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FAQ, FREQUENTLY ASKED QUESTIONS, WITH, ANSWERS, ⌘⌘⌘, , SUBJECT : PHYSICS, P.U.C - I, ⌘⌘⌘, , 1

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CONTENTS, CHAPTER, , TOPIC, , PAGE NO., , 1, , PHYSICAL WORLD, , 004 - 005, , 2, , UNITS AND MEASURMENTS, , 006 - 014, , 3, , MOTION IN A STRAIGHT LINE, , 015 - 034, , 4, , MOTION IN A PLANE, , 035 – 062, , 5, , LAWS OF MOTION, , 063 - 084, , 6, , WORK ENERGY POWER, , 085 – 1 1 1, , 7, , SYSTEM OF PARTICLE AND ROTATARY MOTION, , 1 12 – 129, , 8, , GRAVITATION, , 130 – 157, , 9, , MECHANICAL PROPERTIES OF SOLIDS, , 159 – 167, , 10, , MECHANICAL PROPERTIES OF FLUIDS, , 168 – 173, , 11, , THERMAL PROPERTIES OF MATER, , 174 - 206, , 12, , THERMO DYNAMICS, , 207 – 220, , 13, , KINETIC THEORY OF GASES, , 221 - 231, , 14, , OSCILLATIONS, , 232 – 245, , 15, , WAVES, , 246 – 279, , 2

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PART-1, , 3

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1. PHYSICAL WORLD, Q. What is physics?, Ans: The branch of physical science which deals with matter and energy relation is called, physics., Q. Who discovered X-Rays?, Ans: Roentgen., Q. Name the discovery made by W.K. Roentgen., Ans: X – Rays., Q. Who discovered electron?, Ans: J.J. Thomson., Q. Who discovered neutrons?, Ans: Chadwick discovered Neutrons., Q. Name the principle related to technology of (a) steam engine, , (b) Rocket propulsion., , Ans: (a) steam engine: Conversion of heat in to mechanical work, (b) Rocket propulsion: Newton’s third law of motion, Q. Name two physicists who achieved in unification of electricity and magnetism., Ans: The two Physicists who achieved in unification of electricity and magnetism are Hans, Oersted and Michael Faraday., Q. Name the experiment behind nuclear model of the atom?, Ans: Gold foil experiment, Q. Mention one of major contributions of S. Chandrasekhar in the field of physics., Ans: Evolution of stars., Q. Name any one fundamental force in nature., Ans: (Any one), Gravitational force, (or) Electromagnetic force, (or) Strong nuclear force,, (or) Weak nuclear force., , 4

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Q. Name two basic forces in nature., Ans: (Any two), (i) Gravitational force,, (ii) Electromagnetic force,, (iii) Strong nuclear force,, (iv) Weak nuclear force., Q. Which is the weakest force in nature?, Ans: Gravitational force, Q. Mention strongest and weakest force among fundamental forces in nature., Ans: (i) Strongest Force: Strong nuclear force., (ii) Weakest Force: Gravitational force., Q. Mention the long range and short range fundamental forces in nature., Ans: (i) Short range Force: Strong nuclear force., (ii) Long range Force: Gravitational force., Q. Mention any one conservative law., Ans: Law of conservation of energy, Q. Name any two conservation laws in nature., Ans: (Any Two), (i) Law of Conservation of energy, (ii) Law of Conservation of linear momentum, (iii) Law of Conservation of angular momentum, (iv) Law of Conservation of electric charges, Q. What are conserved quantities? Give an example., Ans: The physical quantity that remains constant in an isolated system is called a conserved, physical, , quantity., , Example: (i) Law of conservation of charge, (ii) Law of conservation of charge, (iii) Law of conservation of linear momentum, (iv) Law of conservation of angular momentum., , 5

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2. UNITS AND MEASUREMENTS, , Q. Name the system of units accepted internationally., Ans: S.I. system, Q. Write the SI unit of momentum., Ans: SI unit of momentum is kg m s-1., Q. Write the SI unit of impulse., Ans: SI unit of impulse is kg m s-1., Q. Write the SI unit of work., Ans: SI unit of work is joule (J)., Q. Write the SI unit of stress., Ans: SI unit of stress is Nm-2 (or) pa., , Q. Write the SI unit of pressure., Ans: SI unit of pressure is pascal(pa) (or) Nm-2., , Q. Write the SI unit of gravitational constant., Ans: SI unit of gravitational constant is Nm2kg-2., , Q. Write the SI unit of luminous intensity., Ans: SI unit of luminous intensity is candela (cd =lumen per steradian)., , Q. Mention the two supplementary units in S.I. system., Ans: radian (rad) and steradian (sr)., , Q. Write the SI unit of solid angle., Ans: Steradian (sr)., , Q. Name the SI unit of plane angle., Ans: radian (rad)., 6

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Q. What is basis?, Ans: The distance between the two points of observation is called basis., Example: The distance between the eyes., Q. Write any two advantages of SI system over a CGS system., Ans: (i) It is a rational system of unit because only one unit is used for a given physical, quantity., (ii) It is a coherent system of unit because all the derived units can be obtained by, fundamental &, , supplementary units without introducing any numerical factors., , Q. What is measurement?, Ans: The comparison of an unknown physical quantity with a known fixed unit quantity is, called measurement., Q. Name the method of measuring long distances such as distance of planets, stars etc, Ans: Parallax method, Q. What is meant by parallax and parallax angle?, Ans: Parallax: An object in front of eyes is looked against sum specific point in the, background, first when the eye and then with the left eye. The position of the object appears to, change with respect to the background. This is called parallax., Parallax angle: The distance between the two (distance between two eyes) observation points is, called basis. The angle subtended by the basis at the object is called parallax (parallactic) angle., [(OR) The angle between the Earths at one time of year, and the Earth six months later, as, measured from a nearby star is called parallax angle. Astronomers use this angle to find the, distance from the Earth to that star]., Q. What is meant by parallax? Mention its use., Ans: Parallax: An object in front of eyes is looked against sum specific point in the, background, first when the eye and then with the left eye. The position of the object, appears to change with respect to the background. This is called parallax., Use: Astronomers estimate the distance of nearby objects in space., O, , Q. Write a note on parallax method for measuring large distances., Ans: An object in front of eyes is looked against sum specific point in the, , θ, , D, , background, first with the Right eye and then with the Left eye., , D, , The position of the object appears to change with respect to, A, , b, , B, 7

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the background. This is called parallax., The distance between the two (distance between two eyes), observation points is called basis., The angle subtended by the basis at the object is called, parallax (parallactic) angle., The relative shift in the position of an object due to a change in the point, Object of observation is called parallax., Consider an object O., Let:, , A and B = two observation points., b=basis, 𝜃 = 𝑝𝑎𝑟𝑎𝑙𝑙𝑎𝑥𝑎𝑛𝑔𝑙𝑒, D= radius., 𝑝𝑎𝑟𝑎𝑙𝑙𝑎𝑥 𝑎𝑛𝑔𝑙𝑒 =, , 𝑙𝑒𝑛𝑔𝑡ℎ𝑜𝑓𝑎𝑟𝑐, 𝑟𝑎𝑑𝑖𝑢𝑠, , 𝑏, , 𝑏, , →𝜃=𝐷→𝐷=𝜃, , Thus, the distance D of an object can be determined by, Knowing basis (b) and parallax angle (𝜃), Q. How many meters make one parsec?, Ans: 1 𝑝𝑎𝑟𝑠𝑒𝑐 = 3.08 x 108 𝑚, Q. Define light year. Give its value in SI Unit., Ans: The distance travelled by light in vacuum in one year is called light year., 𝟏 𝒍𝒚 = 𝟗. 𝟒𝟔 × 𝟏𝟎𝟏𝟓 𝒎., Q. How many kilograms in one unified atomic mass unit?, Ans: 𝟏. 𝟔𝟔𝟏𝐱𝟏𝟎−𝟐𝟕 kg., Q. Which clock is used to measure time accurately?, Ans: Atomic clock., Q. Define accuracy in the measurement. How accuracy does depend on precision in the, measurement?, Ans: Accuracy refers to how close a measurement is to the true value of what is being, measured. Precision refers to how close measurements of the same quantity are to each other,, even if they are not close to the true value. For example, the darts on the dart boards below, represent sets of measurements. A bull's eye represents a perfect measurement-a measurement, exactly the same as the true value., Q. Define error. Mention one type of error., , 8

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Ans: The difference between the actual (true) value and the measured (observed) value of a, physical quantity is called an error., , Q. What is systematic error?, Ans: The errors which are due to known causes acting according to a definite law are called, systematic errors., Q. Write any two systematic errors., Ans: Systematic errors: (any two), (i), , Equipment error, , (ii), , Environment error, , (iii) Processes error, , (iv) Calculation error, , (v), , (vi) Software error, , Calculation error, , (vii) Data sources error, , (viii) Data Processing error, , Q. Define absolute error., Ans: The magnitude of the difference between the actual value and the measured value is called, absolute error. Absolute error = Arithmetic mean – Measured value, Q. What is relative error (Fractional error)?, Ans: The ratio of the mean absolute error to the arithmetic mean (actual value) is called relative error., Q. What are the significant figures? Write the number of significant figures in 𝟔𝟓 × 𝟏𝟎𝟑 𝒌𝒈., Ans: Significant figures: The digits, whose values are accurately known and the first uncertain, digit in a, , particular measurement are known as significant figures., , The number of significant figure for 65 x 103kg is Two (2), Q. What are the significant figures? Write the number of significant figures in 𝟐. 𝟔𝟓 × 𝟏𝟎𝟑 𝒌𝒈., Ans: Significant figures: The digits, whose values are accurately known and the first uncertain, digit in a, , particular measurement are known as significant figures., , The number of significant figure for 2.65 x 103kg is Three (3), , Q. Write any three rules for finding number of significant figures., Ans: Rules for finding number of significant figures (any three), (i). All the non-zero digits are significant., (ii). All zeros between two non-zero are significant., (iii). All zeros to the right of the last non-zero digit are not significant., 9

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(iv). For a number less than 1, the zeros on the right of decimal point but to the left of the, first non-zero digit are not significant., (v). All zeros to the right of decimal point and to the right of a non-zero digit are, significant., Q. Give the Number of significant figures in 6.200 ×100., Ans: The number of significant figure for 6.200 x 100 is Four (4)., Q. Give the Number of significant figures in 5.300 ×103., Ans: The number of significant figure for 5.300 x 103 is Four (4)., Q. Write the number of significant figures in 287.5 m., Ans: The number of significant figure for 287.5 is Four (4)., Q. Write the number of significant figures. (i) 0.000456, , (ii) 4.200 x 10 8, , Ans: The number of significant figure for 0.000456 is Three (3), The number of significant figure for 4.200 x 108 is Four (4), Q. Write the number of significant figures of following (a) 0.007, (b) 1.2340, Ans: The number of significant figure for 0.007 is One (1), The number of significant figure for 1.2340 is Five (5), Q. Write the dimensional formula for wavelength., Ans: The dimensional formula for wavelength: [M0L1T0], Q. Write the dimensional formula for linear momentum., Ans: The dimensional formula for linear momentum: [ 𝑀1 𝐿1 𝑇 −1 ], Q. Write the dimensional formula for impulse of a force., Ans: The dimensional formula for work: [M1L1T-1], Q. Write the dimensional formula for force., Ans: The dimensional formula for force: [M1L1T-2], Q. Write the dimensional formula for tension., Ans: The dimensional formula for tension: [𝑀1 𝐿1 𝑇 −2 ]., Q. Write the dimensional formula for force per unit length., Ans: The dimensional formula for force per unit length: [ML0T-2] (or) [MT-2], , 10

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Q. Write the dimensional formula for work., Ans: The dimensional formula for work: [ML2T-2], Q. Write the dimensions or dimensional formula for torque., Ans: The dimensions or dimensional formula for torque: [ML2T-2], Q. Write the dimensional formula for power., Ans: The dimensional formula for power: [ML2T-3], Q. Write the dimensional formula for pressure., Ans: The dimensional formula for pressure: [ML-1T-2], Q. Mention the physical quantities whose dimensions are [𝑴𝟏 𝑳𝟐 𝑻−𝟐 ]., Ans: [𝑴𝟏 𝑳𝟐 𝑻−𝟐 ] : work (or) torque, Q. Mention the physical quantities whose dimensions are (a) [𝑴𝟏 𝑳−𝟏 𝑻−𝟐 ], (b) [𝑴𝟏 𝑳𝟐 𝑻−𝟑 ], Ans: (a)[𝑴𝟏 𝑳−𝟏 𝑻−𝟐 ] : Pressure (or) stress (or) Young’s modulus, (b)[𝑴𝟏 𝑳𝟐 𝑻−𝟑 ] : Power, Q. Write dimensional formula of (a) Force, (b) Energy, Ans: (a) Dimensional formula of Force: [ 𝑀1 𝐿1 𝑇 −2 ], (b) Dimensional formula of Energy: [ 𝑀1 𝐿2 𝑇 −2 ], Q. Mention a physical quantity which has neither dimensions nor a unit., Ans: Refractive index (or) Relative density (or) Coefficient of friction, Q. Name a dimensionless physical quantity which has no unit., Ans: Refractive index (or) relative density., Q. Mention any two uses (applications) of dimensional analysis., Ans: 1. To check the correctness of an equation., 2. To derive an equation (relation) between different physical quantities., 3. To convert one system of unit into another., Q. Mention any two limitations of dimensional analysis., Ans: Limitations of dimensional analysis., (1) If the physical quantity depend more than three fundamental quantities, then the method, of dimensional analysis cannot be used., , 11

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3. MOTION IN A STRAIGHT LINE, Q. Distinguish between path length (distance) and displacement., Ans:, path length (distance, 1. It is the change in the position of, , displacement, 1. It is the change in the position of the, , the particle in a given time interval, , particle in a given time interval in a, , without reference to direction., , particular direction., , 2. It is a scalar quantity., , 2. It is a vector quantity., , 3. It depends on the actual path, , 3. It does not depend on the actual path, , covered by the particle., 4. It is always positive and cannot be, zero or negative for a moving, , covered by the particle., 4. It can be positive, negative or zero, for a moving particle., , particle., 5. It is equal to or greater than, displacement, but cannot be less, , 5. It is equal to or less than distance, but cannot be greater than distance., , than displacement., , Q. What does the slope of position - time graph represent?, Ans: Velocity., , Q. Draw the Position – time graph for an object moving with a positive velocity., Ans: moving with a positive velocity, , 15

