the walls of the container due to its tendency to flow., Neeraj Toshniwal, AIR 23 , NSO, , 11.2 Pressure in a Liquid, The magnitude of the force exerted by a liquid on a surface inside it does not depend on how the surface, is tilted. Surface is horizontal, vertical or at any angle, the force has the same magnitude. It means the, pressure at a given point in the liquid is uniquely defined. This force due to liquid acts normal to area,, pressing the surface. There are two important rules regarding pressure in a liquid., (i) The pressure in a liquid is the same at all the points at the same horizontal level., (ii) The pressure increases with depth., , www.plancess.com

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Foundation for Physics, , F, , A, F, , A, , (a), , A, , (b), , 3.25, , F, , (c), , Figure 3.7: Pressure on the surface in a fluid, , In the Figure 3.7, a solid rod immersed in a liquid is shown in three cases. Consider the shaded surface, in figures (a), (b) and (c)., In figure 3.7 (a) surface of area ‘a‘ is horizontal and its centre is at point A in the liquid. The force exerted, by the liquid on this surface is in downward direction. Pressure at the point A is p = F/a where F is the, magnitude of the force., In figure 3.7 (b) the centre of the surface is still at the same point A, but the surface itself is now vertical., In this case again, pressure at point p = F/a, where F is the magnitude of force towards right., In figure 3.7 (c) the considered surface is at the bottom. The force exerted by the liquid is now upwards., The centre of the surface is still at the same point A and the magnitude of the force is also same, and, pressure at A is p = F/a., Consider a liquid of density ρ contained in a cylindrical vessel of cross sectional area a. Let h be the height, of liquid column and g be the acceleration due to gravity. The weight of liquid will exert a downward, thrust on the bottom surface of the vessel. Therefore, pressure due to liquid acts on that surface., Weight of liquid inside the vessel = volume × density of liquid × acceleration due to gravity, Thrust of liquid on area a = weight of liquid = a h ρ g, Liquid pressure on the base of vessel is, P=, , thrust ahρg, =, = hρg ...(i), area, a, , , , h, , a, Figure 3.8: Liquid pressure on the base of vessel, , www.plancess.com

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3.26, , Gravitation and Floatation, , NOTE, (i) The liquid at rest exerts equal pressure in all directions at a point inside the liquid., (ii) The liquid at rest exerts equal pressure at all those points which are in one level inside the liquid., (iii) Liquid pressure is independent of shape of the liquid surface, but depends upon the height of, liquid column., (iv) Total pressure at a depth h below the liquid surface = P0 + hρg where P0 = atmospheric pressure., (v) Pressure is a scalar quantity., (vi) Mean pressure on the walls of a beaker containing liquid upto height h is (= hρg /2 ), where ρ is, the density of liquid., (vii) Thrust exerted by liquid on the walls of the vessel in contact with liquid is normal to the wall, surface., , 11.3 Thrust, The total force acting normal to any given area which is immersed inside a liquid is also called thrust on, the area., , 11.4 The Mercury Barometer, Given figure 3.9 shows a very basic mercury barometer, a device used, to measure the pressure of the atmosphere. The long glass tube is, filled with mercury and inverted with its end in a dish of mercury, as, the Figure shows., The space above the mercury column (called Torricellian vacuum), contains only mercury vapour, whose pressure is so small at ordinary, temperatures that it can be neglected (PA = 0)., , PA=0, P atm, C, , H, B, , Hg, , Figure 3.9: Mercury barometer, Let us consider a point C on the mercury surface in the cup and, another point B in the tube at the same horizontal level. The pressure at C is equal to the atmospheric, pressure. As B and C are in the same horizontal level, the pressure at B and C are equal. Thus, the pressure, at B is equal to the atmospheric pressure in the lab. ∴ PC = PB = PA + rgH or PB = rgH [ PA = 0], , where ρ is the density of mercury and H is depth of point B below A. At sea level H = 76 cm, Atmospheric Pressure, , The pressure at any place due to atmosphere is called atmospheric pressure. Its value varies from place, to place. The value of atmospheric pressure at the earth’s surface near the sea level is approximately, 1.01 × 105 Pa, which is also known as 1 atmospheric pressure (atm). Thus atm is the unit of pressure., 1 atm = pressure exerted by a mercury column of length 760 mm = 760 mm of Hg., , www.plancess.com

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Foundation for Physics, , 3.27, , PLANCESS CONCEPTS, The pressure at a point in a fluid in static equilibrium depends on the depth of that point but, not on any horizontal dimension of the fluid or its container., Shivam Agarwal, AIR 3, NSO, Illustration 17: A force of 40 N is applied on a nail, whose tip has an area of cross-section of 0.001 cm2., , Find the pressure on the tip., Solution: Given that, force F = 40 N and area A = 0.001 cm2 = 10–7 m2, F, ∴ Pressure (P), = =, A, , 40, 10−7, , = 4 ×108 N/m2, , Illustration 18: Atmospheric pressure is nearly 100 KPa. How large force does the air in a room exert on, , one side of a window of dimensions 40 cm×80 cm?, Solution: Given that P = 100 KPa = 100 × 103 Pa, , P = 105 Pa, And A = 40 cm ×80 cm = 40 × 80 × 10–4 m2, ∴ F = PA = 105 Pa × 40 × 80 × 10–4 m2; ⇒ F = 32 KPa= 32 N (∵ 1KPa = 1N), Illustration 19: A tank 5 m high is half filled with water and then is filled to the top with oil of density, , 0.85 g cm–3. What is the pressure at the bottom of the tank due to these liquids?, A × lw × ρw × g + A × ll × ρl × g, Solution: Pr essure = Total weight of water & liquid =, = 453250 dyne/cm2, base area of tank, , A, , 11.5 Some Facts Involving Thrust and Pressure, (i) Nails have a flat top but pointed end, A small pressure applied on the flat end of the nail through falling hammer becomes a large thrust., The same thrust acts on the wooden board through the pointed end of the nail. It results in a large, pressure. The nail can easily be fixed in the wooden board., (ii) Sewing needle have pointed tips, A small force of fingers makes the needle pierce into the cloth easily and sewing becomes quicker., (iii) Cutting items (knives and blades) have sharp edges:, Cutting becomes easier., , www.plancess.com

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3.28, , Gravitation and Floatation, , (iv) Reducing Pressure:, (a) Vehicle brakes have flat surface, This reduces pressure on the vehicle tyres and avoids their tearing., (b) Broad sole shoes, Broad sole shoes make walking easier on a soft land., (c) Wide steel belt on army tank, Wide steel belt over the wheels of an army tank, makes its movement easier over marshy land., (d) Tractor tyres are broad, Tractors do not sink in the soft land of the field while operating on them., (e) Camel foot are broad and soft, They walk swiftly on sand., (f ) Hanging bags have wide straps:, They reduce pressure on the shoulders., , 12. Buoyancy and Force of Buoyancy (Buoyant Force), Introduction, , When a body is immersed in a fluid (liquid or gas), it displaces the fluid whose volume is equal to the, volume of the body immersed in the fluid. This displaced fluid exerts an upward force on the body. Due, to this, the object loses weight. This upward force is called buoyant force. Buoyant force is also known, as upthrust., Definition, , This tendency of the displaced fluid (exerting an upward force) is called buoyancy. The upward applied, force, is called the force of buoyancy or upthrust. It is equal to the weight of the fluid displaced by the body., Examples, , 1. When a piece of cork is held below the surface of water and then released, the cork immediately, rises to the surface. It means some upward force is exerted by water on the cork which pushes it to, the surface., 2. When a stone lying at the bottom of a pond is lifted up, it appears to be light as long as it is being lifted, inside water. But as soon as the stone is lifted out of water into air, one feels it to be much heavier. It, means some upward force acts on the stone when it is immersed in water and makes us feel lighter., , www.plancess.com

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3.29, , Foundation for Physics, , 3. When a bucket is pulled out of a well, it is felt lighter so long as it remains immersed in water,, inside the well. It acquires its actual weight when out of water., , 12.1 Cause of Buoyant Force, Consider a rubber ball dipped completely inside a liquid filled in a beaker. The, liquid in contact with the ball exerts a force in perpendicular direction, pressing, it. Therefore, the force on the surface near the topmost point A is downwards,, whereas on the surface near the lowest point, C it is upwards. The force on the, surface near the point B is towards the left and near the point D is towards the, right. Now, the forces acting on it are not equal in magnitude., , A, D B, C, , The point C lies at a greater depth than point A. Therefore pressure at C is, larger than A. Consider equal areas near A and C., It means upwards force near C is larger in magnitude than the downward force, near A., , Figure 3.10: Force, on a body in a liquid, , Therefore, the forces exerted on different parts of the ball by the liquid are not balanced. Hence the, resultant force exerted by the liquid acts in upward direction, which is called buoyant force., , 12.2 Factors Affecting Buoyant Force, The magnitude of buoyant force acting on an object immersed in a liquid depends on two factors:, (i) Volume of the solid object immersed in the liquid, As the volume of object immersed inside the liquid increases, the upward buoyant force increases., When the object is completely immersed in the liquid, the buoyant force becomes maximum., (ii) Density of the liquid in which the object is immersed, As the density of liquid increases, the buoyant force exerted by the liquid also increases., E.g. sea-water has higher density than fresh water, therefore sea-water will exert more buoyant, force on an object immersed in it than the fresh water., Let us perform the following two activities:, (i) Take two wooden blocks of different sizes. Push the small block inside the water in a tub and, release it. You will find that the wooden block rises up and come to the surface of water. It rises up, because upthrust or buoyant force acting on it is more than its weight., Now push the large wooden block inside the water and release it. You will find that the large block, rises up faster than the small block. It means, the upthurst or buoyant force acting on the large, block is more than on the small block., , www.plancess.com

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3.30, , Gravitation and Floatation, , (ii) Now add some salt in water so that the density of solution (water + salt) increases. Push a wooden, block inside the solution and release it. You will find that the block rises up faster in a solution than, in pure water. It means, the upthrust or buoyant force acting on a body is more in a liquid having, more density than in a liquid having less density., Conclusion, , From the above mentioned activities, we conclude that upthrust or buoyant force depends on: (i), The size or volume of the body immersed in a liquid., (ii) The density of the liquid in which the body is immersed., , 12.3 Buoyancy, The tendency of an object to float in a liquid or the power of liquid to make an object float is called, buoyancy., Remember these points, , (i) An object whose weight (i.e., downward gravitational force) is greater than the upthrust of the, liquid (say water) on the object, sinks in the liquid. This is possible if density of object is more than, the density of liquid., (ii) An object whose weight (i.e. downward gravitational force) is less than the upthurst of the liquid, on the object, floats on the liquid. This is possible if density of object is less than the density of, liquid., Illustration 20: A weighing machine on which a beaker containing water is placed reads w. What happens, , to its reading if Prakash keeps his hand so that only fingers are immersed in water?, Solution: Th,  e weight w increases. In this case though the total weight of hand is not acting on water,, the hand apparently loses some weight equivalent to buoyant force i.e. the water gets pushed, down with the same force. Hence w increases (by a magnitude equal to buoyant force), , 13. Archimedes Principle, A Greek scientist Archimedes conducted many experiments and concluded that when a body or an, object is immersed partially or completely in a liquid or gas (i.e. fluid), it experiences an upthrust or, buoyant force. The upthrust or buoyant force is equal to the weight of the fluid displaced by the body. It, is known as Archimedes principle., , www.plancess.com

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Foundation for Physics, , 3.31, , 13.1 Statement of Archimedes Principle, When a body is immersed partially or completely in a fluid (liquid or gas), it experiences an upthrust or, buoyant force which is equal to the weight of the fluid displaced by the body., , 13.2 Proof of Archimedes Principle, Consider a cylindrical body of cross-sectional area ‘a’ submerged in a liquid of density ρ. Let the upper, face of the body is at a depth h1 below the surface of the liquid and the lower face is at a depth h2 below, the surface of the liquid. The pressure exerted by the liquid on the upper surface of the body is given, by p1 = h1 ρg ., Downward thrust on the upper surface of the body is,, , F1, , F1 = p1 × a = h1 ρg × a, Pressure exerted by the liquid at the lower surface of, the body,, , h2, , P2 = h2 ρg, , h, , Upward thrust on the lower surface of the body is,, F2 = P2 × a = h2 ρg × a, The horizontal thrusts acting on the vertical sides of, the body being equal and opposite from all the sides, cancels out., , h1, , F2, , density, p, , Figure 3.11: Net force on a body, immersed in a fluid, , ∴ Resultant upthrust or buoyant force acting on the body is,, F = F2 - F1 = h2 ρg a - h1 ρg a = (h2 - h1) ρg a, Since volume of the body, V = (h2 - h1)a, , ∴ F = V ρg, , Which implies that product of the volume of the body, the density of the liquid and the acceleration due, to gravity, which gives the weight of the liquid displaced, is equal to the buoyant force., Thus, when a body is submerged in a liquid, it experiences an upward thrust equal to the weight of the, liquid displaced by the body., , 13.3 Verification of Archimedes Principle, To verify the Archimedes Principle we take following steps:, (i) Take a small piece of stone and suspend it with a spring balance. Let the weight of the stone, indicated by the spring balance be W1., , www.plancess.com

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3.32, , Gravitation and Floatation, , (ii) Now take an empty beaker and measure its weight by suspending it with the spring balance with, the help of a thread of negligible mass. Let the weight of the empty beaker be W2., (iii) Take a Can having a side tube known as spout. Fill Can with water upto the level of spout., (iv) Lower the stone suspended with a spring balance inside the water. The stone displaces the water, which comes out of the Can through the spout. The water coming out of the Can is collected in, the beaker. When the water stops coming out of the spout, note the reading of the spring balance., This reading shows the weight of the stone inside the water. Let the weight of the stone inside the, water be W3. It is seen that W3 is less than W1., (v) Now measure the weight of the beaker along with the water collected in it. Let this weight be W4., (vi) Now find (W1 - W3). This difference in weight is equal to the loss of weight of the stone immersed, in water (i.e. upthrust or buoyant force)., (vii) Also find (W4 - W2). This difference in weight is equal to the weight of the water displaced by, the stone., (viii) It is found that (W1 - W3) = (W4 - W2). That is upthrust or buoyant force is equal to the weight, of the water displaced. Thus, Archimedes principle is verified., , w2, w1, , Cane, Water, , Spout, Beaker, , Fig: 3.12: Verification of Archimedes principle, , 13.4 Density, The ratio of mass and volume of the body is known as the density of the material of the body., Density =, , mass, M, ; ρ=, volume, V, , If V = 1 m3, then, ρ = M, or the mass per unit volume is known as the density of the material of the object., , www.plancess.com

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Foundation for Physics, , Unit of density: ∴, , ρ=, , 3.33, , M, → g / cm3 (in C.G.S.), V, , ρ → kg/m3 (in S.I. system), , 13.5 Relative Density or Specific Gravity, The ratio of density of the substance to the density of water at 40C is known as the relative density of the, substance., Relative density of substance =, , density of substance, density of water at 40 C, , It is pure number having no unit., Illustration 21: The relative density of silver is 10.5. The density of water is 103 kg/m3. What is the, , density of silver in S.I. unit?, Solution:, , Density of water dw = 103 kg/m3, , Relative density (R.D.) of silver = 10.5, Density of silver dAg = ?, , ∴ R.D. = density of silver ⇒ 10.5 =, density of water, , dAg, 3, , 10, , ⇒ dAg = 10.5 × 103 kg/ m3., , PLANCESS CONCEPTS, Why ships float?, Ship is not a solid block of iron and steel. A ship is a hollow object made of iron and steel, which contains a lot of air in it. Air has a very low density. Due to the presence of a lot of air, in it, the average density of the ship becomes less than the density of water, therefore, a ship, floats in water. In fact, all the hollow objects made of dense materials (like metals) float in, water because due to the presence of a lot of air in them, their average density becomes less, than the density of water., Neeraj Toshniwal, AIR 23 , NSO, , www.plancess.com

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Gravitation and Floatation, , 3.34, , 13.6 Uses of Archimedes Principle, Relative density for solids and liquids can also be determined with the help of Archimedes principle., (i) For solids:, By definition, we have, Relative density =, , Relative density =, , Density of subs tan ce, Density of water, , =, , Weight of certain volume of subs tan ce, Weight of same volume of water, , Weight of a body, loss in weight when fully immersed in water, , Relative density of a solid can be measured by weighing it first in air and then when fully immersed, in water., Let weight of the body in air = W1, Weight of solid body in water = W2, , ∴, , Loss in weight = W1 - W2, , =, R.D., ∴, , W1, Weight of solid body in air, =, Loss in weight in water, W1 − W2, , (ii) For liquids:, To measure relative density of a liquid, choose a body which can be fully immersed in water as, well as in the given liquid. The body is weighed first in air, then fully immersed in water and then, fully immersed in that particular liquid., R.D. =, , Density of liquid, Density of water, , R.D. =, , R.D. =, , Weight of liquid displaced by a body, Weight of water displaced by the same body, , loss of weight in liquid, loss of weight in water, , Let the weight of a body in air = W, Weight of the body fully immersed in water = W’, Weight of the body fully immersed in liquid = W’’, Then loss of weight in liquid = W - W’’, and loss of weight in water = W - W’, , ∴ Relative density of the liquid =, , W − W", W − W', , www.plancess.com

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Foundation for Physics, , 3.35, , (iii) Archimedes principle is used to design:, (A) the ships and submarines., (B) the hydrometers to find the densities of liquids., (C) the lactometers to test the purity of milk., Table 3.3: Table of densities and relative densities of some substances:, , S.No., , Name of Substance, , Density at S.T.P. in (kg m-3), , Relative Density, , 1, , Air, , 1.29, , 1.29 × 10-3, , 2, , Wood, , 800, , 0.80, , 3, , Ice, , 920, , 0.917, , 4, , Water, , 1000, , 1.00, , 5, , Glycerin, , 1260, , 1.26, , 6, , Glass, , 2500, , 2.50, , 7, , Aluminium, , 2700, , 2.70, , 8, , Iron, , 7900, , 7.90, , 9, , Silver, , 10500, , 10.50, , 10, , Mercury, , 13600, , 13.60, , 11, , Gold, , 19320, , 19.32, , 13.7 Physical Meaning of Relative Density, Relative density of a substance is the number of times the given substance is heavier than the equal, volume of water. When we say that the relative density of silver = 10.5, it means, silver is 10.5 times, heavier than equal volume of water., , 13.8 Law of Floatation, , W=B, , W<B, , Law of floatation is an extension of Archimedes principle., When a body is immersed partially or wholly in a fluid, then the various forces, acting on the body are (i) upward thrust (B) acting at the centre of buoyancy, and whose magnitude is equal to the weight of the liquid displaced and (ii) the, weight of the body (W) which acts vertically downwards through its centre of, gravity., , W>B, Floatation, Figure 3.13:, Condition of floating, , www.plancess.com

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3.36, , Gravitation and Floatation, , (i) When W > B, the body will sink in the liquid., (ii) When W = B, then the body will remain in equilibrium inside the liquid., (iii) When W < B, then the body will come upto the surface of the liquid and is immersed inside the, liquid in such a way that the weight of the liquid displaced due to it balances the weight of body., Thus, law of floatation is defined as follows:, Definition, , A body floats in a liquid if weight of the liquid displaced by the immersed portion of the body is equal, to the weight of the body., , 13.9 Relation between Density of Solid and Liquid, Let ρ1 be the density of the solid whose volume is V1. Let ρ2 be the density of the liquid and the volume, of the portion of the solid immersed in the liquid be V2., Now, weight of the floating solid = weight of the liquid displaced., ρ1 V2, =, ρ2 V1, , i.e., , V1ρ1g = V2ρ2g ∴, , or, , Density of solid Volume of the immersed portion of the solid, =, Density of liquid, Total volume of the solid, , = Fraction of volume of body immersed in liquid, , 13.10 Equilibrium of Floating Bodies, From law of floatation, we know that a body will float in a liquid when its weight W is equal to the, weight w of the liquid displaced by the immersed part of the body. But this does not necessarily indicate, that the body will be in equilibrium. A body will be in equilibrium only if the resultant of all the forces, and couples acting on the body is zero. Thus, a floating body can be in equilibrium if no couple acts on, it. It will be so if the line of action of W and w is along the same vertical straight line. Thus, there will be, equilibrium of floating bodies if the following conditions are fulfilled:, (i) A,  body can float if the weight of the liquid displaced by the immersed part of body must be equal, to the weight of the body., (ii) A body can be in equilibrium in the centre of gravity of the body and centre of buoyancy must be, along the same vertical line., (iii) The body will be in stable equilibrium if centre of gravity lies vertically above the centre of, buoyancy., , www.plancess.com