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Q. Draw the Position – time graph for an object (i) at rest and (ii) moving with a negative, velocity., Ans: (i) at rest:, , (ii) moving with a negative velocity:, , Q. What is the nature of position –time graph for uniform motion?, Draw the position –time graph., Ans: The nature of uniform motion for a position –time graph is straight line and inclined to, the time axis., Position –time graph for uniform motion:, , Q. Define velocity., Ans: The velocity is defined as “rate of change in displacement”, Q. Write any two differences between speed and velocity., Ans: Any two of the following, Speed, , Velocity, , 1. It is the distance covered per unit, , 1. It is the displacement made per unit, , time., , time., , 2. It is a scalar quantity., , 2. It is a vector quantity., , 3. It can be positive or zero., , 3. It can be positive, zero or negative., , 4. Average speed cannot be zero., , 4. Average velocity can be zero., , 16

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Q. Define average velocity and instantaneous velocity., Ans: Average velocity: The average velocity is defined as the change in position, or displacement (∆𝑥) divided by the interval (∆𝑡) in which the displacement occurs., 𝑣⃗ =, , 𝑥2 − 𝑥1 ∆𝑥, =, 𝑡2 − 𝑡1, ∆𝑡, , Where 𝑥1 and 𝑥2 are the positions of a body in time 𝑡1 and 𝑡2 respectively., Instantaneous velocity: The instantaneous velocity is defined as the limiting, value of average velocity as the time interval value approaches to zero., If d𝑥⃗ is the small displacement of a body in a very small time dt, then its, ∆𝑥⃗, , Instantaneous velocity is: 𝑣⃗ = 𝐿𝑡 ∆𝑡 =, , 𝑑𝑥⃗, 𝑑𝑡, , ., , Q. Define instantaneous velocity of a body in terms of its average velocity., Ans: Instantaneous velocity: the velocity of an object in motion at a specific point in time., This is determined similarly to average velocity, but we narrow the period of time so that, it approaches zero., Q. What does the slope of velocity - time graph represent?, Ans: Acceleration., Q. What are the significances of velocity - time graph?, Ans: Significances of velocity - time graph:, (i) what speed the object is travelling at that point in time., (ii) calculate any acceleration, the change in speed and the change in time., (iii) calculate distance travelled. The area under a speed-time graph represents the, distance travelled., Q. Draw v-t graph for motion of particle with Variable velocity., Ans: Variable velocity., , 17

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Q. Draw the velocity time graph of a particle moving with uniform velocity starting from, rest., (a) Uniform velocity:, , Q. Draw the velocity time graph of a particle moving with (a) uniform velocity and, (b) Uniform acceleration., Ans:, (a) Uniform velocity, , (b) uniform acceleration, , Q. Define acceleration., Ans: Acceleration of a body is defined as rate of change of velocity of the body., Q. What is retardation? Write its SI unit., Ans: If the velocity of the body decreases with time, then acceleration is negative or retardation, or de-acceleration., Retardation S.I. unit: ms-2., Q. What is the acceleration of the body moving with uniform velocity?, Ans: Zero, , Q. Write the expression for average acceleration when the particle is moving in XY- plane., , Ans: 𝑎𝑎𝑣𝑔 =, , |∆𝑣|, ∆𝑡, , =, , |𝑣𝑥 −𝑣𝑦|, 𝑡2 −𝑡1, , =, , √𝑣𝑥2+𝑣𝑦2, ∆𝑡, , 18

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Q. Draw the velocity-time graph for particle in the following situation., (a) Starting from rest and moving with uniform acceleration., (b) Moving with uniform acceleration., Ans: (a) Starting from rest and moving with uniform acceleration., , (b) Moving with uniform acceleration., , Q. Draw the velocity time graph of a particle moving with Positive acceleration starting with, some initial velocity., Ans: Positive acceleration starting with some initial velocity:, , Q. Draw the velocity – time graph for the body moving with uniform retardation., Ans: Uniform retardation:, , 19

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𝟏, , Q. What is v-t graph? Derive the equation 𝑺 = 𝐯𝟎 𝐭 + 𝟐 𝐚𝐭 𝟐 by graphical method., (Or) What is v-t graph? Show that area under v-t graph is equal to distance., Ans: A graph plotted between the velocity of a body along y–axis and the time along x–axis is, called v-t graph., Consider a body moving with uniform acceleration a. Let u be its initial velocity, v be its, final velocity in time t and S be the distance travelled. The v-t graph for a body having some, initial velocity and moving with uniform acceleration is shown below., OA = CD = u, BC = v, OC = AD = t, Acceleration = Slope of the v-t graph, a=, a=, a=, , BD, AD, BC−CD, OC, v−u, t, , at = v-u, v=u+at, , ----------- (1), , Distance travelled in time t = Area enclosed by the line AB, S = area of trapezium OABC, 1, , S = 2x base x sum of parallel sides, 1, , S = x OCx (OA + BC), 2, , 1, , S = 2x t (u + v), S=, , 1, 2, , x t (u + u + at), , 1, , S = 2x t (2u + at), 𝟏, , S = ut + 𝟐 𝐚𝐭 𝟐 --------------- (2), If u = V0, then equation (2) becomes:, 𝟏, , 𝐒 = 𝐕𝟎 𝐭 + 𝟐 𝐚𝐭 𝟐, If u = V0 and S=X then equation (2) becomes:, 𝟏, , 𝐗 = 𝐕𝟎 𝐭 + 𝟐 𝐚𝐭 𝟐, 20

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Q. What is v-t graph? Derive v2 = vo2 + 2ax using v-t graph., Ans: A graph plotted between the velocity of a body along y–axis and the time along x–axis is, called v-t graph., Consider a body moving with uniform acceleration a. Let u be its initial velocity, v be its final, velocity in time t and s be the distance travelled. The v-t graph for a body having some initial, velocity and moving with uniform acceleration is shown below., OA = CD = u, BC = v, OC = AD = t, , Acceleration = Slope of v–t graph, a=, , BD, AD, , a=, , BC  CD, OC, , a=, , vu, t, , t=, , vu, ………………..(1), a, , Distance travelled in time t = Area enclosed by the v-t graph, , S = area of trapezium OABC, S=, , 1,  OC  (OA  BC ), 2, , S=, , 1,  t  ( u  v), 2, , S=, , vu, 1vu, , ( v  u )  t =, 2 a , a, , S=, , 1 2, (v  u 2 ), 2a, , 2as = v2 – u2,  v2 = u2 + 2as................... (2), , If u = v o and s = x, then equation (2) becomes:, v2 = vo2 + 2ax ………..…… (3), , 21

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Q. A man with a wrist watch on his hand falls from the top of tower. How does the watch, show time during the free fall?, Ans: The wrist watch uses an electronic system or spring system to give the time, which, does not change with acceleration due to gravity. Therefore, watch gives the correct time., 2ℎ, , (𝑡 = √ 𝑔 )., Q. A player throws a ball upwards with an initial speed of 29.4𝒎𝒔−𝟏. What is the direction of, acceleration during upward motion? Find its velocity at the highest point of its motion?, Ans: The direction of acceleration during upward motion is downwards (a = -g)., Velocity at the highest point of its path is v = 0, , Q. What is relative velocity? Mention the expression for relative velocity., Ans: Relative velocity: The velocity of one object with respect to another moving object is, called relative velocity. (or), Relative velocity is the velocity of an object relative to some other object which might be, stationary, moving slowly, moving with same velocity, moving with higher velocity or moving, in opposite direction., Moving in the same direction:, Relative velocity of A with respect to B is: VAB = VA − VB, Relative velocity of B with respect to A is: VBA = VB − VA, , Q. What is relative velocity? Explain with an example., Ans: Relative velocity: The velocity of one object with respect to another moving object is, called relative velocity., Example: A motorcycle travelling on the highway at a velocity of 120km/h passes a car, travelling at a velocity of 90km/h., , 22

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Q .A train accelerates form 36kmph to 72kmph in covering a distance of 50m. Calculate the, acceleration of the train and the time taken to cover the distance., Solution:, Given,, 5, , u=36kmph=36x18 = 10ms-1, 5, , v= 72kmph=72x18 = 20ms-1, S= 50m, v2= u2+2as, a=, a=, a=, , V2 – u2, 2𝑠, 202 – 102, 2(50), 400– 100, 100, 300, , a = 100, a = 3ms-2, v = u + at, t=, t=, t=, , v−u, 𝑎, 20−10, 3, 10, 3, , t = 3.33sec, Q. A player throws a ball upwards with an initial speed of 29.4𝒎𝒔−𝟏., (a) What is the velocity at the highest point of its path?, (b) To what height, does the ball rise and after how long, does the ball return to the, player’s hands? Take g=9.8𝒎𝒔−𝟐 . Neglect air resistance., Solution:, (a), Velocity at the highest point of its path is, v = 0, (b), , u = 29.4𝑚𝑠 −1 , v = 0 and g = 9.8𝑚𝑠 −2 , H =?, , We know that, v 2 − u2 = 2gH, 0 − (29.4)2 = −2(9.8)𝐻, 𝐻=, , (29.4)2, 2𝑥9.8, , 𝑯 = 𝟒𝟒. 𝟏 𝒎., , 26

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𝑻=, , 𝟐𝑼, 𝒈, , ⇒𝑻=, , 𝟐(𝟐𝟗.𝟒), 𝟗.𝟖, , 𝑻 = 𝟔𝒔, Q. A body moving with initial velocity 83 m/s moved under constant retardation of 3 m/s2., Find its velocity after 2 sec. Time after which it will stop and the distance it will travel, before coming to rest., Solution:, Given;, Initial velocity (u) = 83 m/s, Retardation = -a =3m/s2,  a = -3 m/s2, (a) Time (t) = 2 sec, Velocity after 2 sec is, V =?, We have v = v + at, V= 83 + (-3) x 2, V = 83-6 = 77m/s, (b) Time after body will stop (t) =?, V = 0 m/s, Distance it will travel before coming to rest (s) =?, We have V = u + at, 0 = 83+ (-3) x t, 3t = 83, 83, , t=, , 3, , s = 27.66 sec, , and, V2 – u2= 2as,  02 – (83)2 = 2 x (-3) x s, s=, , − 83×83, , s=, , −6, 6889, 6, , = 1148.16m, , s = 1148.16 m, Q. An automobile travelling with an initial velocity of 40 ms-1 comes to rest after travelling a, distance of 100m. Assuming the retardation to be uniformly then find its value. What, time does it take to cover the distance?, Solution:, Given;, 27

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Initial velocity U=40 ms-1,, final velocity V= 0ms-1 and, distance S=100m., Retardation :, By using the concept of equations of motion, 𝑣 2 − 𝑢2 = 2𝑎𝑠, 02 − 402 = 2𝑎(100), 200𝑎 = −1600, 𝑎=, , −1600, 200, , retardation 𝒂 = −𝟖𝒎𝒔−𝟐, (b) Time after body will stop (t) =?, V = 0 m/s, We have V = u + at, 0 = 40 + (-8) x t, 8t = 40, t=, , 40, 8, , s, , t= 5 sec, , Q. A constant retardation force of 50N is applied to a body of mass 20kg moving initially, with a speed of 15 ms-1. How long does the body take to stop?, Solution: Given, Constant retardation force F = 50N, mass of body m = 20 kg, initially velocity u=15m/s, t=?, Body moves with constant retardation so final velocity reached to 0m/s to stop,, , ∴ V = 0m/s., , 𝐹 = 𝑚𝑎, 𝑎=, , 𝐹, 𝑚, 50, , 𝑎 = 20, 𝒂 = 𝟐. 𝟓𝒎𝒔−𝟐, Time after body will stop (t) =?, V = 0 m/s, We have V = u + at, 0 = 15 + (-2.5) x t, 28

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2.5t = 15, 15, , t = 2.5 s, t= 6 sec, Q. A stone dropped from the top of building reaches the ground in 2 second. What is the, height of the building? ( given g=10m/s2 ), Solution:, Given: t=2s, g=10m/s2, h=?, 2ℎ, , 𝑡 = √𝑔, , 2ℎ, , 2 = √ 10, 2ℎ, , 4=, ℎ=, , 10, 40, 2, , 𝒉 = 𝟐𝟎𝒎, Q. A stone dropped from the top of building reaches the ground in 3 second. What is the, height of the building? (given g=10m/s2), Solution:, Given: t=3s,, g=10m/s2,, h=?, 2ℎ, , 𝑡 = √𝑔, , 2ℎ, , 3 = √ 10, 2ℎ, , 9 = 10, ℎ=, , 90, 2, , 𝒉 = 𝟒𝟓𝒎, Q. An iron ball (A stone) is dropped from the top of a tower 100 m height. At the same time, another iron ball is thrown vertically upwards from the base of the, tower with a velocity of 50 m/s. When and where two iron balls, meet? Given g=10m/s2., Solution: Given From equation of motion, Case1: An iron ball is dropped from the top of a tower, 1, , S1 = ut + 2 at2, 29

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Q. Two towns A and B are connected by regular bus service with a bus leaving in either, direction every T minutes. A man cycling with speed of 20kmph in the direction A to B, notices that a bus goes past him every 18minin the direction of his motion, and every 6 min, in the opposite direction. What is the period T of the bus service and with what speed, (assumed constant) do the buses ply on the road?, Solution:, Given;, Case1 :, As both the bus and cyclist are moving in the same direction their relative velocity will be, Relative velocity of bus = VB−VC = VB - (+20), Distance covered =VT= Distance =Speed × Time, (VB−20) × 18 = VBT ............ (1), Case 2 − Bus moving from B to A:, As the bus and cyclist are moving in the opposite direction their relative velocity will be, Relative velocity of bus = VB−VC = VB -(-20), Distance covered =VT= Distance =Speed × Time, (VB + 20) × 6 = VBT ............ (2), Solving equations (1) and (2), (VB−20)×18, VB=40 km / hr, Substituting value of VB in equation (2):, (40+20)×6=40T, T=9minutes, Speed of the bus is 40km/h, and time interval after which each bus leaves is 9 minutes., , 34

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4. MOTION IN A PLANE, Q. Define vector quantity., Ans: A physical quantity having both magnitude and direction is called vector quantity., Q. What is vector quantity? Give an example for it., Ans: A physical quantity having both magnitude and direction is called a vector quantity., Example: Displacement, Q. Distinguish between scalar and vector., Ans:, Scalar, 1.Physical, , quantities, , Vector, having, , only 1.Physical quantities having both, , magnitude are called scalars., , magnitude and direction and that obey, vector law of addition are called vectors, , 2. Ex: Distance, speed, Area, volume, , 2.Ex: Displacement, velocity, force, , ….etc, , …..etc, , 3. They obey ordinary laws of algebra, , 3. They do not obey ordinary laws of, algebra ( follow vector rules ), , Q. When the two vectors are said to be equal?, Ans: The two vectors are said to be equal if they have both magnitude and direction are same., , Q. What is position vector?, Ans: A vector that gives the position of a particle at an instant with respect to the origin of a coordinate system is called position vector., , Q. (a) What is null vector? (b) Is a scalar multiplied by a vector, a vector or a scalar?, Ans: (a) null vector: A vector whose magnitude is Zero and direction is indeterminate, (or arbitrary) is called null vector., (b) Scalar multiplied by a vector gives a vector., , Q. What is a unit vector? (Or) Define unit vector?, 35