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Foundation for Physics, , 3.37, , NOTE, When an ice block is floating in water in a vessel, then the level of water in the vessel will not change, when the whole ice melts into water., When an ice block is floating in a liquid in a vessel and ice completely melts, then the following cases, may arise for the level of liquid in the vessel., (i) If density of liquid is greater than that of water i.e., ρL > ρw the level of liquid plus water will rise., (ii) If density of liquid is less than the density of water i.e., ρL < ρw the level of liquid plus water will, decrease, (iii) If density of liquid is equal to the density of water i.e., ρL =ρw , the level of liquid plus water will, remain unchanged., , 13.11 Conditions for Floatation in Terms of Densities, For bodies with no air space:, , An ice cube, a wooden block etc. are bodies with no air space. The condition for the flotation of such, bodies can be expressed in terms of the densities of the bodies and the concerned liquids., Suppose the volume of a body is V and its density is ρ. It is completely, immersed in a liquid of density σ., Mass of the body (M) = Vρ, Weight of the body (W) = Mg = Vρg, , B, , Volume of the displaced liquid = V, Weight of the displaced liquid = Force of buoyancy (B) = Vσg, , W, , Body will come up and float if,, B > W ⇒ Vσg > Vρg, , Figure 3.14: Condition, for floation, , Or, σ > ρ, ⇒ Thus body floats, if the density of the body is less than that of the liquid., For bodies with Air Space, , The shape of the body plays an important role in deciding whether it sinks or floats, e.g. A small needle, made of stainless steel sinks in water, but a large bowl of same material and same density floats in water., Density of steel is much greater than of water. The steel bowl floats in water. This is because a bowl can, displace a much greater volume of water than the volume of needle., , www.plancess.com

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Gravitation and Floatation, , 3.38, , Illustration 22: A solid floats in water with 3/4th of its volume below the surface of water. What is the, , density of the substance? (Density of water = 1000 kg/m3), Solution: Let the volume and density of the solid be V and ρ respectively., , Weight of the solid = Vρg, Volume of block under water = (3V/4)m3 = volume of water displaced,  3V, , Weight of water displaced =  × 1000 × g  N,  4, , , Since, weight of body = weight of water displaced by it, ∴ Vρg=, , 3V, × 1000 × g, 4, , ⇒ ρ = 750 kg/m3, , We =, can say ⇒Density of body Volume of body immersed × 1000kg / m3, Total volume, , Illustration 23: A solid sphere weighs 2.0 N in air and 1.6 N when totally immersed in water. What is, , the magnitude of the up thrust?, Solution: Weight of sphere in air = 2.0 N, , Weight of sphere in water = 1.6 N, Loss in weight of the sphere = (2.0 –1.6) N = 0.4 N, ∴ Buoyant Force (up thrust) = Loss in the weight of sphere in water = 0.4 N, , 13.12 Apparent Weight, When a body is dipped in a liquid, the liquid exerts an upward force on it which is called buoyant force., Due to which less force is needed to hold the body. This, weight of the body inside a liquid is called the, apparent weight of the body. Apparent weight of a body dipped in a liquid is less than its true weight., Illustration 24: Four fifths of a cylindrical block of wood floats inside a liquid. If the relative density of, , the wood is 0.7, find the density of the liquid., Solution: Let V be the volume of the block., , Density of the block = 0.7 × 1000 kg/m3, Volume of the block under the liquid =, , 4V, 5, , www.plancess.com

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Foundation for Physics, , 3.39, , Weight of the body = Weight of the liquid displaced (since body floats), V × 700 ×=, g, , 5, 4V, × 700= 875 kg / m3, × ρ × g ⇒ ρ=, 5, 4, , Illustration 25: A solid sphere has a radius of 1.5 cm and a mass of 0.038 kg. What is the relative density, , of the sphere?, Solution: Given that, mass of sphere m = 0.038 kg, and radius r = 1.5 cm, 4, 3, , 4, 3, , Volume of the sphere (V) = πr 3 = × 3.14 × (1.5 × 10−2 )3 = 1.413 ×10–5 m3, m, v, , 0.038, Density of the sphere, (d) =, =, =, 2690kg / m3, −5, 1.413 × 10, , subs tance, Relative density = = density of =, density of water, , 2690kg / m3, = 2.69, 1000kg / m3, , Illustration 26: A flat bottomed loaded test-tube sinks to a depth of 10 cm in a liquid and 12 cm in water., , Calculate the relative density of the liquid., Solution: L,  et the cross section of the tube be = A cm2 and the, , density of the liquid be ‘d’ g/cm3 and that of water ‘d1’, g/cm3, , 12 cm, , 10 cm, ,  ass of liquid displaced = volume displaced × density of, M, liquid = A × 10 × d grams, Mass of water displaced = A × 12 × d1 grams, , Water, , liquid, , Mass of liquid displaced = Mass of water displaced (Since test-tube floats), i.e. A × 10 × d = A × 12 × d1 or,, But, , d, d1, , d, 12, = = 1.2, d1 10, , is specific gravity = 1.2, , Hence relative density of the liquid = 1.2, Illustration 27: The relative density of a body weighing 20 N in air is 4. Find its weight in water., , Solution: R.D =, , Wair, Wair − Wwater, , ; 4=, , 20, 20 − Wwater, , ⇒ Wwater = 15 N, , www.plancess.com

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3.40, , Gravitation and Floatation, , Illustration 28: The weight of a solid is 80 g in air and 50 g in a liquid of relative density 1.5. Calculate, , the relative density of the solid., Solution: Weight of solid in air = 80 g, , Weight of solid in liquid = 50 g, ∴ Loss of weight of solid in liquid = (80 – 50) g = 30 g, If the displaced liquid has a volume of V cm3, then its weight = V × 1.5 = 30 g, ∴ V = 20 cm3, Since solid is completely immersed in liquid, so volume of solid = V, The volume of water that will be displaced by same solid = 20 cm3, ∴ Weight of this displaced water = 20 g, So we have specific gravity of solid =, , weight of solid in air=, weight of water displaced, , 80 g, = 4, 20 g, , www.plancess.com

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Foundation for Physics, , 3.41, , SUMMARY, •• The force of gravitation acts even if the two objects are not connected by any means., •• Universal law of gravitation: The gravitational force of attraction between two particles is directly, proportional to the product of the masses of the particles and is inversely proportional to the square, of the distance between the particles., •• The direction of the force is along the line joining the two particles., F=G, , m1m2, r2, , Here G is a constant known as the universal constant of gravitation. In S.I. G = 6.67 × 10-11 Nm2 kg2, •• Gravitational forces between two bodies form an action and reaction pair i.e., the forces are equal in, magnitude but opposite in direction., •• Gravitational force between two bodies is independent of the nature of the intervening medium., •• If gravity is the only acting force (meaning that air resistance is neglected), all objects move with the, same acceleration near the earth’s surface. This acceleration is called the acceleration due to gravity,, whose magnitude of ‘g’ is given by, g=, , GMe, R 2e, 2, , −11 Nm, × 6 × 1024 kg,  6.67 × 10, 2 , , kg , =, 9.8ms −2, 2, 6.4 × 106 m, , (, , g, , (, , ), , ), , •• The direction of this acceleration is towards the centre of the earth, i.e., in the vertically downward, direction., •• When a force acts perpendicular on a surface and is uniformly distributed over an area A of the, surface, then the pressure on the points over the area is defined as, P, =, , F Force, =, A Area, , www.plancess.com

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3.42, , Gravitation and Floatation, , •• Pressure is the force acting perpendicular on a unit area of the object. SI unit of pressure is Newton/, metre2 (N/m2) which is also called Pascal (Pa)., 1 Pascal = 1 N/m2., •• The pressure in a liquid is the same at all the points at the same horizontal level. The pressure increases, with depth., •• When a body is immersed in a fluid (liquid or gas), it displaces the fluid whose volume is equal to the, volume of the body immersed in the fluid., •• The displaced fluid exerts an upward force on the body. Due to this, the object loses weight. This, upward force is called buoyant force. Buoyant force is also known as upthrust., The tendency of the displaced fluid (exerting an upward force) is called buoyancy. The upward applied, force, is called the force of buoyancy or upthrust. It is equal to the weight of the fluid displaced by, the body., •• The ratio of mass and volume of the body is known as the density of the material of the body., •• The ratio of density of the substance to the density of water at 40C is known as the relative density of, the substance., •• A body floats is a liquid if weight of the liquid displaced by the immersed portion of the body is equal, to the weight of the body., , www.plancess.com

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Foundation for Physics, , 3.43, , SOLVED EXAMPLES, Example 1. What is the force of gravitation between two point masses of 1 kg and 2 kg kept 1 m apart?, Solution: m = 1 kg, m2 = 2 kg, r = 1 m, F=G, , m1m2, r2, , = 6.67 × 10 −11 ×, , 1×2, 2, , 1, , =13.34 × 10 −11 N., , This is an extremely small force., Example 2. Calculate the force of gravitation between the earth and the sun., Mass of Earth = 6 ×1024 kg, Mass of Sun = 2 × 1030 kg. The average distance the two is 1.5 × 1011m., Solution: F = Gm12m2, r, , =, , 6.67 × 10−11 × 6 × 1024 × 2 × 1030, (1.5 × 1011 )2, , ,  6.67 × 6 × 2   1024 × 1030, = , ×  11, × 10−11  =35.57 × 1021 N, , 11, ,  1.5 × 1.5   10 × 10, , , Example 3. Write down the expression for acceleration experienced by a particle on the surface of the, moon due to gravitational force on the moon. Find the ratio of this acceleration to that experienced by, the same particle on the surface of the earth. If the acceleration due to gravity on the earth is 9.8 ms-2,, what is the acceleration of a particle on the moon’s surface? Mass of moon = 7.3 × 1022 kg; Mass of Earth, = 6 × 1024 kg. Radius of moon = 1.74 × 106 m, Radius of earth = 6.4 × 106 m., Solution: Acceleration on moon, gm =, , GMm, 2, Rm, , Also acceleration on the surface of Earth, gE =, , GME, RE2, , Dividing (1) by (2), gm GMm RE2, MmRE2, 7.3 × 6.4 × 6.4, 7.3 × 1022 × 6.4 × 106 × 6.4 × 106, ;, =, ×, =, =, =, 2, 2, 6, 6, 24, gE, GME RmME, 1.74 × 1.74 × 6 × 100, RE, 1.74 × 10 × 1.74 × 10 × 6 × 10, , www.plancess.com