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Q. Mention any three methods of vector addition., Ans: Three methods of vector addition, 1). Triangle law of vector addition., 2). Parallelogram law of vector addition., 3). Polygon law of vector addition., , Q. State the law of triangle of addition of two vectors?, Ans: Triangle law of vector addition :It states that , if two vectors can be represented both in, magnitude and direction by the two sides of a triangle taken in the same order, then their, resultant is represented completely both in magnitude and direction by the third side of a, triangle taken in the opposite direction., , Q. State and explain triangle law of vector addition. When will be the resultant of two given, vectors be become maximum., Ans: Triangle law of vector addition :It states that , if two vectors can be represented both in, magnitude and direction by the two sides of a triangle taken in the same order, then their resultant is, represented completely both in magnitude and direction by the third side of a triangle taken in the, opposite direction., ⃗⃗⃗⃗⃗⃗, ⃗⃗ 𝑎𝑛𝑑 ⃗⃗⃗⃗⃗⃗, 𝐴𝐵 = 𝑃⃗⃗, ⃗⃗⃗⃗⃗⃗, 𝐵𝐶 = 𝑄, 𝐴𝐶 = 𝑅⃗⃗, ⃗⃗ be the two vectors represent both in the, Let 𝑃⃗⃗ and 𝑄, magnitude and direction by the sides AB and BC of a, triangle ABC taken in the same order. Then, the resultant R, is given by third side AC of the triangle taken in the opposite, order. From the triangle law of vectors addition:, ⃗⃗⃗⃗⃗⃗, ⃗⃗, 𝐴𝐶 = ⃗⃗⃗⃗⃗⃗, 𝐴𝐵 + ⃗⃗⃗⃗⃗⃗, 𝐵𝐶 ⇒ 𝑅⃗⃗ = 𝑃⃗⃗ + 𝑄, , The resultant of two given vectors be become maximum:, , ⃗⃗, 𝑅⃗⃗ = 𝑃⃗⃗ + 𝑄, , Q. What is the minimum number of vectors to give zero, resultant?, Ans: As the magnitude of the vectors is not equal so, two vectors cannot give zero resultant. According to, , 37

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the Triangle Law of vector addition, a minimum of three vectors are needed to get zero, resultant., , Q. State and explain law of parallelogram of vectors., Ans : Parallelogram of vectors: It states that , if two vectors acting at a point can be, represented both in magnitude and direction by the adjacent sides of a parallelogram , then, their resultant is represented completely both in magnitude and direction by the diagonal of the, parallelogram drawn from the same point., ⃗⃗ and Q, ⃗⃗⃗ acting at a point O are the two vectors inclined at an angle"𝜃". Let them be, Let P, represented both in magnitude and direction by the adjacent sides OA and OB of the, ⃗⃗, parallelogram OACB. According to the parallelogram law of vectors addition, the resultant R, of ⃗P⃗ and ⃗⃗⃗, Q is represented both in magnitude and direction by the diagonal OC of the, parallelogram drawn from the point O., , Magnitude of the resultant: |𝑅| =, , √𝑃2 + 𝑄2 + 2𝑃𝑄 cos 𝜃, Q. State the law of parallelogram of vectors. Derive an expression for magnitude of resultant, of two concurrent forces using parallelogram law of addition. OR, Derive an expression for magnitude and direction of resultant of two vectors P and Q, acting at an angle  ., Ans: Parallelogram of vectors: It states that , if two vectors acting at a point can be, represented both in magnitude and direction by the adjacent sides of a parallelogram , then, their resultant is represented completely both in magnitude and direction by the diagonal of the, parallelogram drawn from the same point., Let P and Q be the two vectors acting at a point O and  be the angle between them. Let them, be represented by the adjacent sides OA and OB respectively of a parallelogram OACB. The, diagonal OC represents the resultant R which makes an angle, , , , with P., , A perpendicular CD is drawn to OA produced., From the right angled  ODC :, OC2 = OD2 + DC2= (OA + AD)2 + DC2, OC2 = OA2 + AD2 + 2.OA.AD + DC2 ……(1), From the right angled  ADC:, sin  =, , DC, AC, , DC = AC .sin  ………………….………...(2), 38

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𝑅 = √𝑃2 + 𝑄2 + 2𝑃𝑄 cos 𝜃, If 𝜃 = 180°(anti parallel) so that cos 180° = −1 then 𝑅 = √𝑃2 + 𝑄2 + 2𝑃𝑄(−1), 𝑅 = √(𝑃 − 𝑄)2, R = 𝑷 − 𝑸., Q. Define Scalar product of two vectors. Give an example., Ans: The scalar product (dot product) of two vectors is defined as the product of magnitudes, of two vectors and cosine of the angle between them., , ⃗⃗ = |𝐴⃗||𝐵, ⃗⃗| cos 𝜃, 𝐴⃗ . 𝐵, , Ex: Work (W) is equal to the scalar product (dot product) of force (𝐹⃗ ) and displacement (𝑆⃗)., ⃗⃗⃗. 𝑺, ⃗⃗ = 𝑭𝑺 𝐜𝐨𝐬 𝜽, 𝑾= 𝑭, , Q. When will be the dot product of two vectors maximum?, Ans: If two vector are in same direction(𝜃 = 0°) then the dot product of two vectors, maximum., , Q. Give an example of dot product and cross product of two vectors., Ans: Dot product: Ex: Work (W) is equal to the dot product of force (𝐹⃗ ), and displacement (𝑠⃗ ), , 𝑊 = 𝐹⃗ . 𝑠⃗ = Fs cos 𝜃, , Cross product: Ex: Torque(𝜏) is equal to the cross product of position vector (𝑟⃗), and force (𝐹⃗ )., , 𝜏⃗ = 𝑟⃗ × 𝐹⃗, , Q. What is Projectile?, Ans: Projectile: An object thrown into space which moves under the effect of gravity alone is, called projectile., Q. Can a body possess velocity at the same time in horizontal and vertical direction?, Ans: Yes. (Note: projectile), Q. What is the angle between velocity and acceleration at the peak point of a projectile, projected for maximum range?, Ans: 90°, , Q. Which component of velocity of a projectile remains constant?, Ans: The horizontal component of velocity of a projectile (𝑢𝑥 = 𝑢 cos 𝜃) remains constant, , 40

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Q. Give an example for two dimensional motions., Ans: (i) projectile., (ii) A body released from an aero plane., (iii) A bullet fired from a gun., Q. Give an example for a projectile., Ans: (i) A body released from an aero plane., (ii) A bullet fired from a gun., Q. What is the nature of trajectory of projectile?, Ans: The nature of trajectory of projectile is parabola., , Q. What is meant by projectile? Mention the expression for trajectory of the projectile., Ans: An object thrown in air (or space) other than 90o then the object moves with under action, of gravity is called projectile., 𝑔, , Expression for trajectory of the projectile: 𝑌 = (𝑡𝑎𝑛𝜃)𝑥 − (2𝑈 2 𝑐𝑜𝑠 2𝜃 ) 𝑥 2, Or 𝑌 = 𝐴𝑋 − 𝐵𝑋 2 ., , Q. What is a projectile? Show that the path traced by the projectile is a parabola., Ans: Projectile: An object thrown into space which moves under the effect of gravity, alone, is called projectile., The path traced by the projectile is a parabola:, Consider a projectile projected from the point O with a velocity u making an angle𝜃 with the, horizontal. The velocity u is resolved into two components: 𝑢𝑥 = ucos 𝜃 along the horizontal, direction OX and 𝑢𝑦 = usin 𝜃 along the vertical direction OY. The horizontal component, 𝑢𝑥 remains constant and the vertical component 𝑢𝑦 changes with time. Let P, (x ,y) be the position of the projectile in time t., , The distance travelled can be calculated using the equation,, 41

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Where: 𝑇𝑓 =, , 2𝑢 sin 𝜃, 𝑔, , 𝑅 = 𝑢 cos 𝜃 ×, 𝑅=, 𝑅=, , 2𝑢 sin 𝜃, 𝑔, , 𝑢 22 sin 𝜃 cos 𝜃, 𝑔, 𝑢 2 sin 2𝜃, 𝑔, , (∵sin 2𝜃 = 2 sin 𝜃 cos 𝜃), , NOTE: For maximum range, 𝜃 = 450 ⇒ 𝑅𝑚𝑎𝑥 =, , 𝑢2, 𝑔, , Q. Write SI unit for range of projectile., Ans: m (meter), , Q. When is the range of a projectile maximum? Or, For what angle of projection, does the range of a projectile become maximum?, Ans: If the angle of projection is 450 then the range of a projectile becomes Maximum., For maximum range, 𝜃 = 450 ⇒ 𝑅𝑚𝑎𝑥 =, , 𝑢2, 𝑔, , ., , Q. Define time of flight of projectile motion, write an expression for it., Ans: It is the total time during which the projectile is in flight., Time of flight (Tf) =, , 2𝑢 sin 𝜃, 𝑔, , Q. Obtain the expression for time of flight of a projectile., Ans: Expression for time of flight of a projectile: Consider a projectile projected from the, point O with a velocity u making an angle 𝜃, with a velocity u making an angle θ with the, horizontal direction. The velocity u is, resolved into two components: 𝑢𝑥 = 𝑢 cos 𝜃, along the horizontal direction OX and, uy = usinθ along the vertical direction OY., The horizontal component, 𝑢𝑥 (𝑢𝑥 = 𝑈 cos 𝜃) remains constant and, the vertical component 𝑢𝑦 changes with time., Time of flight (T): The time of ascent (ta) can be calculated using the equation,, vy = uy + at, 44

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For vertical motion:vy = 0, uy = usinθ, a= -g and t= t a ., From equation, 0 = usinθ − gt a, gt a = usinθ, ta =, , u sin θ, g, , Time of flight = 2x Time of ascent., 𝑇 = 2(t a), 𝑇 = 2(, 𝐓=, , 𝑢 sin 𝜃, 𝑔, , ), , 𝟐𝒖 𝐬𝐢𝐧 𝜽, 𝒈, , Q. Give an example for three dimensional motions., Ans: Motion of an aeroplane., Q. What is uniform circular motion?, Ans: Uniform Circular Motion: The motion of a body along a circular path with constant, speed is called uniform circular motion., , Q. What is uniform circular motion? Write an expression for angular velocity., Ans: When a body moves in a circular path with uniform speed or constant speed its motion is, known as uniform circular motion., Angular velocity: 𝑣⃗ = 𝜔, ⃗⃗ × 𝑟⃗ (𝑜𝑟) 𝑣 = 𝑟𝜔., , Q. Write the relation connecting angular velocity and linear velocity., Ans: The relation connecting angular velocity and linear velocity: 𝑣⃗ = 𝜔, ⃗⃗ × 𝑟⃗ (𝑜𝑟) 𝑣 = 𝑟𝜔., , Q. Obtain the relation between linear velocity and angular velocity in uniform circular, motion., Ans: Relation between linear velocity and angular velocity:, Consider a particle moving along a circle of radius r with an angular velocity (𝜔)., 45

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and the instantaneous linear velocity (𝑣) is:, 𝑑𝑠, , 𝑣 = 𝑑𝑡, 𝑣=, , (𝑑(𝑟𝜃)), 𝑑𝑡, 𝑑𝜃, , 𝑣 = 𝑟 𝑑𝑡, 𝑣 = 𝑟𝜔, , ⃗⃗ = 𝝎, ⃗⃗⃗⃗ × 𝒓, ⃗⃗, Vector form: 𝒗, Q. Distinguish between linear and angular velocity., Ans:, Linear velocity, , Angular velocity, , 1. Velocity of a body is defined as, , 1. The ratio of angular displacement, , rate of change of position of a body, in a particular direction, , ( or ), , to the time interval is called, angular velocity., , Velocity of a body is defined as, the displacement of a body per, unit time., 2. SI unit is m/s, , 2. SI unit is rad /s, , Q. Define linear and angular acceleration write the relation between them., Ans: Linear acceleration: Acceleration of a body is defined as rate of change of velocity, of the body., Angular acceleration: The rate of change of angular velocity is called angular acceleration., Relation between linear acceleration and angular acceleration: Consider a particle, moving along a circle of radius r with an angular acceleration𝛼 ., The instantaneous linear acceleration is:, 𝑎=, , 𝑑𝑣, , 𝑎=, , 𝑑(𝑟𝜔), , 𝑑𝑡, , 𝑎=𝑟, , 𝑑𝑡, 𝑑𝜔, 𝑑𝑡, , 46

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𝑎 = 𝑟𝛼, ⃗⃗ = 𝜶, ⃗⃗⃗ × 𝒓, ⃗⃗, Vector form: 𝒂, Q. What is centripetal acceleration? How it is related with radius of the circular path., Ans: The acceleration produced in the particle in uniform circular motion which acts along the, radius and towards the centre of the circular path is called centripetal acceleration., The Centripetal acceleration decreases as the distance from the axis of rotation increases (if, Angular velocity remains constant), Q. Mention any two factors on which the centripetal acceleration depends., Ans: 1. Velocity and Angular velocity, 2. Radius, Q. What is centripetal acceleration? Give an expression for it and explain the terms., Ans: The acceleration produced in the particle in uniform circular motion which acts along the, radius and towards the centre of the circular path is called centripetal acceleration., Expression for centripetal acceleration:, 𝒂=, , 𝒗𝟐, 𝒓, , = 𝒓𝝎𝟐, , Where 𝒂 is centripetal acceleration,, 𝒗 is linear velocity (or) tangential velocity, 𝒓 is radius of circular path, 𝝎 is angular velocity, , Q. What is uniform circular motion? Derive an expression for centripetal acceleration., Ans: Uniform Circular Motion: The motion of a body along a circular path with constant, speed is called uniform circular motion., Expression for centripetal acceleration: Consider a particle moving in a circular path of, radius r with constant speed v and uniform velocity𝛚. Let the particle moves from A to B in, time t and θ be the small angular displacement made by the radius, vector., The velocity at B is resolved into two components: v𝐜𝐨𝐬 𝛉 parallel, to AC and v𝐬𝐢𝐧 𝛉 parallel to AO, When the particle moves from A to B:, , 47

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(b) The time taken by the ball to reach the maximum height:(time of ascent), 𝒖𝒔𝒊𝒏𝜽, , 𝒕=, , 𝒈, , 𝑡=, , (25)𝑠𝑖𝑛30, , 𝑡=, , 9.8, 1, 2, , (25)( ), 9.8, , t =1.275sec, Q. A cricket ball is thrown at a speed of 28ms-1 in the direction 300 above the horizontal., Calculate (a) maximum height reached by the projectile and (b) time taken by it to return, to the same level (c) horizontal range of the projectile. ( g = 9.8ms-2) ., Solution:, Given;, Initial velocity is 𝑢 = 28𝑚𝑠 −1, Angle of projectile is 𝜃 = 300, (a) The maximum height up to which a projectile can go is by the formula:, 𝑯=, 𝐻=, 𝐻=, , 𝒖𝟐𝒔𝒊𝒏𝟐 𝜽, 𝟐𝒈, (28)2 𝑠𝑖𝑛 2 (30), 2×9.8, (28)2 ×(0.5)2, 2×9.8, , 𝑯 = 𝟏𝟎𝒎, (b) the time taken to return to the same level ( time of flight(T)):, T is Time of flight 𝑻 =, , 𝟐𝒖 𝐬𝐢𝐧 𝜽, 𝒈, , Substituting the values in it,, 𝑇=, , 2×28×sin 30, 9.8, , 𝑻 = 𝟐. 𝟖𝟔𝒔, 51

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Q. A cricket ball is thrown at a speed of 50m/s in a direction 𝟑𝟎𝟎 above the horizontal., , Calculate (a) The maximum height, (b) The time of flight (g=9.8m/s2)., Solution:, Given;, Initial velocity (u) = 50m/s, Angle (𝜃) =300 , g = 9.8 𝑚𝑠 −2, (a) Maximum height:, H=, , 𝐻=, , 𝒖𝟐 𝑺𝒊𝒏𝟐 𝜽, 𝟐𝒈, 1 2, 2, , 50×50×( ), 2×9.8, 2500×, , 𝐻=, , 1, 4, , 2 × 9.8, , 𝐻=, , 625, 19.6, , 𝑯 = 𝟑𝟏. 𝟖𝟗𝒎, (b) Time taken to return to same level:, 𝑻=, 𝑇=, 𝑇=, 𝑇=, , 𝟐𝒖 𝐬𝐢𝐧 𝜽, 𝒈, 2 × 50 × sin 300, 9.8, 1, 2, , 2 × 50x( ), 9.8, , 2 ×25, 9.8, , 𝑻 = 𝟓. 𝟏𝒔𝒆𝒄, Q. A cricket ball is thrown at a speed of 56m/s in a direction 𝟑𝟎𝟎 above the horizontal., , Calculate (a) The maximum height, (b) The time taken by the ball to return to the same, level (c) The distance from the thrower to the point where ball returns to the same level., Solution:, Given;, Initial velocity (u) = 56m/s, 54

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u 2 sin 2, R, g, , ………………………………(1), , Height of projectile is given by:, , H, , u 2 sin 2 , 2g, , ………………………………(2), , Dividing equation (2) by equation (1):, , H 1 sin 2 ,  , R 2 sin 2, H 1 sin  sin ,  , R 2 2 sin  cos , H 1,   tan , R 4, , 𝐭𝐚𝐧 𝜽 =, , 𝟒𝑯, 𝑹, , tan  , , 4  10, 50, , tan  , , 40, 50, , tan   0.8, ,   tan 1 0.8  38037 ., Q. A cricketer can throw a ball to maximum horizontal distance 100m. How much high, above the ground can the cricketer throw the same ball? And for how long the ball, remains in the air in this case. (g=10m/s2), Solution:, Given;, Maximum horizontal distance travelled by the ball:, 𝑅𝑚𝑎𝑥 =, , 𝑈2, 𝑔, , = 100𝑚 (𝑤ℎ𝑒𝑛 𝜃 = 45°), , ∴ 𝑈 2 = 100x10 = 1000, 𝑈2, , 1000, , 𝐻𝑚𝑎𝑥 = 2𝑔 = 2x10, , 𝐇𝐦𝐚𝐱 = 𝟓𝟎𝐦 (𝑤ℎ𝑒𝑛 𝜃 = 90°), Time flight: (How long the ball remains in the air), Here 𝑈 2 = 100x10 = 1000, 59

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𝑎=(, 𝑎=(, , 43.99 2, 13, , ) 0.8, , 1935.12, 169, , ) 0.8, , 𝑎 = (11.45)0.8, Magnitude of acceleration: a= 9.16 m/s² ( Direction of acceleration of stone: towards, the centre of circle., , 62

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5. LAWS OF MOTION, Q. State Aristotle’s fallacy., Ans: An external force is required to keep a body in uniform motion is known as Aristotle’s, fallacy., , Q. What is inertia of a body?, Ans: Inertia of a body: The property of body to remain in its state of rest or of uniform, motion along a straight line., , Q. State principle of inertia. Explain in brief the inertia due to motion with an example., Ans: Principle of inertia: Inertia resists change in motion. Objects want to stay in rest, or motion unless an outside force causes a change., Example: if you roll a ball, it will continue rolling unless friction or something else, stops it by force., Q. State Newton’s First Law of Motion. Hence define force and inertia., Ans: Newton’s First Law of Motion: It states that a body is continues to be in its state of rest, or of uniform motion along a straight line unless an external force acts on it., Force: It is that external agent which changes or tends to change the state of rest or of, uniform motion of a body along a straight line., Inertia: The property of body to remain in its state of rest or of uniform motion along a, straight line., Q. State Newton’s first law and explain the concept of inertia with an example., Ans: Newton’s first law of motion: It states that a body continues to be in its state of rest or, of uniform motion along a straight line unless an external force acts on it., Inertia: The property of a body to remain in its state of rest or of uniform motion along a, straight line is called inertia., Example: (1) a book on a table continues to be at rest due to inertia., (2) A passenger in a bus pushed backwards, when the bus suddenly starts moving, this because the lower part of the passenger moves forward with the bus while, the upper part continues to be at rest due to inertia., , 63

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Q. State and explain Newton’s second law of motion., Ans: Newton’s second law of motion: It states that the rate of change of momentum of a body, is directly proportional to the force applied on it and the change in momentum takes place in, the direction of force., 𝐹𝑜𝑟𝑐𝑒 ∝ 𝑅𝑎𝑡𝑒 𝑜𝑓 𝑐ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚, 𝐹∝, , 𝑑𝑝, 𝑑𝑡, , Where: 𝐹 = 𝐹𝑜𝑟𝑐𝑒, 𝑑𝑝 = 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚, 𝑑𝑡 = 𝑠𝑚𝑎𝑙𝑙 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑡𝑖𝑚𝑒 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙, Note: 𝐹 = 𝑚𝑎, Where: 𝐹 = 𝐹𝑜𝑟𝑐𝑒, 𝑚 = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑜𝑑𝑦, 𝑎 = 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛, Q. State Newton’s second law of motion and hence derive F=ma., Ans: Newton’s second law of motion: It states that the rate of change of momentum of a body, is directly proportional to the force applied on it and the change in momentum takes place in, the direction of force., Force, F, , , ,  Rate of change of momentum, , dp, dt, , Consider a body moving along a straight line., Let:, m = mass of the body., u = initial velocity of the body., v = final velocity of the body in time t., F = Force applied on the body., a = acceleration of the body., Initial momentum of the body is :, p1 = mu, Final momentum of the body is :, p2 = mv, 64

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Change in momentum of the body :, p2 – p1 = mv–mu, dp= m(v – u), Rate of change of momentum =, , change in momentum, time, , v  u  ……………….. (1), dp, =m, dt, t, dp, = ma …………………….. (2), dt, , , , vu, a, t, , From Newton’s second law of motion:, dp, dt, , F, , , , F, ,  ma, , F =k ma ……….……………... (3), Where: k = constant of proportionality., The value of k is made equal to 1 by defining unit force., Unit force: Unit force is defined as that force which when acts on a body of unit mass, produces unit acceleration (If m = 1 and a = 1, then F = 1)., Equation (3) becomes:, F = ma ………………………... (4), Force = Mass  acceleration, , Q. Define Unit force., Ans: Unit force: Unit force is defined as that the force which when acts on a body of unit mass, produces unit acceleration (If m = 1kg and a = 1m/s2, then F = 1N)., , Q. Write the SI unit of force., Ans: newton (N), , Q. Define one newton., Ans: One newton: A force which produces an acceleration of 1𝑚𝑠 −2 in a body of mass 1𝑘𝑔, is called one newton., , 65

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Q. A car manufacturer claims that the breaks of his cars are so perfect that they can be, stopped instantaneously. Do you agree? Explain., Ans: It in not possible to have perfect brakes which can stop the car instantaneously, because it the car is to stop instantaneously. It as velocity is to be zero in an extremely small, interval of time. As a result of it, the car needs infinite retardation, which is not possible., Q. What is an impulsive force? Give examples for impulsive force., Ans: Impulsive force: An impulsive force is a large force that acts for a short duration of, time on, , an object. (An impulsive force is mainly generated in a collision that results in a, , change in velocity or momentum of the one or all objects involved in the collision.), Example:, 1. When a batsman plays a shot for six, a force acts on a ball through a bat for short interval of, time. We can call that force, an impulsive force, 2. When a bullet strikes a glass, an impulsive force acts on the glass., 3. Force acting by cue ball (white ball) on to other coloured billiard ball., Q. Define terms (a) Impulsive force (b) Impulse of a force., Ans: (a) Impulsive force: An impulsive force is a large force that acts for a short duration of, time on an object. (An impulsive force is mainly generated in a collision that results in a, change in velocity or momentum of the one or all objects involved in the collision.), (b) Impulse of a force: The product of force and the time for which it acts on a body is called, impulse of a force. (or) Impulse is the change of momentum of an object when the object is, upon by a force for an interval of time., , 𝑖𝑚𝑝𝑢𝑙𝑠𝑒 = 𝑓𝑜𝑟𝑐𝑒 × 𝑡𝑖𝑚𝑒 (𝒐𝒓) 𝐽 = 𝐹 × ∆𝑡, , Q. Define Impulsive force? Show that the impulse force is equal to change in momentum, produced in the body., Ans: Impulse of a force: The product of force and the time for which it acts on a body is, called impulse of a force. (or) Impulse is the change of momentum of an object when the, object is upon by a force for an interval of time., , 𝑖𝑚𝑝𝑢𝑙𝑠𝑒 = 𝑓𝑜𝑟𝑐𝑒 × 𝑡𝑖𝑚𝑒 (𝒐𝒓) 𝐽 = 𝐹 × ∆𝑡, , Let:, m = mass of the body., F = a large force acting on the body., 66

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t = small time for which the force acts., u = initial velocity of a body., v = final velocity of the body., a = acceleration produced in the body., 𝑖𝑚𝑝𝑢𝑙𝑠𝑒 = 𝑓𝑜𝑟𝑐𝑒 × 𝑡𝑖𝑚𝑒, J = 𝐹 × ∆𝑡, (∵ 𝐹 = 𝑚𝑎), , J = 𝑚𝑎𝑡, J = 𝑚(, , 𝑣−𝑢, 𝑡, , )𝑡, , (∵ 𝑎 =, , 𝑣−𝑢, 𝑡, , ), , 𝐉 = 𝒎𝒗 − 𝒎𝒖, Impulse force is equal to change in momentum., , Q. State Newton’s third law of motion. Give illustrations or examples., Ans: Newton’s third law: It states that for every action there is always an equal and, opposite reaction., Illustrations: (1) when a bullet is fired from a gun, the bullet moves forward and the gun, moves backward., (2) While walking on the floor, we push the floor backward and the floor pushes us forward., (3) Rocket propulsion., , Q. What is the direction of motion When a bullet is fired from a gun?, Ans: When a bullet is fired from a gun, the bullet moves forward and the gun moves backward., , Q. Which laws of motion is used to explain rocket propulsion?, Ans: Newton’s third law, , Q. Why don’t action and reaction forces cancel each other?, Ans: Action and reaction forces pair don’t cancel because they act on different objects. Forces, can cancel only if they act on the same object., , 67

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Q. State Newton’s laws of motion., Ans: Newton’s First Law of Motion: It states that a body is continues to be in its state of rest, or of uniform motion along a straight line unless an external force acts on it., Newton’s second law of motion: It states that the rate of change of momentum of a body is, directly proportional to the force applied on it and the change in momentum takes place in the, direction of force., Newton’s third law: It states that for every action there is always an equal and opposite, reaction., , Q. How linear momentums vary with its mass and velocity?, Ans: Linear momentum 𝑃⃗⃗ = 𝑚𝑣⃗, Linear momentum is directly proportional to mass., Linear momentum is directly proportional to velocity., , Q. State and prove law of conservation of linear momentum in case of collision of two, bodies. (OR), Q. Define law of conservation of momentum and prove it in case of two bodies collide along, the straight line., Ans: Law of conservation of momentum: It states that the total momentum of an isolated, system of interacting bodies remains constant. OR it states that, if the resultant external force, acting on the system is zero, then the total momentum of the system remains constant., Proof: Consider two bodies A and B of masses 𝑚1 𝑎𝑛𝑑 𝑚2 moving with velocities 𝑢, ⃗⃗1 𝑎𝑛𝑑 𝑢, ⃗⃗2, respectively .The bodies collide for a time interval ∆𝑡. After collision the bodies move with, velocities 𝑣⃗1 𝑎𝑛𝑑 𝑣⃗2 respectively., Initial linear momentum of A is:, 𝑃⃗⃗𝐴 =𝑚1 𝑢, ⃗⃗1 ------------- (1), Initial linear momentum of B is:, 𝑃⃗⃗𝐵 =𝑚2 𝑢, ⃗⃗2 ------------ (2), Final linear momentum of A is:, 𝑃⃗⃗𝐴′ =𝑚1 𝑣⃗1 ------------- (3), 68

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During the collision, A exerts a force on B and B exerts an equal and opposite force on A., Force on A exerted by B = Mass of A  Acceleration of A, ,  v  u1 , FAB = m1  1,  …..………….(1),  t , , (Action), , Force on B exerted by A = Mass of B  Acceleration of B, ,  v  u2, FBA = m2  2,  t, , ,  ………….….(2), , , (Reaction), , According to Newton’s third law of motion:, FAB = - FBA, , v u ,  v  u2 , m1  1 1   m2  2, ,  t ,  t , m1 (v1 – u1) = -m2 (v2 – u2), m1 v1 – m1u1 = - m2v2 + m2u2, m1 v1 + m2 v2 = m1u1 + m2u2, m1u1 + m2u2 = m1v1 + m2v2 …….(3), This equation shows that total linear momentum of the system before collision is equal to, total linear momentum of the system after collision. This proves the law of conservation of, linear momentum., Q. State law of conservation of linear momentum and apply it to explain the recoil of a gun., Ans: Law of conservation of momentum: It states that the total momentum of an isolated, system of interacting bodies remains constant .OR It states that, if the resultant external force, acting on the system is zero, then the total momentum of the system remains constant., Recoil of a gun: When a bullet is fired from a gun, the bullet moves forward with a very high, velocity called muzzle velocity and the gun moves in the opposite direction with a velocity, called the recoil velocity. Before firing the bullet, both the bullet and the gun are at rest and, the total initial momentum is zero. According to the law of conservation of momentum, the, total momentum after firing must be equal to zero., Total momentum before firing = Total momentum after firing, 0 = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑏𝑢𝑙𝑙𝑒𝑡 × 𝑚𝑢𝑧𝑧𝑙𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 + 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑔𝑢𝑛 × 𝑟𝑒𝑐𝑜𝑖𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦, 0 = 𝑚𝑣 + 𝑀𝑉, 𝑀𝑉 = −𝑚𝑣, 70