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Gravitation and Floatation, , 3.44, , gm, = 0.16, gE, gm= 0.16 × gE= 0.16 × 9.8= 1.57ms−2, , Example 4. Find the value of acceleration due to gravity at a height of (a) 6400 km, (b) 12,800 km from, the surface of the earth. Radius of earth is 6400 km., B, , Solution: We know that, Also g =, , GM, R2, , ∴, , gh =, , gh, ,  R , ∴ gh =, g, , R +h, , g, , =, , GM, A, , (R + h)2, , R2, , 2R, , R, , (R + h)2, 6400km, , 2, , O, , (a) g = 9.8 ms-2, R = 6400 km, R + h = 6400 + 6400 = 12800 km., 2, , 2, ,  6400 , 1, 9.8, ∴ (gh )A =, 9.8 , 9.8  =, =, = 2.45 ms −2, , , 12800, 2, 4, , ,  , , As h increases gh decreases. This is because gh and h are inversely related., , , 6400, , , , 2, , 1, , 2, , −2, (b) ∴ (gh )B= 9.8 ,  = 9.8  3  = 1.09ms, +, 6400, 12800, , ,  , , Example 5. A particle is thrown up vertically with a velocity of 50 m/s. (a) What will be its velocity at, the highest point of its journey? (b) How high would the particle rise? (c) What time would it take to, reach the highest point., Solution: At the highest point the velocity will be zero., Considering activity A to B, Using v = u + at, 0 = 50 – 9.8 × t, ∴ t=, , 50, = 5.1 s, 9.8, , www.plancess.com

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3.45, , Foundation for Physics, Also v2 – u2 = 2as, , v=0, t=?, a = -9.8 ms, s=?, , B, , 02 – (50)2 = 2 (-9.8) × s, ∴=, s, , 50 × 50, = 127.5m, 2 × 9.8, A, , u = 50 m/s, , Example 6. With reference to the above sample problem, (a) Find the time the particle takes from the, highest point back to the initial point (b) Find the velocity with which the particle reaches the initial, point., Solution: Using the data given in above problem and considering activity B to A., As vu, tu = vd , td i.e the velocity with which the, particle goes up in time tu is equal to velocity of, particle in downward motion vd with the time td., , A, , A, vu, tu, , Using v2 – u2 = 2as, vd, td, , v2 – 02 = 5(9.8) (127.5), v = 50 m/s, , B, , B, , Also v = u + at, ∴ 50 = 0 + 9.8 (t), ∴ t = 5.1 s, , Example 7. A ball is dropped from the top of a tower 40 m high. What is its velocity when it has covered, 20 m? What would be its velocity when it hits the ground? Take g = 10 m/s2., Solution: Let the point B be at a height of 20 m., , K, , Activity from K to L:, u1 = 0, a1 = 10 ms-2, s1 = 20 m, v1 = ?, v12 − u12 =, 2a1s1, ∴, , v12, , 40m, , 2, , L, , −0 =, 2 (10)(20), , ∴ v12 =, 202, , v1 = 20 ms‒1, , 20m, M, , www.plancess.com

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3.46, , Gravitation and Floatation, , Activity from K to M : M is a point on the ground, u2 = 0, a2 = 10 ms-2, s2 = 40 m, v2 = ?, ∴ v 22 − u22 =, 2a2s2, ∴ v 22 − 02 =, 2 (10) (40), ∴ v 22 =, 800 ∴ v 2 =, 28.28 ms−1, , Example 8. A body is thrown up with a speed 29.4 ms-1, (a) What is the speed after (i) t = 1s, (ii) t = 2 s and (iii) t = 3s?, (b) What is its height after (i) t = 1 s, (ii) t = 2 s and (iii) t = 3s?, Solution: (a) (i) u = 29.4 ms-1, a = - 9.8 ms-2, t1 = 1 s, v1 = ?, v1 = u + at1, , (ii), , = 29.4 + (-9.8) × 1 = 19.6 ms-1, u = 29.4 ms‒1, a = -9.8 ms-2, t2 = 2 s, v2 = ?, , v2 = u + at2, , (iii), , = 29.4 + (-9.8) × 2 = 9.8 ms-1, u = 29.4 ms-1, a = - 9.8 ms-2, t3 = 3 s, v3 = ?, , v3 = u + at3= 29.4 + (-9.8) × 3 = 0, (b) (i) u = 29.4 ms-1, a = -9.8 ms-2, s1 = h1, t1 = 1s, 1, 1, h1= ut1 + at12= 29.4 × 1 + ( −9.8) × 1= 24.5 m, 2, 2, , , (ii), , u = 29.4 ms-1, a = - 9.8 ms-2, s2 = h2, t2 = 2 s, , 1, 1, h2 = ut2 + at22= 29.4 × 2 + ( −9.8) × 22= 38.2m, 2, 2, , , (iii), , u = 29.4 ms-1, a = - 9.8 ms-2, s3 = h3, t3 = 3s, , 1, 1, h3= ut3 + at32= 29.4 × 3 + ( −9.8) × 32= 44.1m, 2, 2, , , Example 9. What is the weight of a person whose mass is 50 kg?, Solution: The weight of the person, W = mg = 50 × 9.8 = 490 N, , www.plancess.com

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Foundation for Physics, , 3.47, , Note: The gravitational unit of force is kg-f (kilogram force) or kg-wt (kilogram weight), 1 kg-wt = 9.8 N = 1 kg-f, , ∴ 490 N = 50 kg-f, Example 10. Weight of a girl is 294 N. Find her mass., Solution: W = mg, 294 = m × 9.8, =, m, , 294, = 30 kg, 9.8, , Example 11. Weight of an object is 294 N on the surface of the earth. What is its weight at a height of, 200 km from the surface of the earth? (Radius of the earth = 6400 km), Solution: Weight at a height of 200 km,  R  ,  R , =, =, Wh = mgh mg, ,   gh g , ,  R + h  , R +h, 2, , 2, , , , , , Here mg = 294 N, R = 6400 km, h = 200 km, 2, , 2, ,  6400 ,  6400 , =, Wh 294  =, =,  294,  6600  276.45N, +, 6400, 200, , , , , , Note : Weight decreases with increase of height from the surface of the earth., Example 12. The gravitational force between two objects is F. How will this force change when, (i) Distance between than is reduced to half?, (ii) the mass of each object is quadrupled?, Solution: (i) According to Newton’s law of gravitation, gravitational force F between two objects distance, r apart is F ∝ 12, r, , When distance between them is reduced to half, i.e., r’ = r/2, the force, F' ∝ 12, r, , www.plancess.com

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3.48, , Gravitation and Floatation, , F' r 2, r2, = 4 or F’ = 4F, Thus, = =, F r'2 (r / 2)2, , i.e., force becomes 4 times its previous value., (ii) A,  gain, according to Newton’s law of gravitation, the gravitational force F between two object, of masses m1 and m2 is, , F ∝ m1m2, When mass of each object is quadrupled,, m’1 = 4m1, and m’2 = 4m2, The force, F' ∝ m'1 m'2, F' m'1 m'2 (4m1 )(4m2 ), = =, = 16, Thus,, F m1m2, m1m2, , Or F’ = 16 F, i.e., force becomes 16 times its previous value., Example 13. A sphere of mass 40 kg is attracted by a second sphere of mass 15 kg when their centres are, 20 cm apart, with a force of 0.1 milligram weight. Calculate the value of gravitational constant., Solution: Here, m1 = 40 kg, m2 = 15 kg, From r = 20 cm =, , 20, = 2 × 10 −1 m, 100, , F = 0.1 milligram weight = 0.1 × 10-3 gram weight= 10-4 × 10-3 kg wt = 10-7 × 9.8 N, (1 kg wt = 9.8 N), From F = Gm12m2, r, , =, G, , F × r 2 10−7 × 9.8 × (2 × 10−1 )2, =, m1m 2, 40 × 15, , G = 6.53 × 10-11 Nm2/kg2, , www.plancess.com

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Foundation for Physics, , 3.49, , Example 14. Calculate the force of gravity acting on your friend of mass 60 kg. Given mass of earth = 6, × 1024 kg and radius of Earth = 6.4 × 106m, Solution: Here, m = 60 kg, M = 6 × 1024kg, R = 6.4 × 106m, F = ?, G = 6.67 × 10-11 Nm2/kg2, Thus, F = GMm, 2, R, , F=, , 6.67 × 10−11 × 6 × 1024 × 60, (6.4 × 106 )2, , Or F = 58.62 N, Example 15. A particle is thrown up vertically with a velocity of 50 m/s. What will be its velocity at the, highest point of the journey? How high would the particle rise? What time would it take to reach the, highest point? Take g = 10 m/s2., Solution: Here, initial velocity, u = 50 m/s, Final velocity, v = ?, Height covered, h = ?, Time taken, t = ?, g = 10 m/s2, At the highest point, final velocity v = 0, From v2 – u2 = 2 gh,, where g = - 10 m/s2 for upward journey,, 0 – (50)2 = 2 (-10) h, =, h, , −2500, = 125m, −20, , From v = u + gt,, or 0 = 50 + (-10) t, t = 50/10 = 5s, , www.plancess.com

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3.50, , Gravitation and Floatation, , Example 16. A force of 15 N is uniformly distributed over an area of 150 m2. Find the pressure in, pascals., Solution: Here, force, F = 15 N, Area, A = 150 cm2 = 150 × 10-4 m2, (1 cm = 10-2 m, 1 cm2 = 10-4 m2), Thus, pressure, P= F=, A, , 15N, = 1000 Pa, 150 × 10 −4 m 2, , Example 17. How much force should be applied on an area of 1 cm2 to get a pressure of 15 Pa?, Solution: Here, area, A = 1 cm2 = 10-4 m2, Pressure, P = 15 Pa = 15 N/m2, As, F, , F =P × A =(15N / m2 ) × (10−4 m2 ), A, = 1.5 × 10−3 N, P=, , Example 18. A block weighting 1.0 kg is in the shape of a cube of length 10 cm. It is kept on a horizontal, table. Find the pressure on the portion of the table where the block is kept., Solution: Here, force acting on the table, F = 1.0 kg = 10 N, Area of the table on which this force acts, A = 10 cm × 10 cm = 100 cm2, = 100 × 10-4 m2 = 10-2 m2 (1 cm2 = 10-4 m2), Pressure on the table,, , P, =, , 10N, F, =, = 1000 Pa, A 10 −2 m2, , Example 19. The pressure due to atmosphere is 1.013 × 105 Pa. Find the force exerted by the atmosphere, on the top surface of a table 2.0 m long and 1.0 m wide., Solution: Here, pressure due to the atmosphere,, P = 1.013 × 105 Pa = 1.013 × 105 N/m2, Area on which atmospheric pressure acts,, A = 2.0 m × 1.0 m = 2.0 m2, Thus, force exerted by the atmosphere,, F = PA = (1.013 × 105 N/m2) × (2.0 m2) = 2.026 × 105 N, , www.plancess.com