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𝑚, , 𝑉 = − (𝑀 ) 𝑣, This is the expression for recoil velocity of the gun., The negative sign indicates that recoil velocity and muzzle velocity are oppositely directed., Q. What is friction?, Ans: The component of contact force parallel to the surfaces of two bodies in contact is called, friction., , Q. What is friction? Mention the types of friction., Ans: The component of contact force parallel to the surfaces of two bodies in contact is called, friction., Types of friction: (i) Static friction,, (ii) dynamic (kinetic) friction and, (iii) rolling friction, , Q. Define coefficient of static friction?, Ans: The coefficient of static friction is defined as the ratio of limiting friction to the, normal reaction., , 𝜇𝑠 =, , 𝐹𝑠, 𝑅, , Q. What is kinetic friction? Write the expression., Ans: The frictional force which appears when a body actually moves on the surface of, another body is called kinetic friction., 𝐹𝑘 ∝ 𝑅, , ⇒, , 𝐹𝑘 = 𝜇𝑘 𝑅, , Q. State laws of friction., Ans: laws of friction:, 1. The direction of frictional force is opposite to the direction of motion., 2. The frictional force acts tangentially to the surface in contact., 3. The frictional force depends upon the nature of the surfaces in contact., 4. The frictional force does not depend upon the area of the surfaces in contact, if normal, reaction between the surfaces remains same., 5. The magnitude of frictional force is directly proportional to the normal reaction between, the surfaces., 71

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Q. Write advantages of friction., Ans: Advantages of friction:, 1. Walking is possible due to friction between the ground and the feet., 2. Writing with the chalk on the black board is possible to due to friction., 3. Nails can be fixed on account of friction., 4. Vehicles can run only when there is friction between the road and the tyre., 5. Stopping of moving objects is possible due to friction., , Q. Write disadvantages of friction., Ans: Disadvantages of friction:, 1. Friction always opposes relative motion., 2. Friction causes wear and tear of the machine parts., 3. Friction between different parts of the machine produces heat and causes damage to, them., Q. Give the methods of reducing friction., Ans: Methods of reducing friction:, (i). By polishing and making the surfaces smoother., (ii). By lubrication with oil , grease etc., (iii). By using ball bearings., (iv). By streamlining., Q. Derive an expression for maximum speed of circular motion of a car on a level road., Ans: Motion of a car on a level circular road: consider a car is moving on a level circular, road with a uniform speed., Let:, m = mass of the car., v = speed of the car., r = radius of the circular path., 𝐹𝑐 =, , 𝑚𝑣 2, 𝑟, , = 𝑐𝑒𝑛𝑡𝑟𝑖𝑝𝑒𝑡𝑎𝑙 𝑓𝑜𝑟𝑐𝑒., , g = acceleration due to gravity., mg = weight of the car., R = normal reaction., 𝐹𝑠 = 𝜇𝑠 𝑅 = 𝑙𝑖𝑚𝑖𝑡𝑖𝑛𝑔 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛., 72

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𝜇𝑠 = coefficient of limiting friction., The forces acting on the car its weight are its weight mg, normal reaction R & frictional, force 𝐹𝑠 ., As there is no vertical motion of the car, the weight of the car and the normal reaction balance, each other., Weight of the car = Normal reaction, mg = R ............ (1), The necessary centripetal force is provided by the frictional force between the tyres and the, road., Centripetal force = Frictional force, 𝐹𝑐 = 𝐹𝑠, 𝑚𝑣 2, 𝑟, 𝑚𝑣 2, 𝑟, 𝑣2, 𝑟, , = 𝜇𝑠 𝑅, = 𝜇𝑠 𝑚𝑔, , ∵ 𝑅 = 𝑚𝑔 𝑓𝑟𝑜𝑚 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛(1), , = 𝜇𝑠 𝑔, , 𝑣 2 = 𝜇𝑠 𝑟𝑔, 𝑣 = √ 𝜇𝑠 𝑟𝑔 ............ (2), This gives the maximum speed with which the car can move., , Q. What is banking of roads? Why banking is necessary for roads?, Ans: Raising the outer edge of a curved road above the inner edge to provide the necessary, centripetal force is called banking of roads., Necessity of banking of the road:, (i) When a vehicle moves along horizontal curved road, necessary centripetal force is supplied, by the force of friction between the wheels of vehicle and surface of road., (ii) Frictional force is not enough and unreliable every time as it changes when road becomes, oily or wet in rainy season., (iii) To increase the centripetal force the road should be made rough. But it will cause wear and, tear of the tyres of the wheel., (iv) Thus, due to lack of centripetal force vehicle tends to skid., 73

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Q. Derive an expression for maximum safe speed of a car moving on banked circular road., (or), Derive the expression for maximum safe speed of a vehicle on a banked road in circular, motion., Ans: Expression for maximum safe speed of a car moving on banked circular road:, Consider a vehicle moving along a smooth banked circular road., Let :, m = mass of the vehicle ., g = acceleration due to gravity., mg = weight of the vehicle., r = radius of the circular path ., 𝜃 = angle of banking, R= normal reaction., v = speed of the vehicle, 𝐹𝑐 =, , 𝑚𝑣 2, 𝑟, , = Centripetal force., , The normal reaction R is resolved into two components: Rcos 𝜃 and Rsin 𝜃., The components Rcos 𝜃 balances the weight of the vehicle., Rcos 𝜃 = 𝑚𝑔 ------- (1), The component Rsin 𝜃 provides the required centripetal force., Rsin 𝜃 =, , 𝑚𝑣 2, 𝑟, , ------- (2), , Dividing equation (2) by (1), sin 𝜃, , 𝑣2, , = 𝑟𝑔, cos 𝜃, 𝑣2, , tan 𝜃 = 𝑟𝑔 ----------- (3), 𝑣2, , 𝜃 = tan−1 (𝑟𝑔) ---------- (4), This is the expression for angle of banking., , 74

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Note: from equation (3), 𝑣 2 = 𝑟𝑔 tan θ, 𝑣 = √𝑟𝑔 tan 𝜃 ---------- (5), This gives the expression for maximum safe speed of the vehicle on the banked circular, path., NUMERICALS, Q. A body of mass 5kg is acted upon by two perpendicular forces 8N and 6N. Calculate the, magnitude and direction of the acceleration of the body., Solution:, Given;, Mass of the body, 𝑚 = 5𝑘𝑔, The given situation can be represented as follows:, The resultant of two forces is given by,, 𝑅 = √(8)2 + (−6)2, 𝑅 = √64 + 36, 𝑅 = √100, 𝑅 = 10𝑁, 𝜃 is the angle made by R with the force of 8N, Therefore,, 6, , 𝜃 = tan−1 (− 8) = −36.87°, The negative sign indicates that 𝜃 is in the clockwise direction with respect to the force of, magnitude 8N., According to Newton’s second law of motion, Acceleration (a) of the body is given as,, 𝐹 = 𝑚𝑎, 𝐹, , ∴𝑎=𝑚=, , 10, 5, , = 2 𝑚 ⁄𝑠 2, 75

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Q. The velocity of body of mass 10 kg changes from 5 m/s to 10 m/s .Find the force acting on, it and acceleration in it., Solution:, Given;, Body of mass m= 10 kg, Initial velocity u = 5 m/s, Final velocity v = 10 m/s, 𝐹 = 𝑚(, , 𝑣−𝑢, , 𝐹 = 10 (, , 𝑡, , ), , 10−5, 1, , ), , 𝐹 = 10(5), 𝑭 = 𝟓𝟎𝑵, Q. A bullet of mass 0.015kg strikes a metal plate of thickness 10cm with a velocity of 400m/s, and emerges from it with velocity of 260m/s. Find the average resistance offered by the, plate to the motion of bullet., Solution:, Given;, Mass of the bullet, m = 0.015 kg, Initial speed of the bullet u = 400 m/s, Finally, it emerges, v = 260m/s, Thickness or distance travelled it is 10, 𝑣 2 − 𝑢2 = 2𝑎𝑠, (260)2 − (400)2 = 2𝑎(10x10−2 ), 67600 − 160000 = 0.2𝑎, 𝑎=, , −92400, 0.2, , = −462000𝑚𝑠 −2, , 𝒂 = −𝟒𝟔𝟐𝟎𝟎𝟎𝒎/𝒔𝟐, We need to find the force exerted by the bullet on the plank, 𝐹 = 𝑚𝑎, 𝐹 = 0.015 x (−462000)𝑁 ⇒ F = -6930N. Or F = -6.93kN., 76

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Q. A bullet of mass 0.04kg moving with a speed of 60m/s enters heavy wooden block and is, stopped after a distance of 60cm. What is the average resistance force exerted by the, block on the bullet?, Solution:, Given;, Mass of the bullet, m = 0.04 kg, Initial speed of the bullet u = 60 m/s, Finally, it stopped so that v = 0., Thickness or distance travelled it is 60cm=60x10 -2 m, 𝑣 2 − 𝑢2 = 2𝑎𝑠, (0)2 − (60)2 = 2𝑎(60x10−2 ), 0 − 3600 = 120x10−2 𝑎, −3600, , 𝑎 = 120x10−2 = −3000𝑚𝑠 −2, 𝒂 = −𝟑𝟎𝟎𝟎𝒎/𝒔𝟐, We need to find the force exerted by the bullet on the plank, 𝐹 = 𝑚𝑎, 𝐹 = 0.04 x (−3000)𝑁, F = -120N., , Q. Jet plane of mass 50,000 Kg requires 1500m runway to attain a velocity of 50 m/s, assuming uniform acceleration, Find (a) acceleration, (b) time of take off and (c) the force, on the engine., Solution:, Given; mass of jet plane (m) = 50,000 kg, Distance travelled (S) = 1500 m, Velocity (v) = 50 m/s, (a) Acceleration (a) =?, (b) Time of take off (t) =?, (c) The force on the engine (F) =?, (a) We have V2 – u2 = 2as, 50 x 50 = 2 x a x 1500, 77

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50×50, , 5, , 𝑎 = 2 ×1500 = 6 ms-2, (b) We know that, V = a t  t=, t=, , 50, 5, , 𝑉, 𝑎, , × 6 = 60 sec, , (c) Force on the engine F= ma, 5, , F = 5x 104 x 6 =, , 25, 6, , x 104 N, , F = 4.16 x 104 N, Q. A driver of a car moving at 20 ms-1 sees a child on the road 50m ahead and stops the car, 10m earlier to the child. If the mass of the car with the driver is 1000 kg . Calculate the, force exerted by the brakes on the car and the time taken to stop the car., Solution:, Given ;, Initial speed of the car, u = 20 m/s, Finally, it stops, v = 0, The child on the road 50 m ahead and stops the car 10 m ahead., The distance covered by the car, d = 50 - 10 = 40 m, Mass of the car, m = 1000 kg, We need to find the force exerted by the brakes on the car and time taken to stop the car. It, can be calculated using third equation of motion as:, 𝑣 2 − 𝑢2 = 2𝑎𝑠, −𝑢2 = 2𝑎𝑑, 𝑢2, , 𝑎 = − 2𝑑, (20)2, , 𝑎 = − 2 x 40, 𝒂 = −𝟓𝒎/𝒔𝟐, Force exerted by the brakes is:, 𝐹 = 𝑚𝑎, 𝐹 = 1000 x (−5)𝑁, F = -5000 N, 78

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Solution:, Given;, We know that Force = Impulse/ Time, Impulse = mass x change in velocity, Time = 0.01 s, Mass m = 0.2kg = 2 x 10 ̄ 1 kg, Change in velocity = Final velocity - initial velocity, Change in velocity = 16 - (-12) = 28 m/s, ∴ Force = mass x change in velocity / time, ∴ 𝐹 = 𝑚(, , 𝑣−𝑢, 𝑡, , ), 28, , ∴ 𝐹 = 2x10−1 (0.01 ), 𝐹 = 20x28, 𝐅 = 𝟓𝟔𝟎𝐍, Average force on the ball is 560 N, , Q. A cricket ball moving horizontally with a velocity of 12m/s is brought to rest by a player, in 0.1s. if the cricket ball weighs 0.15kg. Calculate impulse of a force and the average, force applied., Solution:, Given;, Time =0.1s, Mass m =0.15kg, m =1.5×10-1 kg, Change in velocity = final velocity – initial velocity, Change in velocity = 20 - (-12) = 32 m/s, Impulse = mass x change in velocity, J = 1.5×10-1×32, Impulse J = 4.8 Ns, ∴ force = (mass x change in velocity) / time, 𝐹 = (1.5×10-1×32) ÷ 0.01, F = 15×32, F = 480N, 80

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Force F = 480N, Q. A batsman deflects a ball by an angle of 450without changing its initial speed which is, equal to 54kmph. What is the impulse imparted to the ball? If the time of contact between, the ball and the bat is 0.01s, what is the average force exerted by the bat on the ball?, (Mass of the ball is 0.15kg), Solution:, Given;, 5, , Velocity of ball u=54 km/h=54×18 =15 m/s, The ball struck by the bat is deflected back such that, Total angle =450, Now the initial momentum of the ball is m u cosθ = 0.15 × (54 ×, , 5, 18, , ) × cos22.50, , =0.15×15×0.9239 along NO,, FInal momentum of the ball is mucosθ Along ON,, Thus Impulse is equal to the change in momentum,, J=Fxt=mucosθ−(−mucosθ)=2mucosθ, J =2mucosθ =2×0.15×15×0.9239, J = 4.16kgm/s, , Q. Two masses 8 Kg and 12 Kg are connected at the two ends of a light inextensible string, that goes over a frictionless pulley. Find the acceleration of masses and tension in the, string when the masses are released. (g = 10m/s2), Solution:, Given;, The given system of two masses and a pulley, can be represented as shown in the following figure:, Smaller mass, m1 = 8 kg, Larger mass, m2 = 12kg, Tension in the string=T, Mass m2, owing to its weight, moves downward with acceleration,, and mass m1 moves upward., , 81

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Q. A block of mass 4kg rest on horizontal plane. The plane is gradually inclined until at, angle 150 with the horizontal, the block just begins to slide. Find the c0-efficient of static, friction between the block and the surface., Solution:, Given;, We know that,, Mass m=4kg, angle θ=150, Now, the normal force is, N=mg, N=4×10, N=40N, Now, frictional force always opposes direction of motion, So,, Ff=mgcosθ, μN=4×10×cos150, μ×40=40×0.96, μ=0.96, Hence, the friction is 0.96, , 84

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6. WORK, ENERGY AND POWER, Q. Define unit of work., Ans: It is defined as the work used by a force of one newton through a displacement of a metre., Q. Define work done by the force., Ans: Work done is defined as the dot product of force and displacement., Q. Define work done by a force. Under what conditions, the work done by a force is, maximum and minimum?, Ans: Work is sad to be done , when the force acting on a body produces a displacement in it ., 𝑊 = 𝐹𝑠 cos 𝜃 = 𝐹⃗ . 𝑠⃗, Conditions:, 1) If = 𝟎𝟎 , then 𝑾 = +𝑭𝒔, Work is said to be maximum (positive), if the force acts in the same direction of, displacement., 2 ) If 𝜽 = 𝟗𝟎𝟎 then 𝑾 = 𝟎, No work (minimum) is said to be done, if the force acts perpendicular to the direction of, displacement., Q. The force F acting on a body displaces it through a distance S perpendicular to it, then, what will be the work done?, Ans: W = 0., Q. When is the work done negative? Give an example., Ans: Negative work is performed if the displacement is opposite to the direction of the force, applied., Example: work was done the gravity on the rocket going perpendicular upwards., Q. Mention the conditions in which work is not done., Ans: The conditions in which work is not done:, (i) when the angle between the displacement and force applied is 900., i.e the force applied is perpendicular to displacement. No work is said to be done., (ii) If there is no change in position then No work is said to be done., (iii) If there is no motion even after application of force then No work is said to be done., Q. Mention the expression for work done in vector form., 85

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Q. What is the unit of measurement of energy used in our electricity bill?, Ans: kilowatt hour (kwh), Q. What is the value of One kilowatt hour (kwh) in joule?, Ans: 1kwh = 3.6 x 106 J, Q. What is the magnitude of work when the particle completes one revolution on the circular, path?, Ans: zero., Q. State and prove law of conservation of linear momentum in case of collision of two, bodies., Ans: Law of conservation of momentum: It states that the total momentum of an isolated, system of interacting bodies remains constant. OR it states that, if the resultant external force, acting on the system is zero, then the total momentum of the system remains constant., Proof: Consider two bodies A and B of masses 𝑚1 𝑎𝑛𝑑 𝑚2 moving with velocities 𝑢, ⃗⃗1 𝑎𝑛𝑑 𝑢, ⃗⃗2, respectively .The bodies collide for a time interval ∆𝑡. After collision the bodies move with, velocities 𝑣⃗1 𝑎𝑛𝑑 𝑣⃗2 respectively., Initial linear momentum of A is:, 𝑃⃗⃗𝐴 =𝑚1 𝑢, ⃗⃗1 ------------- (1), Initial linear momentum of B is:, 𝑃⃗⃗𝐵 =𝑚2 𝑢, ⃗⃗2 ------------ (2), Final linear momentum of A is:, 𝑃⃗⃗𝐴′ =𝑚1 𝑣⃗1 ------------- (3), Fina linear momentum of B is:, 𝑃⃗⃗𝐵′ =𝑚2 𝑣⃗2 ----------- (4), During the collision, A exert a force on B and B exert an equal and opposite force on A., Change in momentum of A is:, 𝐹⃗𝐴𝐵 ∆𝑡 = 𝑃⃗⃗𝐴′ − 𝑃⃗⃗𝐴, ′, , 𝑃⃗⃗ −𝑃⃗⃗, 𝐹⃗𝐴𝐵 = 𝐴∆𝑡 𝐴--------------- (5), , 87

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⃗⃗ ∙ s⃗, W =F, W = F s cos θ, W = F s cos 00, ∵ ⃗F⃗and s⃗ act in the same directionθ=00, , W = Fs, , ∵ F=ma, , W = mas, v2, , W = ma∙ 2a, 1, , W =2 mv 2 ∵ W = KE, 𝟏, , KE = 𝐦𝐯 𝟐, 𝟐, , Q. A lighter body and a heavy body have same momentum. Which one will have, greater kinetic energy?, Ans: since momentum of heaver body and lighter body are same, thus velocity of lighter, body is greater than that of heavy body. Also kinetic energy is proportional to square of, velocity hence, lighter body has greater kinetic energy., , Q. Define potential energy., Ans: The energy possessed by a body by virtue of its state or position is called potential energy., Q. State work – energy theorem., Ans: Statement: The work on a particle by the net force is equal to the change in its kinetic, energy., , 1, , 1, , 2, , 2, , 𝑊 = 𝑚𝑣 2 − 𝑚𝑢2, , Q. State and prove work – energy theorem for a constant force. Or, Prove that for a particle in rectilinear motion under constant acceleration, the change in, kinetic energy of the particle is equal to work done on it by the net force?, Ans: Statement: The work on a particle by the net force is equal to the change in its kinetic, energy., , 1, , 1, , 𝑊 = 2 𝑚𝑣 2 − 2 𝑚𝑢2, , Consider a body of mass m is moving with initial velocity u. Let the body be subjected to a constant, force F. Then it produces acceleration “a” for a distance S and attains final velocity v., , Equation of motion is:, v 2 = u2 + 2as ------------- (1), 89

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restoring force tries to bring the spring back to its initial position. Therefore, to stretch or, compress the spring, work must be done against the restoring force. This work done is stored in, the spring in the form of elastic potential energy., Consider an elastic spring of negligible mass. The one end of the spring is attached to a rigid, support and the free end to a rectangular block of mass m on a smooth horizontal surface. If no, force is applied to the free end, then the spring is in its initial position A (X=0). When a force, is applied, then the spring is displaced from the position A to a position B through a, displacement X., The restoring force is directly proportional to displacement., 𝐹′ ∝ 𝑥, 𝐹 ′ = 𝑘𝑥 ………………… (1), The negative sign indicates that the restoring force and displacement are oppositely directed., Therefore, to stretch the spring by X , an external force F equal to restoring force 𝐹 ′ must be, applied., Applied force = - Restoring force, 𝐹 = −𝐹 ′, 𝐹 = −(−𝑘𝑥), 𝐹 = 𝑘𝑥 ………………….. (2), If dw is the small work done in stretching the spring through a small displacement dx, then:, 𝑑𝑤 = 𝐹. 𝑑𝑥, 𝑑𝑤 = 𝑘𝑥𝑑𝑥 ……………… (3), Total work done in stretching the spring from 0 to X is given by :, 𝑥, , ∫ 𝑑𝑤 = ∫0 𝑘𝑥𝑑𝑥, 𝑥2, , 𝑥, , 𝑊 = 𝑘[2], , 0, , 𝑥 2 −0, , 𝑊 = 𝑘[, , 2, , ], , 1, , 𝑊 = 2 𝑘𝑥 2 ………………. (4), This work done is stored in the spring in the form of elastic potential energy., 92

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Elastic potential energy = work done, 𝑈=𝑊, 1, , 𝑈 = 𝑘𝑥 2, 2, , This gives the expression for elastic potential energy., Q. What are Conservative and Non – Conservative forces? Give examples to each., Ans: Conservative force: A force is said to be conservative, if the work done by or against a, force in moving a body depends only on the initial and final positions of the body and not on the, actual path followed between the initial and final positions of the body., Example: Gravitational force, electrostatic force, spring force etc., Non – conservative force: A force is said to be non – conservative, if the work done by or, against a force in moving a body depends on the actual path followed between the initial and, final positions of the body., Example: Frictional force, viscous force, air resistance, tension in a string etc., Q. Distinguish between conservative force and non – conservative force., Ans:, Conservative, , Non-conservative, , 1) If work done by the force around a 1) If work done by the force around a, closed Path is zero and it is independent closed path is not equal to zero and it, of the path, Such a force is called depend on the path, such a force is, conservative force., , called non-conservative force., , 2) Work done by the conservative force 2) Work done by the non conservative, is stored in the form of potential energy., , force will not be stored in the form of, potential energy., , 3) Gravitational force, electrostatic force 3) Frictional force and viscous force., and spring force., , Q. Verify law of conservation of mechanical energy for a freely falling body., Ans: Verification of law of conservation of mechanical energy, for a freely falling body: Consider a body of mass m falling, freely under gravity from a height h above the ground., , At the position A: As the body is at rest at A., its kinetic, 93

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energy is zero., Total energy of the body at A is:, EA = UA + KA, EA = mgh + 0, EA = mgh …………………….(1), , At the position B: When the body falls from A to B, it travels a distance x and reaches the, point B with a velocity vB, For the motion from A to B:, , u = 0, v = vB, a = g and, , s=x, , Equation of motion:, v2 = u2 + 2as, , becomes,, , v 2B  2gx, When the body falls A to B, its height decreases and velocity increases. Therefore, its potential, energy decreases and kinetic energy increases., Total energy of the body at B is:, EB = UB +KB, EB = mg (h-x) +, , 1, mv 2B, 2, , EB = mgh – mgx +, , 1, m 2gx, 2, , EB = mgh – mgx + mgx, EB = mgh ……………………… (2), At the position C: When the body reaches the ground at C with velocity vC, it travels the, distanceh., For the motion from A to B :, , u = 0, v = vC, a = g and, , s = h., , Equation of motion :, v2 = u2 + 2 as, , becomes,, , v C2  0  2gh, v C2  2gh, , When the body reaches the ground at C with velocity vC, its height becomes zero., Therefore, its potential energy becomes zero., Total energy of the body at C is :, EC = UC + KC, EC = 0 +, , 1, mv C2, 2, 94

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EC =, , 1, m 2gh, 2, , EC = mgh ………………………(3), It follows from equations (1) , (2) and (3) that :, EA = EB = EC,  Total energy of the body remains constant at all points in the path of the body. This, , proves the law of conservation of mechanical energy for a freely body., Q. What is meant by collision?, Ans: A collision is an event in which two bodies come together for a short time and thereby, produces impulsive forces on each other. (or) An interaction between two bodies is, called collision., Q. What is head on collision (one dimensional collision)?, Ans: The collision is said to be one dimensional, if the colliding bodies move along the same, straight line before and after the collision., Q. What is elastic collision?, Ans: It is a collision in which both momentum and kinetic energy are conserved., Q. What type of energy is present in a compressed spring?, Ans: Elastic potential energy., Q. Name the type of energy stored in a stretched or compressed spring., Ans: Elastic Potential Energy., Q. Distinguish between elastic collision and inelastic collision., Ans:, Elastic collision, , Inelastic collision, , 1.It is a collision in which both 1. It is a collision in which only, momentum and kinetic energy are momentum is conserved but not, conserved., , kinetic energy., , 2. Mechanical energy is not converted 2. Mechanical energy is converted, into any other form of energy, 3., , The, , forces, , conservative., , involved, , into other form of energy., are 3. The forces involved are nonconservative., , 95

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NUMERICALS, Q. A force of 10N acts for 10s on a body of mass 1Kg initially at rest. Calculate the energy, acquired by the body and distance travelled by the body in the direction of the applied, force, Solution:, Given;, Initial velocity u =0m/s, F = ma, a=F÷m, So, a = 10 ÷ 1, a = 10 m/s2, S=ut+1/2at×t, S =1/2 ×10×16, S = 80m, Energy or work W = F S, W = 10 × 80, W = 800J, Q.A body of mass 20 Kg is initially at rest is subjected to a force of 40N. What is the kinetic, energy at the end of 25 s., Solution:, Given,, Mass of a body (m), , = 20 kg, , Initial velocity (u), , = 0 m/s, , Force (F), , = 40 N, , Time (t), , = 25 Sec, , Kinetic energy (KE), , = ?, , We have F= ma, 99

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Density = 1000kgm-3, g = 9.8ms-2, Volume of the well v= πr2h, V = 3.14(3.5)2(20), V = 769.3m3, Mass of water in the well = Volume (Density), M=769.3(1000)kg, Average height through which water is lifted, havg =, , 20, 2, , = 10m, , Work done = Change in Potential Energy of water., W= mghavg, W = (769.3)(1000)(10)(10), W = 769.3 (10)5, W = 7.69(10)7 joule, , Q. Find the height from which the body of mass 50kg should fall in order to have the kinetic, energy of a car of mass 200kg travelling at 10 ms-1? Given; g = 10 ms-2?, Solution: M = 200kg, v = 10 ms-1, g = 10 ms-2 and m = 50kg, 1, , The kinetic energy of a car, 𝐾. 𝐸. = 2 𝑀𝑣 2, 1, , 𝐾𝐸 = 2 (200)(10)2, 𝐾. 𝐸. = 10,000 𝐽, Potential Energy = mgh, 10,000 = (50)(10)h, H=, , 10000, 500, , 𝐡 = 𝟐𝟎𝐦., , Q. A body of mass 5kg initially at rest is subjected to a force of 20N. What is the kinetic, energy acquired by the body at the end of 10s., Answer:, Given :m= 5kg , u=0, F=20N and t=10s., To find: K, Solution:, Force is given by: F  ma, , 101

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a, , F, m, , a, , 20, 5, , a  4ms 2, Velocity acquired by the body at the end of time t is given by:, , v  u  at, v  0  4 10, , v  40ms 1, Kinetic energy acquired by the body at the end of time t is given by:, K, , 1, mv 2, 2, , K, , 1,  5  40 2, 2, , K  4000 J  4  103 J  4kJ, , Q. A body of mass 20kg initially at rest is subjected to a force of 40N. What is the kinetic, energy acquired by the body at the end of 25s., Answer:, Given: m= 20kg, u=0, F=40N and t=25s., To find: K, Solution:, Force is given by: F  ma, a, , F, m, , a, , 40, 20, , a  2ms 2, Velocity acquired by the body at the end of time t is given by:, , v  u  at, v  0  2  25, , v  50ms 1, Kinetic energy acquired by the body at the end of time t is given by:, K, , 1, mv 2, 2, 102