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Foundation for Physics, , 3.51, , EXERCISE 1 – For School Examinations, Fill in the Blanks, Directions: Complete the following statements with an appropriate word/term to be filled in the blank, space(s)., Q.1., , The force of, , is the centripetal force on the moon., , Q.2., , The acceleration due to gravity at the surface of a planet depends on the, and the, of the planet., , Q.3., , ge and gp denotes the acceleration due to gravity on the surface of the earth and another planet, whose mass and radius are twice that of the earth. The relation that holds is, ., , Q.4., , Every object inside the satellite is, , Q.5., , Acceleration due to gravity is, , ., proportional to the density of the planet., , velocity of the planet is constant when it revolves in an elliptical orbit, , Q.6., around the sun., Q.7., , Acceleration due to gravity, , Q.8., , S.I. unit of gravitational potential is, , Q.9., , The weight of an object at, , with depth below the surface of the earth., ., is zero., , True / False, Directions: Read the following statements and write your answer as true or false., Q.10. Copernicus discovered that the earth moves around the sun., True, , False, , Q.11. Weightlessness experienced while orbiting the earth in a spaceship is the result of zero gravity., True, , False, , Q.12. The gravitation force between a spherical shell and a point mass inside it is negative and finite., True, , False, , Q.13. It is possible to put on artificial satellite in to orbit in such a way that it will remain directly over, New Delhi., True, , False, , Q.14. Weight of body at centre of the earth is zero., True, , False, , www.plancess.com

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Gravitation and Floatation, , 3.52, , Q.15. Escape velocity of a body is independent of direction of projection., True, , False, , Q.16. If both mass and radius of the earth decrease by 1%, the value of acceleration due to gravity, would increase by nearly 1%., True, , False, , Q.17. An object in a satellite experiences weightlessness., True, , False, , Match the Following Columns, Directions: Each question contains statements given in two columns which have to be matched., Statements (A, B, C, D) in column I have to be matched with statements (p, q, r, s) in Column II, Q.18. Match the columnColumn I, , Column II, , (A), , A Newton’s law of gravitation, , (p), , zero, , (B), , At height h=R, , (q), , inverse square law, , (r), , decreases by a factor ¼, , (s), , decreases by a factor ½, , value of g, (C), , At height h=R/2, value of g, , (D), , At depth h=R, value of g, , Very Short Answer Questions, Directions: Give answer in one word or one sentence., Q.19. The mass of two bodies are doubled and the distance is halved, how will gravitational force, change?, Q.20. What is the height at which the value of g is the same as at a depth of R/2?, Q.21. Two artificial satellite, one close to the surface, other away are revolving around earth. Which has, larger speed?, Q.22. Will 1 kg sugar weight more at poles or at the equator?, Q.23. What is the gravitational potential energy of two point masses infinite distance away from each, other?, , www.plancess.com

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Foundation for Physics, , 3.53, , Q.24. What is the gravitational field strength of a planet, where the weight of a 60 kg astronaut, is 300 N?, Q.25. Why moon has no atmosphere?, Q.26. A block of wood floats in a bucket of water in a lift. Will there be any change in the equilibrium, of wood if the lift starts accelerating upwards?, Q.27. A piece of ice is floating in a vessel containing water and inside the ice there is an air bubble., What will be the effect on the level of water, when the ice melts?, , Short Answer Questions, Directions: Give answer in two to three sentences., Q.28. A planet reduces its radius by 1% keeping its mass unchanged. How does its acceleration due to, gravity change?, Q.29. What should be the angular velocity at any point on the equator so that the body feels, weightlessness?, Q.30. The value of ‘g’ at the moon is 1/6th of the value of ‘g’ at the surface of the earth and the diameter, of the moon is 1/4th of the diameter of the earth. Compare the ratio of the escape velocities., Q.31. The radius of the earth is reduced by 4%. The mass of the earth remains unchanged. What will, be the change in escape velocity?, Q.32. A high jumper can jump 1.5 m on earth. With the same effort, how high will he be able to jump, on a planet whose density is 1/3rd and radius 1/4th of that of the earth?, Q.33. Explain, why a tennis ball bounces higher on hills than in plains., Q.34. The earth’s gravitational field at a certain point out in space accelerates a 1 kg mass at 5 m s-2., How much will it accelerate a 3 kg mass?, Q.35. Does the escape velocity of a body from the earth depend upon, (a) the mass of the body?, (b) the location from where it is projected?, (c) the direction of projection?, (d) the height of the location from where the body is launched?, Q.36. A sphere of relative density, , σ, , and diameter d has concentric cavity of diameter D. What is the, , ratio of d/D if it just floats on water in a tank?, , www.plancess.com

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3.54, , Gravitation and Floatation, , Long Answer Questions, Directions: Give answer in four to five sentences., Q.37. What do you mean by satellite? What is a geostationary satellite? What are the three conditions, which must be met by a satellite to be a geostationary satellite?, Q.38. What is the difference between gravitational and inertial masses? Are they always equal?, Q.39. What do you mean by buoyancy? What is Archimedes principle?, , www.plancess.com

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Foundation for Physics, , 3.55, , EXERCISE 2 – For Competitive Examinations, Multiple Choice Questions, Directions: This section contains multiple choice questions. Each question has 4 choices (a), (b), (c) and, (d) out of which ONLY ONE is correct., Q.1., , The value of the universal gravitational constant, (a) Change with change of place, (b) Does not change from place to place, (c) Becomes more during night, (d) Becomes more during day, , Q.2., , Q.3., , Q.4., , Q.5., , Two stars of masses m and 5m are 3000 parsec apart. If the force on the large one is F, the force, on the small star is, (a) F/25, , (b) F/5, , (c) F, , (d) 5F, , An astronaut in a spacecraft goes into circular orbit above a strange planet that has a radius of, 2100 km. The craft is going at 1200 meters per second at an altitude of 6500 km. The astronaut’s, life-support pack has a mass of 60 kg. What will the life-support pack weight on the surface of, the planet, (a) 150 N, , (b) 170 N, , (c) 90 N, , (d) 190 N, , The radius of the earth is 6400 km. What is its mass?, 24, (a) 6.0 × 10 kg, , 24, (b) 5.0 × 10 kg, , 22, (c) 1.0 × 10 kg, , 14, (d) 6.5 × 10 kg, , When a satellite is in the synchronous orbit above the equator, it stays in one place with reference, to the earth by making each revolution in just the same time as it takes the earth rotate once., What is the altitude of the synchronous orbit, (a) 20000 km, , (b) 30000 km, , (c) 32500 km, , (d) 36000 km, , www.plancess.com

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3.56, Q.6., , Gravitation and Floatation, , The radius of earth is about 6400 km. and that of Mars is about 3200 km. The mass of earth is, about 10 times the mass of mars. An object weights 200 N on earth’s surface. Then its weight on, the surface of Mars will be, (a) 8 N , , Q.7., , (b) 20 N , , g  4 πR 3 , , , G  3 , , (b), ,  g  4 2 , =, (c) D   πR ,  G  3, , , D=, , (d) D =, , 3g, 4 π RG, , 3g, 4GπR 3, , The gravitational force of attraction between two bodies is F N. If the mass of each body and the, distance between them are doubled, then the gravitational force between them in Newton is, (a) 16 F , , Q.9., , (d) 80 N, , If G is gravitational constant and R is the radius of the earth, then acceleration due to gravity `g’, and the mean density of earth D are related by the equation, (a) D =, , Q.8., , (c) 40 N , , (b) F/16 , , (c) F/4 , , (d) F, , If the earth stops rotating about its axis, the acceleration due to gravity will remain unchanged at, (a) equator, , (b) latitude 45o, , (c) latitude 60o, , (d) poles, , Q.10. An object weight 10N when measured on the surface of earth. What should be its weight on, moon?, (a) 1.67 N, , (b) 4.18 N, , (c) 2.20 N, , (d) 16.7 N, , Q.11. The force of gravitation between objects of ordinary size kept at a distance of 1 meter from each, other is of the order of?, (a) 10 N, , (b) 100 N, , (c) 10-8 N, , (d) 10-11 N, , Q.12. A stone is thrown vertically upwards and caught at the point of projection after 10 seconds. The, time taken by the stone to reach the highest point is, (a) 5 sec., , (b) 10 sec., , (c) 9.8 sec., , (d) 4.9 sec., , Q.13. A wooden cube floating in water supports a mass m=0.2 kg on its top. When the mass is removed, the cube rises by 2cm. The side of the cube is – (density of water 103 kg / m3), (a) 6 cm , , (b) 12 cm , , (c) 8 cm , , (d) 10 cm, , www.plancess.com

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Foundation for Physics, , 3.57, , Q.14. Two point masses each equal to 1 kg attract one another with a force of 10-10 N. The distance, between the two point masses is (=, G 6.6 × 10 −11 MKS units ), (a) 8 cm, , (b) 0.8 cm, , (c) 80 cm, , (d) 0.08 cm, , Q.15. Two bodies of masses 10 kg and 100 kg are separated by a distance of 2 m, =, G 6.67 × 10−11 Nm 2 kg −2 . The gravitational potential at the mid-point on the line joining the, two is, , (, , ), , −7, (a) 7.3 × 10 J / kg, −9, (c) −7.3 × 10 J / kg, , −9, (b) 7.3 × 10 J / kg, −6, (d) 7.3 × 10 J / kg, , Q.16. The potential energy of a satellite of mass m and revolving at a height Re above the surface of, earth where Re =radius of earth, is, (a) −mgR e, (c), , −mgR e, 3, , (b), , −mgR e, 2, , (d), , −mgR e, 4, , Q.17. The kinetic energy of a satellite in its orbit around the earth is E. What should be the kinetic, energy of the satellite so as to enable it escape from the gravitational pull of the earth?, (a) 4 E, , (b) 2E, , (c) 2E, , (d) E, , Q.18. A planet revolves in an elliptical orbit around the sun. The semi-major and semi-minor axes are, a and b. Then the square of time period, T is directly proportional to, (a) a3, , (b) b3, 3, , a+b, , (c) , ,  2 , , 3, ,  a−b, , (d) , ,  2 , , Q.19. The time period of an earth satellite in circular orbit is independent of, (a) Both the mass and radius of the orbit, (b) Radius of its orbit, (c) The mass of the satellite, (d) Neither the mass of the satellite nor the radius of its orbit., , www.plancess.com