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P=, , 294×104, 30×60, , P = 1.633× 𝟏𝟎𝟑 Watt, P =1.633KW, Q. A pump at the ground floor of a building can pump water to fall a tank of volume 30𝒎𝟑 in, 15 minutes.If the rank is 40m above the ground and the efficiency of the pump is 30 %, then how much electric power is consumed by the pump?, Solution:, Given,, Volume of water: 30𝑚3, t = 15min = 15 × 60 = 900sec, Density of water = 103 kg/𝑚3, h=40m, g=10m/𝑠 2, ∴ Mass of water pumped = volume × density, m = 30× 103, m = 3× 104 kg, 𝑃𝑜𝑢𝑡𝑝𝑢𝑡 =, , 𝑤, 𝑡, , =, , 𝑚𝑔ℎ, 𝑡, , Efficiency, 𝜂 =, 𝜂 = 𝑃𝑖𝑛𝑝𝑢𝑡 =, , (3×104 )(10)(40), 900, , =, , 4, 3, , × 104w, , 𝑃𝑜𝑢𝑡𝑝𝑢𝑡, 𝑃𝑖𝑛𝑝𝑢𝑡, , 𝑃𝑜𝑢𝑡𝑝𝑢𝑡, 𝜂, , =, , =, , 4×104, 3×, , 30, 100, , 4, , 𝜂 = 9 × 105, 𝜼 = 44.4KW, , Q. A pump on the ground floor of a building can pump up water to fill a tank of volume, 30𝒎𝟑 in 20 min. If the tank is 40m above the ground and the efficiency of the pump is, 30% how much electric power is consumed by the pump?, Solution:, Given,, Volume of tank V=30m3, Density of water d=103kgm-3, Time T = 20min = (20)60 = 1200Sec, The efficiency of the pump η = 30%, Mass of water M = (Volume)(Density) = (30)103kg, The height of tank from the ground h=40m, 107

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Solution:, mass of the object 1, 𝑚1 = 1𝑘𝑔, , Given,, , Initial speed of the object 1,𝑢1 = 0𝑚𝑠 −1, Let v be the velocity of third fragment (3/5kg), By momentum conservation,, ⇒Vx (3/5) − (1/5)(30) = 0, ⇒Vx= 10 m/s, Similarly Vy=10m/s, ⇒𝑣 = √𝑣𝑥2 − 𝑣𝑦2, 𝑣 = 10√2 𝑚/𝑠, 𝑣 = 10√2 𝑚/𝑠 ., Hence, the answer is 10√2 𝑚/𝑠., , Q. A Person trying to lose weight (dieter) lifts 10kg mass, one thousand times, to a height of, 0,5m each time. Assume that the potential energy lost each time she lowers the mass is, dissipated. (a) How much work does she do against the gravitational force? (b) Fat, supplies 3.8 x 107 J of energy per kilogram which is converted to mechanical energy with, a 20% efficiency rate. How much fat will the dieter use up?, Solution:, Given,, (a) Total work done = m g h × no. of times = 10 × 9.8 × 0.5 × 1000 = 49000J, (b) Energy equivalent of 1 kg of fat =3.8×107J, Efficiency rate = 20%, Mechanical energy supplied by the person’s body:, = (20/100) × 3.8 × 107 J, = (1/5) × 3.8 × 107 J, Equivalent mass of fat lost by the dieter:, =, , 1, 1, ( )x 3.8 x107, 5, , x49x103, , = (245/3.8) × 10 −4, = 6.45 × 10−3kg, , 111

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Q. Give an example for torque or moment of couple., Ans: 1. Forces applied to the handle of a screw press., 2. Turning a cap of a pen., 3. Opening or closing a water tap., 4. Opening or closing a bottle cap., Q. When is the moment of force (torque) positive?, Ans: Torque tends to rotate the body in anti-clockwise direction and it is positive., Q. When is the moment of force (torque) negative?, Ans: Torque tends to rotate the body in clockwise direction and it is negative., Q. The mechanical advantage of a lever is greater than one. What does it mean?, Ans: A Small effort is enough to lift a large load., Q. What is the use of mechanical advantage of a lever?, Ans: A lever provides mechanical advantage. Mechanical advantage refers to how much a, simple machine multiplies an applied force., Q. When does the rigid body said to be in mechanical equilibrium? Write the conditions for, the equilibrium of a rigid body., Ans: A rigid body is said to be in mechanical equilibrium, if it’s both linear momentum and, angular momentum remains constant. This means that the body has neither linear, acceleration nor angular acceleration. In other words, the rigid body is said to be in, mechanical equilibrium, if it is both in translational and rotational equilibrium., Conditions for equilibrium of a rigid body:, 1. Condition for translational equilibrium: A rigid body is said to be in translational, equilibrium, if, , it remains in its state of uniform motion along straight line., , For translational equilibrium, the vector sum of all the forces acting on the body must be zero., ⃗⃗⃗⃗⃗, ⃗⃗⃗⃗2 + 𝐹, ⃗⃗⃗⃗3 ...............𝐹, ⃗⃗⃗⃗, 𝐹1 + 𝐹, 𝑛 =0, In translation equilibrium, the linear momentum of the system remains constant., ⃗⃗ = constant., 𝒑, 2. Condition for rotational equilibrium: A rigid body is said to be in rotational equilibrium,, if its state of rest or of uniform rotational motion., For rotational equilibrium, the vector sum of all the torques acting on the body must be zero., 𝑟⃗⃗⃗⃗1 + ⃗⃗⃗⃗, 𝑟2 + ⃗⃗⃗⃗---------𝑟, 𝑟3, ⃗⃗⃗⃗, 𝑛 =0, In rotational equilibrium, the angular momentum of the system remains constant., 114

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⃗⃗ = constant., 𝑳, Q. Define angular velocity., Ans: Angular velocity: The ratio of angular displacement to the time interval is called, If 𝜃 is the angular displacement in time t, then angular velocity 𝜔 is, , angular velocity., given by: 𝜔 =, , 𝜃, 𝑡, , Q. Define angular velocity and linear velocity and mention the relation between them., Ans: Angular velocity: The rate change in angular displacement is called angular velocity., If 𝜃 is the angular displacement in time t, then angular velocity 𝜔 is given by: 𝜔 =, , 𝜃, 𝑡, , Linear velocity: The rate change in linear displacement is called linear velocity., 𝑠, , If 𝑆 is the linear displacement in time t, then linear velocity 𝑣 is given by: 𝑣 = 𝑡, Relation between linear velocity and angular velocity: v = r𝜔 or 𝑣⃗ = 𝜔, ⃗⃗ × 𝑟⃗, , Q. Write the relation connecting angular velocity and linear velocity., Ans: v = rω, Where, v = velocity, r = perpendicular distance from that point to axis of rotation,, ω = Angular velocity, Q. Derive relation between linear velocity and angular velocity., Consider a particle moving along a circle of radius r. Let ds be the very small linear, displacement in a very small time dt., Instantaneous linear velocity is:, v=, v=, , 𝑑𝑠, 𝑑𝑡, 𝑑(𝑟𝜃), , v=r, , 𝑑𝑡, , ∵s = r𝜃, , 𝑑𝜃, 𝑑𝑡, , v = r𝜔 (∵𝜔 =, , 𝑑𝜃, 𝑑𝑡, , ), ,  Linear velocity = Radius × 𝐴𝑛𝑔𝑢𝑙𝑎𝑟 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦, Vector form of linear velocity is 𝑣⃗ = 𝜔, ⃗⃗ × 𝑟⃗, Q. Define angular acceleration and linear acceleration and mention the relation between them., Ans: Angular acceleration: The rate of change angular velocity is called angular acceleration., If 𝜔 is the angular velocity in time t, then angular acceleration 𝛼 is given by: 𝛼 =, , 𝜔, 𝑡, , Linear acceleration: The rate of change in linear velocity is called linear acceleration., , 115

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If 𝑣 is the linear velocity in time t, then linear acceleration 𝑎 is given by: 𝑎 =, , 𝑣, 𝑡, , Relation between linear acceleration and angular acceleration: 𝑎 = 𝑟𝛼 or 𝑎⃗ = 𝛼⃗ × 𝑟⃗, Q. Write the statement for axis of rotation of a rigid body., Ans: The fixed line about which the rigid body rotates is called axis of rotation., Q. What is meant by moment of inertia?, Ans: Moment of inertia of a body about an axis of rotation is defined as the sum of the product, of the mass of each particle and square of its perpendicular distance from the axis of rotation., 𝐼 = 𝑀𝑅2, Q. Is moment of inertia a scalar or a vector?, Ans: Tensor (not a scalar, not a vector), Q. Define radius of gyration., Ans: The perpendicular distance from the axis to a point at which the entire mass of the body, is supposed to be concentrated to have the same moment of inertia as that of the entire body, is called radius of gyration of a rigid body about an axis of rotation., Q. Derive the expression for the moment of inertia., Ans: Moment of inertia: It is defined as the sum of the product of the mass of each particle, and square of its perpendicular distance from the axis of rotation, Suppose the body consists of n number of particles each, of masses m1, m2, m3 ........mn and the perpendicular, distances r1, r2, r3 ........rn respectively from the axis of, rotation YY1 ., I= 𝑚1 𝑟12 + 𝑚2 𝑟22 + 𝑚3 𝑟32 + − − − − − − +𝑚𝑛 𝑟𝑛2, I =𝑚𝑛(, , 𝑟12+𝑟22+𝑟32+⋯……………..𝑟𝑛2, 𝑛, , ), , I= (𝑚1 𝑟12 + 𝑚2 𝑟22 +𝑚3 𝑟32 + − − − − − − +𝑚𝑛 𝑟𝑛2 )=∑𝑛𝑖=1 𝑚𝑖 𝑟𝑖2, Q. Derive an expression for rotational kinetic energy., Ans: Consider a rigid body rotating with uniform angular velocity 𝜔 about a given axis, YY1.Suppose the body is made up of a large number of particles of masses m1, m2, m3 ----mn, situated at perpendicular distances r1, r2, r3 ---rn respectively from the axis of rotation. Let v1,, v2, v3 ----vn be their respective linear velocities. As the body rotates, each particle of it moves in, a circular path around the axis., 116

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Alternative method:, Consider a rigid body of moment inertia I rotating with uniform angular acceleration 𝛼⃗ about an, axis under effect of constant torque 𝜏⃗ given by:, 𝜏⃗ = I𝛼⃗, dω, ⃗⃗⃗, , 𝜏⃗ = I ( dt ) (∵ 𝛼⃗ =, 𝜏⃗ =, 𝜏⃗ =, , dω, ⃗⃗⃗, dt, , ), , d⃗⃗⃗⃗⃗⃗⃗⃗, (Iω), dt, dL, dt, , (∵ 𝐿 = 𝐼𝜔), , Q. State parallel and perpendicular axes theorem., Ans: Parallel axes theorem: The moment of inertia of a rigid body about an axes passing, through a point is equal to the sum of its moment of inertia about a parallel axis passing through, its centre of mass (Ic). The product of mass of the body and square of the perpendicular distance, between the axes (Mr2). 𝐈 = 𝐈𝐜 + 𝐌𝐫 𝟐, , (OR), , It states that the moment of inertia of a body about any axis is equal to the sum of the moment, of inertia of the body about a parallel axis passing through its centre of gravity and product of, mass of the body and square of the perpendicular distance between the two parallel axes. 𝐈 =, 𝐈𝐆 + 𝐌𝐫 𝟐, Perpendicular axes theorem: The moment of inertia of a plane lamina about an axis, perpendicular to its plane is equal to the sum of the moments of inertia of the plane lamina, about perpendicular to each other in its own plane and intersecting each other at the point,, where the axis perpendicular to the plane passes. 𝐈𝐙 = 𝐈𝐗 + 𝐈𝐘, Q. State and explain a) parallel and b) perpendicular axes theorems., Ans: Theorem of parallel axes: It states that the moment of inertia of a body about any, axis is equal to the sum of the moment of inertia of the body, about a parallel axis passing through its centre of gravity and, product of mass of the body and the square of perpendicular, distance between the two parallel axes., Where I= moment of inertia of a body about an axis PQ passing, through C., 𝐈𝐝 = 𝐈𝐆 + 𝐌𝐝𝟐, , 118

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𝐼𝑔 = Moment of inertia of a body about an axis AB passing through the centre of gravity G., M= mass of the body., d = perpendicular distance between two parallel axes AB and PQ., , Theorem of perpendicular axes: It states that if there are three, mutually perpendicular axes through a point the body, then the, sum of moments of inertia of plane lamina about the two axes in, its plane is equal to its moment of inertia about the third axis, perpendicular to its plane., 𝑰𝒁 = 𝑰𝑿 + 𝑰𝒀, , Where, 𝑰𝑿 , 𝑰𝒀 𝒂𝒏𝒅 𝑰𝒁 = Moment of inertia of the body about the axes OX, OY and OZ respectively., Q. a) Write the kinematic equations for rotational motion of a body and explain the term., b) Write the expression for the moment of inertia of a solid sphere rotating about its, diameter and explain the terms., a) Equations of rotational motion:, 1. 𝜔 = 𝜔0 + 𝛼𝑡, 2. 𝜔2 = 𝜔02 = 2𝛼𝜃, 1, , 3. 𝜃 = ω0 t + 2 𝛼𝑡 2, α, , 4. θn = ωo + 2 (2n − 1), Where:, 𝜔0 = initial angular velocity., ω = final angular velocity., α=uniform angular acceleration., 𝜃= angular displacement in time t., 𝜃𝑛 = angular displacement in the nth second., b) Write the expression for the moment of inertia of a solid sphere rotating about its, diameter and explain the terms., Ans: Consider a solid sphere of mass M and radius R. Its moment of inertia about an axis of, rotation passing through its diameter is, I=, , 2, 5, , 𝑀𝑅2, 119

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Where I is moment of inertia of solid sphere along its axis, M is mass of the solid cylinder, R is distance between axes of rotation to center of the sphere., Q. The moment of inertia of solid sphere about a tangential axis., 𝟕, , 𝐀𝐧𝐬: 𝐈 = 𝟓 𝐌𝐑𝟐, Where M is mass of sphere and, R is the perpendicular distance between axes of rotation to center of the sphere., Q. Write the expression for moment of inertia of solid cylinder along its axis. Explain the, terms used., 𝟏, , Ans: The expression for moment of inertia of solid cylinder along its axis: 𝑰 = 𝟐 𝒎𝑹𝟐, Where I is moment of inertia of solid cylinder along its axis, M is mass of the solid, cylinder and R is distance between axes of rotation to center of the cylinder., Q. Mention the expression for moment of inertia of circular disc of radius R about its, diameter., Ans: The moment of inertia of circular disc about an axis passing through its diameter:, 1, , I = 4 MR2 ., Q. State the law of conservation of angular momentum and explain it with two illustrations., Ans: Law of conservation of angular momentum: It states that the angular momentum of the, system remains constant, if no torque acts on the system., We know that:, 𝜏=, , 𝑑𝐿, 𝑑𝑡, , If 𝜏 = 0 , 𝑡ℎ𝑒𝑛, , 𝑑𝐿, 𝑑𝑡, , =0, , L= constant, , Illustrations or examples:, 1. A diver jumping from a spring board folds his arms and legs inwards. Due to this, his, moment of inertia (I) decreases, angular velocity (𝜔) increases and L = I 𝜔 remains, constant., 2. Ballet dancers use the conservation of angular momentum to increase their speed. They start, springing with their arms stretched then bring their arms closer. As a result , moment of, inertia (I) decreases and angular velocity (𝜔) increases . thus,, L = I 𝜔 remains constant, 120