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3.58, , Gravitation and Floatation, , Q.20. An egg when placed in ordinary water sinks but floats when placed in brine. This is because, (a) Density of brine is less than that of ordinary water., (b) Density of brine is equal to that of ordinary water, (c) Density of brine is greater than that of ordinary water, (d) None of these, Q.21. A man is sitting in a boat which is floating in pond. If the man drinks some water from the pond,, the level of water in the pond will, (a) Rise a little, , (b) Fail a little, , (c) Remain stationary, , (d) None of these, , Q.22. The total weight of a piece of wood is 6 kg. In the floating state in water its 1/3 part remains, inside the water. On this floating piece of wood the maximum weight to be put so that it is, completely drowned in the water will be?, (a) 15 kg, , (b) 14 kg, , (c) 10 kg, , (d) 12 kg, , More than One Correct, Directions: This section contains multiple choice questions. Each question has 4 choices (a), (b), (c) and, (d) out of which ONE OR MORE may be correct., Q.23. What is the weight and mass of a man weighing 600 newton on earth?, (a) 10 kg, , (b) 60 kg, , (c) 600 kg m, , (d) 1.66 kg, , Q.24. The mass of an object is, (a) Measured by its inertia, , (b) Constant at all places, , (c) Scalar quantity, , (d) Vector quantity, , Q.25. In case of an orbiting satellite. If the radius of orbit is decreased, (a) Its KE decreases, , (b) Its ME decreases, , (c) Its PE decreases, , (d) Its speed decreases, , Q.26. Choose the correct statements from the following:, (a) Gravitational constant (G) is scalar but acceleration due to gravity (g) is a vector, (b) G and g are constant everywhere, (c) The gravitational force between two particles are an action and reaction pair., (d) The value of G and g are to be determined experimentally., , www.plancess.com

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Foundation for Physics, , 3.59, , Q.27. An object is weighed at the north pole by a beam balance and a spring balance, giving reading,, W’B and W’S respectively. Assume that the acceleration due to gravity is the same everywhere and, that the balances are quite sensitive:, (a) WB=WS, , (b) W’B<W’S, , (c) W’B=W’S, , (d) WB=W’B, , Q.28. The buoyant force acting on a body immersed in a liquid depends on, (a) Atmospheric pressure, , (b) Weight of the body, , (c) Weight of the liquid displaced, , (d) Density of the body, , Q.29. A body is floating in a liquid in a breaker. If the whole system falls freely under gravity, the, upthrust on the body due to the liquid is, (a) Equal to the weight of the immersed portion of the body, (b) Zero, (c) Equal to the weight of the body in air., (d) Equal to the weight of the liquid displaced., , Passage Based Questions, Directions: Study the given passage (s) and answer the following questions., Passage I, The quantity of matter contained in an object is called mass. It remains constant whether the, object is on earth, moon or even in the outer space. Weight on the other hand is the force of, attraction of earth with which an object is attracted towards the earth. Now, suppose a man, weighs 600 N on earth, his weight on moon would be 100 N., Q.30. The mass of man on earth, if g is 10 m/s2 is, (a) 60 kg, , (b) 10 kg, , (c) 6000 kg, , (d) 1000 kg, , Q.31. The mass of man on moon is, (a) 60 kg, , (b) 10 kg, , (c) 6000 kg, , (d) 1000 kg, , Q.32. Acceleration due to gravity on moon is, (a) 10 m/s2, , (b) 9.8 m/s2, , (c) 1.66 m/s2, , (d) 1 m/s2, , www.plancess.com

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Gravitation and Floatation, , 3.60, Passage II, , Motion of ships, boats in sea, river, etc. can be studied on the basis of laws of flotation of bodies., Laws of floatation depends on Archimedes’ principle. For body to float in equilibrium, different, conditions are to be maintained. On the basis of Archimedes’ principle, answer the following., Q.33. The body will not float in equilibrium if, (a) Centre of gravity and buoyancy lie along same line, (b) Weight of body is equal to weight of liquid displaced by sinking part of body, (c) Centre of buoyancy coincides with the centre of gravity, (d) Weight of body is equal to weight of liquid displaced by whole volume of the body, Q.34. A ship do not sink in sea water because, (a) Density of material of ship is less than that of water, (b) Density of material of ship is equal to that of water, (c) Density of material of ship is more than that of water, (d) Space occupied by a ship is much more than the actual volume of material of the ship., Q.35. If d and d0 represent density of material of body floating on surface of water and that of water, respectively, then the fraction of volume of body inside the water is, (a), , d, d0, , (b), , d0, d, , (c), , d0 − d, d, , (d), , d, d0 − d, , Assertion and Reason, Directions: Each of these questions contains an Assertion followed by Reason. Read them carefully and, answer the questions on the basis of following options. You have to select the one that best describes the, two statements., (a) If both Assertion and Reason are correct and Reason is the correct explanation of Assertion., (b) If both Assertion and Reason are correct but Reason is not the correct explanation of, Assertion., (c) If Assertion is correct and Reason is incorrect., (d) If Assertion is incorrect and Reason is correct., Q.36. Assertion: The value of acceleration due to gravity does not depend upon mass of the body., Reason: Acceleration due to gravity is a constant quantity., , www.plancess.com

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Foundation for Physics, , 3.61, , Q.37. Assertion: A person sitting in an artificial satellite revolving around the earth feels weightlessness., Reason: There is no gravitational force on the satellite:, Q.38. Assertion: According to Newton’s law of motion, if earth attracts a stone towards itself, then the, stone should also attract the earth., Reason: Earth attracts towards stone but with a very small acceleration., Q.39. Assertion: The value of ‘g’ is greater at the equator than at the poles., Reason: Radius is more at the equator than at the poles., Q.40. Assertion: The time period of geostationary satellite is 24 hours., Reason: Geostationary satellite must have the same time period as the time taken by the earth to, complete one revolution about its axis., , Subjective Questions, Directions: Answer the following questions., Q.41. The moon falls around the earth rather than straight into it. If the tangential velocity were zero,, how would the moon move?, Q.42. A friend says that astronauts in orbit are weightless because they are beyond the pull of earth’s, gravity. Correct your friend’s ignorance., Q.43. If you were in a car that drove off the edge of a cliff, why would you be momentarily weightless?, Would gravity still be acting on you?, Q.44. The figure shows four arrangements of three particles equal masses. (a) Rank the arrangements, according to the magnitude of the net gravitational force on the particle labeled m, greatest first., (b) In arrangement 2, is the direction of the net force, closer to the line of length d or to the line, of length D?, , P, , d, m, , P, , D, Q, , R, , m, , (1), , P, , d, , m D, Q, (3), , D, Q, , d, , R, , (2), , R, , Q m D, d, P, , R, , (4), , Q.45. If there is an attractive force between all objects, why do we not feel ourselves gravitating towards, massive buildings in our vicinity?, , www.plancess.com

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3.62, , Gravitation and Floatation, , Q.46. Would the springs inside a spring balance be more compressed or less compressed if you weighed, yourself in an elevator that accelerated upward or downward?, Q.47. Two particles m1 and m2 are at initially at rest at infinite separation. They start moving towards, each other due to mutual force of gravitation. Find their relative velocity of approach when, separation between them is ‘d’., Q.48. When you are bathing on a stony beach, why do the stones hurt your feet less when you get in, deep water?, Q.49. When an ice cube in a glass of water melts, does the water level in the glass rise, fall, or remain, unchanged? Does your answer change if the ice cube contains many grains of heavy sand?, Q.50. Would it be slightly more difficult to draw soda through a straw at sea level or on top of a very, high mountain? Explain., Q.51. Atmospheric pressure varies from day to day. Does a ship float higher in the water on a highpressure day than on a low-pressure day?, Q.52. A pound of Styrofoam and a pound of lead have the same weight. If they are placed on an equalarm balance surrounded by air, will it balance?, Q.53. Suppose an office party is taking place on the top floor of a tall building. Carrying an iced soft, drink, you step on the elevator, which begins to accelerate downward. What happens to the ice, in the drink? Does it rise farther out of the liquid? Sink deeper into the liquid? Or is it unaffected, by the motion?, , www.plancess.com

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Foundation for Physics, , 3.63, , SOLUTIONS, EXERCISE 1 – For School Examinations, Fill in the Blanks, ge, , 1. gravity, , 2. mass, radius , , 3. gp =, , 5. directly , , 6. Areal , , 7. decreases , , 11. False , , 12. False, , 2, , 4. Weightless, 8. J/kg, , 9. centre of the earth, , True / False, 10. True , , 13. True, put geostationary satellite (a special case of geo-synchronous satellite) in orbit over New Delhi., 14. True 15. True, it doesn’t depend on direction of propagation of the orbiting body, 16. True , , 17. True, , Match the Following Columns, 18., , ( A ) → (q) ; (B) → (r ) ; (C) → (s) ; (D) → (p), , Very Short Answer Questions, 19., , F=, , GM1M2, R2, , New force= F=, , G ( 2M1 )( 2M2 ), , (, , GM1M2, =, 16, =, 16F, 2, R2, R/2, , ), , ∴ Force will become 16 times., 20., ,  2h ,  d, gh g  1 −  and =, gd g  1 − , =, R, ,  R,  2h ,  R /2, R, 2h, 1 1, As d = ; ⇒ g  1 −  =g  1 −,  1 − =1 − =, 2, R, R, R, 2 2, , , , , , www.plancess.com

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Gravitation and Floatation, , 3.64, , 1 2h, 1 2h, R, ⇒1− =, ⇒ =, ⇒ h=, 2 R, 2 R, 4, , 21., , As orbital speed, ν = GM ; therefore higher the value of h, lesser will be the speed., R +h, , ∴ The satellite close to the surface will have larger speed., 22., , The value of ‘g’ is larger at poles than at the equator., , ∴1, , kg sugar will weigh more at poles when weighed by a spring balance calibrated at the, equator., , 23., , Zero, because gravitational potential energy (U) of two point masses separated at larger distance,, lowers the energy and for infinite distance, U is zero., , 24., , The gravitational field strength, is the gravitational acceleration of the planet acting on body, calculated as,, F 300 N, = = 5 ms−2, m 60 kg, , 25., , Escape velocity for any object from the surface of the moon is very small. (it is only 2.354 km/, sec). For the temperature on the surface of the moon, the calculated r.m.s velocity of all gas, molecules is more than this escape velocity. Hence all of them have escaped from moon’s surface., , 26., , No – because equilibrium of floating bodies remain unaffected by variation of acceleration due, to gravity., , 27., , The level of water will not change because for a floating ice, the weight of ice is equal to the, weight of the displaced water by the immersed part of ice. When the ice melts, the water so, formed will be equal in volume which was displaced by the immersed part of the ice., , Short Answer Questions, 28., , ∆R, ∆g, ∆R, When mass is same, g ∝ 12 ∴ ∆ g =, 2, ⇒, × 100 =, 2, × 100 =, 2 × 1% =, 2%., g, , R, , R, , g, , R, , Therefore the acceleration due to gravity increases by 2%., 29., , At any point on the surface earth the acceleration due to gravity is,  Rω2, , g' =, g 1 −, cos2 α , , , g, , , ,  α =latitude  For equator α =0, , , For weightlessness g' = 0; g  1 −, , , , Rω2 , g,  = 0 ⇒ω =, g , R, , www.plancess.com

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Gravitation and Floatation, , 3.66, 33., , Suppose that the tennis ball bounces with a velocity u. It will go up, till its velocity becomes zero., If h is the height up to which it rises on the hill, then using kinematical equation, we get, , ( 0 ) − u2 = 2 ( −g') h, where, 2, , u2, g’ is acceleration due to gravity on the hill. ∴ h = Since the, 2g', , acceleration due to gravity on the hill (g’) is less than that on the surface of earth (effect of, height), it follows that tennis ball will bounce higher on hills than in plains., 34., , Gravitational acceleration of any body at some height above earth is gh ∝ 1, which is, (R + h)2, independent of mass of the body. Therefore irrespective of mass at the, same height the acceleration on the body of 3 kg will experience the same gravitational acceleration, of 5 ms‒1., , 35., , (a) The escape velocity of a body does not depend on its mass,, (b) The escape velocity depends upon the location from where the body is projected., Explanation. A body of mass m on the surface of the earth has a gravitational potential energy, −GMm, ., R, , Since earth in not a perfect sphere, value of R will be different at different locations. It makes, escape velocity dependent of location., (c) The escape velocity of a body does not depend upon the direction of its projection., (d) The escape velocity depends upon the height (altitude) from where the body is projected., Explanation. For a body to be projected from a height above the surface of the earth, the escape, velocity is given by, , 2Gm where the height of the body above earth is a considerable factor., R+h, , For h = R the escape velocity will be lowest than body projected from the surface of earth., 36., , Let ρ → density of the sphere, ρ ' → density of water. The sphere will float if weight of the sphere, = weight of the displaced water, ⇒, , 3, 3, 3, 4  d   D  , 4  d, π   −    ρg = π   ρ 'g, 3  2   2  , 3 2, , , , ⇒ (d3 − D3 ), , ρ, =, d3, ρ', , ⇒ (d3 − D3 )σ = d3 ⇒ d3 ( σ − 1) = σD3, , www.plancess.com

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Foundation for Physics, , 3.67, , 1/3, , ⇒, , d  σ , =, D  σ − 1 , , Long Answer Questions, 37., , Just as the planet revolves around the sun, in the same way few celestial bodies revolve around, these planets. These bodies are called ‘satellites’. For example moon is a natural satellite of Earth., Artificial satellites are launched from the Earth. Such satellite are used for telecommunication,, weather forecast and other applications. The orbit of these satellites are considered as nearly, circular for practical purpose., Geo-Stationary Satellite, A satellite which appears to be stationary for a person on the surface of the earth is called, geostationary satellite. It is also known as communication satellite or synchronous Satellite., These kind of satellites must meet following conditions:, (i) The orbit of the satellite must be circular and in the equatorial plane of the Earth., (ii) The angular velocity of the satellite must be in the same direction as the angular velocity of, rotation of the earth i.e. from west to east., (iii) The period of revolution of the satellite must be equal to the period of rotation of earth, about its own axis., i.e. 24 hour= 24 × 60 × 60 =, 86400 sec., The height above which this satellite orbit can be calculated as follow,, 1, ,  GM T2  3, 4π 3, 2, e,  but GMe gR 2e, =, T =, r , or r =, 2, , , GMe,  4π , 2, , , 9.8 × 6.38 × 106, ∴=, r, , , , (, , ), , 2, , ( 86400 ), , 2, , ×, , 4 π2, , 1, , 3, , , , , = 42237 × 103 m = 42, 237 km. ≈ 42000 km. approximately., h = r – R = 42000 – 6400 = 35600 km., A geostationary satellite always stays over the same place above the earth. Such a satellite is never at, rest. Such a satellite appears stationary due to its zero relative velocity w.r.t. that place on the earth., The orbit of a geostationary satellite is called ‘Parking Orbit’., , www.plancess.com

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Gravitation and Floatation, , 3.68, 38., , Inertial Mass, Inertial mass of a body is related to its inertia of linear motion, and is defined by Newton’s second, law of motion., Let a body of mass mi moves with acceleration a under the action of an external force F. According, to Newton’s second law of motion, , =, F m=, i a or m i, , F, a, , The mass mi of the body in this sense is the inertial mass of the body. If a=1, then mi=F., Thus inertial mass of a body is equal to the magnitude of external force required to produce unit, acceleration in the body., In fact inertial mass of a body is the measure of the ability of the body to oppose the production, of acceleration in its motion by an external force., Properties of Inertial Mass, 1. Inertial mass of a body is proportional to the quantity of matter contained in the body., 2. Inertial mass of a body is independent of size, shape and state of the body., 3. Inertial mass of a body does not depend upon the temperature of the body., 4. Inertial mass of a body is not affected by the presence or absence of other bodies near it., 5. Inertial mass is conserved when the two bodies combine physically or chemically., 6. Inertial mass can be added by simple laws of algebra, irrespective of the materials of the bodies., 7. Inertial mass of a body increases with the speed of the body. When a body moves with a, velocity v, its inertial mass m is given by m =, , mo, 1 − v 2 / c2, , where mo is the rest mass of the body,, , c is the velocity of light in vacuum. The mass m is affected only when the velocity of the body is, comparable with the velocity of light., Gravitational Mass, Gravitational mass of a body is related to gravitational pull on the body and is defined by, Newton’s law of gravitation. If a body of mass mG is placed on the surface of the earth of radius, R and mass M, then gravitational pull on the body is given by, F, =, , GMmG, R, , 2, , ⇒ m=, G, , F, , F, =, 2, I, GM / R, , The mass mG of the body in this sense is this sense is the gravitational mass of the body. The inertia, of the body has no effect on the gravitational mass of the body. If I=1 then from equation, mG=F., , www.plancess.com

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Foundation for Physics, , 3.69, , Thus gravitational mass of a body is defined as the magnitude of gravitational pull experienced, by the body in a gravitational field of unit intensity., Inertial mass are not always equal to gravitational mass., 39., , Buoyancy, If a body is partially or wholly immersed in a fluid, it experiences an upward force due to the, fluid surrounding it. The phenomenon of force exerted by fluid on the body is called buoyancy, and the force is called buoyant force., A body experiences buoyant force whether it floats or sinks, under its own weight or due to other, forces applied on it., ARCHIMEDES’ PRINCIPLE, A body immersed in a fluid experiences an upward buoyant force equivalent to the weight of the, fluid displaced by it. This principle is known as the Archimedes’ principle., The proof of this principle is very simple. Imagine a body of arbitrary shape completely immersed, in a liquid of density ρ as shown in the figure., A body is being acted upon by the forces from all direction. Let us consider a vertical element of, height h and cross-sectional area dA as shown in the figure 3.22 (b), F, , h, h, , h, , dA, , F, (a), , (b), , The force acting on the upper surface of the element is F1 (downward) and that on the lower, surface is F2 (upward). Since F2 > F1, therefore, the net upward force acting on the element is dF, = F2 – F1, It can be easily seen from the figure (b), that, Fb = ρh × (b) = ρbh and, F2 = (ρgh2)dA so dF = ρg(h)dA, Also, h2 – h1 = h and d (hA) = dV, , ∴ The net upward force is F = ∫ ρgdV = ρVg, , Hence, for the entire body, the buoyant force is the weight of the volume of the fluid displaced., , www.plancess.com

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3.70, , Gravitation and Floatation, , EXERCISE 2 – For Competitive Examinations, Multiple Choice Questions, 1. (b), 2. (c) Because gravitational force F. For 3000 parsec distance the force of interaction is zero but the, gravitational force on the star remains the same as of the first star., 3., , (b) the spacecraft is going in a circle of radius 6500+2100=8600 km. Its centripetal acceleration, is thus, v2, g a=, =, =, c, r, , (1200 m / s )=, 2, , 6, , 8.6 × 10 m, , 0.167 m / s2 ., , At the surface, value of gravitational acceleration will be, 2, , r , =, g0 g=,  ,  ro , , (, , 2, , ), ,  8600 km , =, 0.167 m / s2 , 2.8ms−2, , 2100, km, , , , Then the weight of the pack at the surface is w=mg= (60 kg) (2.8 N/kg) = 170 N, 4., , (a) At the earth’s surface, 6400 km from its centre, a kilogram weighs 9.8 N., Fgravr 2, =, m1, =, Then, Gm2, , (9.8 N) (6.4 × 106 m), ;, (6.67 × 10−11 Nm2 / kg2 ) (1 kg), 2, , 24, m=, 1 6.0 × 10 kg, , 5. (d) The period of revolution of the satellite must be exactly one day, or 86400s. The centripetal, 2, 2, 2, acceleration of the satellite must be 4π r / T , the gravitational field must be g = go (R / r ) . In, free Fall, a=g, so, 2, , R , R , 4 π2r, 4 π2r, so,, go  , = g=, , , o, 2, 2, T, T, r, r, , (9.8 m / s )(6.4 × 10 m), 2, , 2, , r=, , 3, , 6, , 2, , (86400s ), , 2, , 4 π2, , = 4.23 × 107 m ., , To get altitude, subtract the radius of the earth. The satellite must be at an altitude of 36000 km., 6., , (c) the value of g is g = GM, weight of object on Mars, Wm=mgm=, 2, R, , W, =, ∴ m, We, , m/R ), (GM =, (GM m / R ), m, , 2, m, , e, , 2, e, , Mm  R e , , , Me  Rm , , GMmm, 2, Rm, , 2, , www.plancess.com

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Foundation for Physics, , , , , , 3.71, , 2, , or Wm = Mm   R e   We, , , Me  Rm , , Me, =, We 200 N, Given Re = 6400 km, Rm = 3200 km= 10,, Mm, 2, , 1  6400 km , 4, ∴ Wm = × , 40N,  × 200 N = × 200 N =, 10  3200 km , 10, 4, , G  πR 3D , 3, , , ⇒ g=, R2, , 3g, ⇒D=, G ( 4π R ), , 7., , (b) g=, , GM, , 8., , (d) F ∝, , M1M2, , 9., , (d) because position of pole doesn’t change while earth is rotating, while that of any point on, equator changes its direction with earth’s rotation., , 10., , (a) Weight on earth We = 10 N Weight on moon Wm = ?, , R, , 2, , R2, , both masses and distance doubled still force remains same, , We = mg .......(1) ; Wm = mgm ..................(2), On dividing eq. (2) by eq. (1), Wm mgm Wm gm, =, =, ;, We mge We ge, , , , gm 1 Wm 1, We 10, =, = ; W=, = = 1.67 N, m, ge 6 We 6, 6, 6, , 11., , (d) If both masses are same i.e. 1 kg weight than the force will be of the order of G., , 12., , (a) Time taken to reach highest point of trajectory = time for downfall, , 13., , (d) The buoyant force gAh = mg the weight immersed in fluid, As A = and h = 2 cm = 0.02 m,, =, d, , m, =, ρh, , 0.2, =, 10 × 0.02, 3, , −2, =, 10, 0.1m, , www.plancess.com

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Gravitation and Floatation, , 3.72, , 14., , (c) According to Newton’s Gravitation Law Fg =, , GM1M2, r2, , here Fg = 10−10 Newton,, , −11, m=, =, 1 m, 2 1 kg, G = 6.6 × 10, , GM1M2 6.6 × 10−11 × 1 × 1, =, = 0.66, =, or r 0.8125, =, metre 81.25 cm ≈ 80 cm, Fg, 10−10, , =, So r 2, , −6.67 × 10 −11 × 10 6.67 × 10−11 × 100, =, −6.67 × 10 −10 − 6.67 × 10 −9, −, 1, 1, , 15., =, (c) Vg, , =, −6.67 × 10−10 × 11 =, −7.3 × 10−9 j / kg, 16., , At a height h above the surface of earth the gravitational potential energy of the satellite of mass, m is Uh = −, , GMem, where Me and R e are the mass and radius of earth respectively., Re + h, , In this question, since h=Re, GMem −mgR e, −, =, So, Uh =, 2R e, 2, , 17., , (b) We know that v e = 2 v 0 , where v 0 is orbital velocity., 1, 2, , K.E. in the orbit, E = Mv 20, 1, 2, , K.E. to escape the orbit,, =, E =, Mv 2e, 18., , (a) T 2 ∝ (major axis )3 ⇒ T 2 ∝ a3, , 19., , 2, (c) mν = GmM, 2, , R+x, , ( ), , 1, 1, 2, M 2v, =, M v 20=, × 2 2E, 0, 2, 2, , (R + x ), , x=height of satellite from earth surface, M=mass of satellite, =, ⇒ v2, , GM, =, or v, (R + x ), , GM, R+x, , www.plancess.com

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Foundation for Physics, , =, T, , 3.73, , 2 π (R + x ) 2 π (R + x ), =, which is independent of mass of satellite., v, GM, R+x, , 20., , (c) Brine due to its high density exerts an upward thrust which can, balance the weight of the egg., , 21., , (c) volume consumed by man << volume of water in pond, , 22., , (d) Weight of submerged part of the block W=, Excess weight = weight of water having, , W'=, , 1, v × g × (Density of water) ….(i), 3, , 2, volume of the block., 3, , 2, v × g × (Density of water) ….(ii), 3, , Dividing (i) and (ii), W' 2 / 3, =, W 1/3, ∴ W ' = 2W ⇒ W ' = 2 × 6 = 12kg, , More than One Correct, 23., , (b, c) W = mg,, ∴m =, , W 600, =, = 60 kg = 600 N = 600kgms−2, g, 10, , 24., , (a, b, c) mass of any object is a constant scalar quantity but weight of that body may change due, to different gravitational acceleration, , 25., , (b, c) P.E. =, −, , K.E. =, ,, , GMm, r, , GMm, 2r, , GMm, GM, M.E. =, −, & Vo =, r, r, , Hence if r is decreased, P.E. and M.E. decrease but K.E. and V0 increase., , www.plancess.com

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Gravitation and Floatation, , 3.74, 26., , (a, c, d) G is a scalar quantity whereas g is vector quantity., , 27., , (a, b, d), , 28., , (c, d), , 29., , (a, b, d) In freely falling lift, weight of every object is zero., , Passage Based Questions, 30., , (a) W= m × g 6000= m × 10, , 31., , (a) Mass remains same everywhere. So mass on moon =60 kg, , 32, , (c) W= m × g 100, = 60 × g, , m = 60 kg, , =, g 100, =, / 60 1.66 m / s2, 33., , (b), , 34., , (a) Due to the shape of ship, its effective density is much less than the actual density., , 35., , (a) If V be the actual volume of body and V’ the volume of body inside water, then, FB = d0V’g and m = dV, For equilibrium, FB = mg or d0V’g = dVg, Or, , V' d, =, V d0, , Assertion and Reason, 36., , (c) because g depends on mass of planet and mass of each planet is not constant, , 37., , (c) The person feels weightlessness because the orbiting speed of satellite and itself are same but, the force acting on both bodies is similar at all the places, , 38., , (a) It is an action reaction pair of both the bodies, , 39., , (d) g =, , 40., , (a) Time period of the Geostationary satellite is 86400 sec. i.e. similar to that of the earth., , GM, Re, , at poles the radius of earth is small than that at equator., , www.plancess.com

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3.75, , Foundation for Physics, , Subjective Questions, 41., , Moon orbits around earth with the constant tangential velocity which is equal to critical velocity, necessary for its motion around earth. If the moon’s tangential velocity becomes zero, the, necessary centripetal acceleration required for motion of the moon vanishes and it cannot sustain, the uniform motion around earth. As result it would fall straight down and crash into earth., , 42., , The astronaut feels weightlessness because he is not supported by anything. In this case, there, would be no compression in the springs of a scale placed beneath the astronaut’s feet because, the scale is falling as fast as the astronaut is. In other words the astronaut is orbiting with least, acceleration, same for the weighing instrument. So the weight of astronaut is recorded to zero., , 43., , I would be momentarily in weightless situation because the car is falling as fast as I am falling, towards centre of earth with gravitational acceleration g. Thus, all the effects of gravity are, eliminated. This gives the sensation of weightlessness. Gravity still acts on me but it is not felt as, weight as there is no supporting force., , 44., , (a) (i) → (1) , (ii) → (2) and (4) have the same magnitude, (iii) → (3), Justification for ranking against each case:, , d, , m, P, , 2, Using formula F = G m1m, 2, , D, , , , F, , , , F, R, , Q, , r, , R, , We find that the force is maximum for, (1), =, F G, , mm, d2, , mm, + G=, D2, , 1 , Gm  2 + 2 , d D , , For cases (2) and (4)=, F, F1 =, , G mm, d2, , F2 =, , 2, , 1, , F12 + F22, , Xm, Q, , F, , F, m, , Q, , F, , B, , R, , F, , A, , For case (1), , F, F, , P, , P, , ;, , , , F m, d, P, Q, , G mm, d2, , , , F, , R, , D, , R, , Net force, , For case (3):, , F, , D, m, , Q, , d, , P, , www.plancess.com

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Gravitation and Floatation, , 3.76, , =, F, , Gm  m, d2, , Gm  m, −=, D2, , 1 1, Gm2  2 − 2 , d D , , (b) In case (2), the direction of the net force is closer to the line of length d:, , 45., , The building in our surrounding do impart an gravitational force of attraction as any other, body would. And from formula F =, , GM1 M2, R2, , . Comparing the force of building with that of the, , gravitational attraction of earth on us, the force of massive buildings are so small that its effect, escape our notice and we don’t feel the attraction of buildings., 46., , Springs inside a spring balance will be more compressed if weighed in on upward accelerating, elevator., Springs inside a spring balance will be less compressed if weighed in a downward accelerating, elevator. If an elevator accelerates downwards, the supporting force of the floor is less and a body, weigh less. In an elevator that accelerates upward, the supporting force of the floor is more and a, body weigh more., , 47., , V, m, , V, d, , m, , Using conservation of energy principle at infinite separation and at distance ‘d’,, 0=, +0, , Gm1m2, 1, m1 v12 + m2 v 22 −, .......(i), 2, d, , As there is no external force acting so using conservation of linear momentum, , 0 +=, 0 m1v1 − m2v 2, , ……..(ii), , , , Using (i) and (ii) to solve for v1 and v2 then relative velocity of approach, = v1 − ( − v 2 )=, , 48., , 2G (m1 + m2 ), d, , when we are in deep water the effective weight of our body decreases due to the buoyant force., , www.plancess.com

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Foundation for Physics, , As pressure =, , 3.77, , Force or Weight, , if the force is less, pressure on the feet due to stone will be less in, Area, , deep water when the two are in contact. So the stones will hurt our feet less in deep water than, on a stony beach., 49., , When the ice cube in glass water melts, the water level remains unchanged. It is so because when, ice floats, the weight of ice is equal to the weight of the water displaced by the immersed part of, ice in water. The answer will not change if the ice cubes contain air bubbles or grains of heavy, stand because in both case volume of displaced water will be equal to the volume of water formed, by melting ice., , 50., , Atmospheric pressure is more at sea level than on the top of a very high mountain. When the, straw is sucked, the pressure inside the straw decreases, so the atmospheric pressure outside, pushes the liquid in the straw. As atmospheric pressure is less outside the straw on top of a high, mountain, so it will be difficult to suck the soda through a straw on mountain than at sea level., , 51., , The level of floating of a ship in the water is unaffected by the atmospheric pressure. The buoyant, force results from the pressure difference in the fluid. On a high-pressure day, the pressure at all, point in the water is higher than on a low-pressure day. Because water is almost incompressible,, however, the rate of change of pressure with depth is the same, resulting in no change in the, buoyant force., , 52., , The balance will not be in equilibrium. The lead side will be lower. Despite the fact that the, weights on both sides of the balance are the same. The Styrofoam, due to its lower density and, larger surface area feels an upward thrust by surrounding air than lead block of same mass. Due, to this buoyant force the side of Styrofoam block will be higher than that of lead block in equalarm balance., , 53., , The acceleration of the elevator is equivalent to a change in the gravitational field. According to, the principle of equivalence. If the elevator accelerates downward, one might be tempted to say, that the effect is the same as if gravity decrease-the weight of the ice cube decreases, causing it to, float higher in the liquid. Recall, however that the magnitude of the buoyant force is equal to the, weight of the liquid displaced by the ice cube. The weight of the liquid also decreases with the, effectively decreased gravity. Because both weight of the ice cube and buoyant force decreases by, the same factor, the level of the ice cube in liquid is unaffected., , www.plancess.com