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Q. When sand is poured on a rotating disc, what happens to its angular velocity? Explain., Ans: When sand is poured on a rotating disc the moment of inertia of the disc will increase, but, the angular momentum of disc will remain constant, and hence its angular velocity will, decrease., , Q. Derive the expression for the total kinetic energy of a rolling body., Ans: Kinetic energy of rolling motion: Consider a wheel rolling on a horizontal surface, without slipping. The wheel possesses two types of motion translational motion of the Centre, of the mass in the horizontal direction and rotational motion about an axis passing through its, Centre. Therefore, the total kinetic energy of the wheel is equal to the translational kinetic, energy of the Centre of the mass and rotational kinetic energy about the Centre of mass., Total KE = Translational KE + Rotational KE, 𝐾𝐸 =, , 1, 2, , 1, , 2, 𝑀𝑣𝑐𝑚, + 2 𝐼𝜔2 ……………….(1), , Where:, M = mass of the wheel, 𝑣𝑐𝑚 = velocity of Centre of mass, I = moment of inertia about the rolling axis of the wheel, 𝜔 = angular velocity of the wheel about an axis passing through its Centre., If R = radius of the wheel and K = radius of the gyration of the wheel, then, 𝐼 = 𝑀𝐾 2 ………………………………………. (2), And 𝜔 =, , 𝑣𝑐𝑚, 𝑅, , …………………………………... (3), , ∴ Equation (1) becomes:, 𝐾𝐸 =, , 1, , 𝐾𝐸 =, , 1, , 2, , 1, , 2, 𝑀𝑣𝑐𝑚, + 𝐼(, , 𝑣𝑐𝑚 2, , 2, , 𝑅, , ), , 𝐾2, , 2 [1, 𝑀𝑣𝑐𝑚, + 𝑅2 ], 2, , This gives the expression for kinetic energy of rolling body (without slipping)., , 121

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Q. A circular disc of mass 100kg and radius of 0.8m makes 120 revolutions per minute., Calculate (a) angular velocity and (b) moment of inertia about the axis of rotation?, Solution:, Given,, Mass of circular disc m = 100kg,, Radius of circular disc r = 0.8m, No of rotations per minute n=120rpm, (a) Angular velocity 𝜔 =, (b) Moment of inertia, 𝐼 =, 𝐼=, , 100 𝑥 (0.8)2, 2, , 120, 60, , = 2 𝑟𝑎𝑑𝑠 −1, , 𝑀𝑟 2, 2, , = 32 𝑘𝑔𝑚2, , 𝑰 = 𝟑𝟐 𝒌𝒈𝒎𝟐, Q. A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius, 1.5m with a speed of 40 revolutions per minute in a horizontal plane. What is the tension, in the string?, Solution:, Given,, Mass of the stone, m = 0.25kg, Radius of the circle, r=1.5 m, 40, , 2, , Number of revolution per second, 𝑛 = 60 = 3 𝑟𝑝𝑠, 2, , Angular velocity, 𝜔 = 2𝜋𝑛 = 2 x 3.142 x (3) = 4.19 rad𝑠 −1, The centripetal force for the stone is provided by the tension T, in the string, i.e..,, 𝑇 = 𝑚𝜔2 𝑟, 𝑇 = (0.25)(2𝜋𝑛)2 (1.5) = (0.25)[4.19]2 (1.5), 𝑻 = 𝟔. 𝟓𝟖 𝑵, Q. A constant torque of 50Nm turns a wheel of moment of inertia 100kgm 2 about an axis, through its centre. Find the gain in angular velocity in 2s., Solution:, Given:, ,   50 Nm , I  100 Nm2 and t  2s ., Torque is given by:   I, , , , , I, 125

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, , 50, 100, ,   0.5 rad / s 2, Angular velocity acquired by the wheel at the end of time t is given by:, ,   0  t,   0  0.5  2, ,   1 rads 1, Q. A solid cylinder of mass 20kg rotates about its axis with angular speed 10rad/s. The radius, of the cylinder is 0.25m. What is the kinetic energy associate with the rotation of the, cylinder? What is the magnitude of angular momentum of the cylinder about its axis?, Solution:, Given,, M = 20kg, 𝜔 = 10𝑟𝑎𝑑/𝑠 and 𝑅 = 0.25𝑚, (a) Moment of inertia of a solid cylinder about its axis is:, 1, , 𝐼 = 2 𝑀𝑅2, 1, , 𝐼 = 2 × 20 × (0.25)2, 𝐈 = 𝟎. 𝟔𝟐𝟓𝐤𝐠𝐦𝟐, Kinetic energy of the cylinder is:, 1, , 𝐾 = 2 𝐼𝜔2, 1, , 𝐾 = 2 × 0.625 × (10)2, 𝐊 = 𝟑𝟏. 𝟐𝟓𝐉, (b) Angular momentum of the cylinder is :, 𝐿 = 𝐼𝜔, 𝐿 = 0.625 × 10, 𝐿 = 6.25𝐽𝑠, 𝐋 = 𝟔. 𝟐𝟓𝐤𝐠𝐦𝟐 𝐬, Q. A solid cylinder of mass 20 Kg rotates about its axis with angular speed 100rad s -1. The, radius of the cylinder is 0.25m. What is the kinetic energy associated with the rotation of, the cylinder? What is the magnitude of angular momentum of the cylinder about its, axis?, Solution:, 126

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8.GRAVITATION, Q. Draw the schematic diagram of Cavendish experiment to determine the universal, gravitational constant and write the formula used in the experiment., , Or, , How do you determine the gravitational constant value, describe briefly with diagram., Ans: The experimental arrangement consists of two small lead spheres attached to the two end, of bar AB. The bar is suspended from a rigid support by means of a fine wire. Two big lead, spheres are placed near the small spheres at the required positions., Working: The big spheres are brought close to, the small spheres at equal distances on opposite, sides. The big spheres attract the small ones by, equal and opposite forces. This system of forces, produces a gravitational torque in the bar. As a, result of gravitational torque, the bar is rotated, about the wire as an axis and the wire gets, twisted. Due to this twist in the wire, the restoring torque comes into play which opposes the, gravitational torque. The wire gets twisted till the gravitational torque and restoring torque, becomes equal to each other., Gravitational force between m1 and m2 is: 𝐹 = 𝐺, , 𝑚1 𝑚2, 𝑟2, , Gravitational torque (deflecting torque) in the bar is equal to product of gravitational force and, length of bar 𝜏𝑑𝑒𝑓 = 𝐺, , 𝑚1 𝑚2, 𝑟2, , 𝑙.............. (1), , Restoring torque in the wire is equal to the product of torque per unit twist and angle of twist, 𝜏𝑟𝑒𝑠 = 𝐶𝜃.......... (2), For equilibrium position of the rod, gravitational torque and restoring torque are equal., 𝜏𝑑𝑒𝑓 = 𝜏𝑟𝑒𝑠 ., 𝐺, , 𝑚1 𝑚2, 𝑟2, , 𝐶𝜃𝑟 2, , 𝑙 = 𝐶𝜃 → 𝐺 = 𝑚, , 1 𝑚2 𝑙, , ................. (3), , Hence, the value of G can be determined, if C, 𝜃, r, m1, m2 and l are known. The presently, accepted value of universal gravitational constant is: G = 6.67x10-11Nm2kg-2., Q. Name the experiment to measure the value of gravitational constant., Ans: Cavendish experiment., Q. Write the S.I unit of gravitational constant G., 130

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Ans: 𝑁𝑚2 𝑘𝑔−2, Q. Write the value of gravitational constant G., Ans: G = 6.67 × 10−11 𝑁𝑚2 𝑘𝑔 −2 ., Q. State Newton’s law of gravitation., Ans: The magnitude of gravitational force of attraction between two bodies is directly, proportional to the product of their masses and inversely proportional to square of the, distance between them., , Q. State Newton’s law of gravitation and write it’s mathematical form., Ans: The magnitude of gravitational force of attraction between two bodies is directly, proportional to the product of their masses and inversely proportional to square of the, distance between them., It’s mathematical form: F = G, , m1 m2, r2, , Q. State and explain Newton’s law of gravitation., Ans: Newton’s law of gravitation: It states that, everybody in the universe attracts every, other body with a force,, , which is directly proportional to the product of their masses and, , inversely proportional to the square of the distance between them., Consider two bodies separated by a certain distance. If F be the force of attraction between, two bodies of masses 𝑚1 and 𝑚2 separated by a distance r, then:, 𝐹 𝛼 𝑚1 𝑚2, And 𝐹 𝛼, 𝑭=𝑮, , 1, 𝑟2, , 𝒎𝟏 𝒎𝟐, 𝒓𝟐, , Where G = gravitational constant = 𝟔. 𝟔𝟕 × 𝟏𝟎−𝟏𝟏 𝑵𝒎𝟐 𝒌𝒈−𝟐, Q. Why gravitational constant is called as universal gravitational constant?, Ans: G is called the universal gravitational constant since it has a constant value all throughout, the universe., Q. Write properties of gravitational force., Ans: properties of gravitational force:, 131

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1) It is always attractive in nature, 2) It is a central force., 3) It is a conservative force, 4) It is a long-range force, 5) It is the weakest basic force in nature., 6) It is an action and reaction pair., 7) It obeys inverse square law of distance, 8) It does not depend on the nature of the medium between the two masses., 9) It is universal., , Q. Write the relation between g and G and explain the terms., Ans: The relation between g and G: g =, , Gm, R2, , Where: M = mass of the earth (planet),, R = radius of the earth (planet) and, G = universal gravitational constant., g = accelerating due to gravity of earth (planet)., , Q. Derive an expression for accelerating due to gravity in terms of mass and radius of the, earth., Ans: expression for accelerating due to gravity in terms of mass and radius of the earth:, Consider a body of mass m on the surface of the earth of mass M and radius R Gravitation, force between the earth and the body is:, 𝑭=𝑮, , 𝑴𝒎, 𝑹𝟐, , --------------------------(1), , Weight of the body is:, F=mg--------------------------------(2), From equation (1) and (2):, , 132

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mg = 𝑮, g=, , 𝑴𝒎, 𝑹𝟐, , 𝑮𝑴, 𝑹𝟐, , Q. Obtain the expression for variation of acceleration due to gravity above the surface of the, earth (altitude or h)., Ans: The acceleration due to gravity on the surface of earth:, , g=, , Gm, R2, , …………(1), , The acceleration due to gravity at a height h above the surface of the earth of radius R is:, Gm, , g h = (R+h)2 ………(2), Compare equations (2) and (1) then, R2, , gh, , = (R+h)2, , g, , gh = g, gh =, , R2, h 2, R, , R2 (1+ ), g, h 2, R, , (1+ ), , h −2, , g h = g (1 + R), g h = g (1 −, , 2h, R, , h, , + terms containing higher power of R), , 𝑔ℎ = 𝑔 (1 −, , 2ℎ, ), 𝑅, , This gives the expression for acceleration due to gravity at a small height (h<<R), Q. How does ‘g’ vary with height above the surface of earth?, Ans: 𝒈𝒉 = 𝒈 (𝟏 −, , 𝟐𝒉, 𝑹, , ), , The value of acceleration due to gravity decreases with increase in height above the surface of, the earth., , Q. What is the value of acceleration due to gravity at the Centre of the earth?, Ans: Zero., , 133

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Substituting for, 𝑔𝑑, , 𝑅2, , =, ×, 𝑔 (𝑅−𝑑)2, , 𝑔𝑑 =𝑔, , 𝑀′, 𝑀, , in equation (5):, , (𝑅−𝑑)3, 𝑅3, , (𝑅−𝑑), 𝑅, 𝑑, , 𝑔𝑑 = (1 − 𝑅)---------------(9), , Q. Write equation for the variation of acceleration due to gravity below the surface of the, earth., 𝑑, , Ans: 𝑔𝑑 = 𝑔 (1 − 𝑅)., Q. Derive an expression for gravitational potential energy of particle at a point due to the, earth., Ans: Consider a body (particle) of mass m in the gravitational field of another body of mass M., To derive the expression for the gravitational potential energy of the particle at a point P,, the work done in moving the particle of mass m from infinity to P is to be calculated. For, this, the mass m is displaced from A to B through a small displacement dx., Gravitational, , force, , of, , attraction between the body of, mass M and the particle of, mass, , m, , separated, , by, , a, , distance x is:, F, , GMm, ----------------------------- (1), x2, , Work done dw to move the mass m through a small displacement dx is:, , Work done = Force x Displacement, , dw  Fdx, dw , , GMm, dx ------------------------ (2), x2, , Total work done in moving the mass m from infinity to the point P is:, r, , GMm, dx, x2, , ,  dw  , , 135

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r, ,  dw  GMm x, , 2, , dx, , , , r, ,  x  21 , w  GMm, ,   2  1 , r, ,  x 1 , w  GMm ,  1 , r, ,  1, w  GMm ,  x, r, , 1, w  GMm ,  x, 1 1 , w  GMm  , r , , w, , GMm, ------------------------- (3), r, , This work done w is stored in the particle as its gravitational potential energy U., U=w, , U , , GMm, …………….……..(4), r, , This gives the expression for gravitational potential energy at a point due to a particle, (body) of mass m., , Q. Give difference between gravitational potential and gravitational potential energy., Ans:, gravitational potential, , gravitational potential energy, , Gravitational potential is the work done, , Gravitational potential energy is the work, , for a unit mass when it is brought from, , done when you move a body from infinity, , infinity to a certain point., , to the point. It manifests itself in form of, stored energy or the potential energy., , Gravitational potential is the potential, , Gravitational potential energy is the work, , energy per unit mass, , done by the gravitational field of a, moving body to its given position in space, from infinity, 136

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Q. Write the expression for escape velocity? Explain the terms used in the equation., Ans: Expression for escape velocity: 𝑣𝑒 = √2𝑔𝑅 𝑜𝑟 ve = √, Where, , 2GM, R, , 𝑣𝑒 = escape velocity of the body, g = acceleration due to gravity, R = radius of the earth, G = universal gravitational constant., , Q. How does the escape speed of a body from the earth depend on height of the location from, where the body is launched?, Ans: it depends upon the gravitational potential at the point from where the body launched., Since this potential depends on the height of the point of projection, the escape velocity, depends on it., , Q. How does the escape speed of a body from the earth depend on its direction of projection?, Ans: The escape velocity is independent of mass of the body and the direction of projection., It depends upon the gravitational potential at the point from where the body is launched. Since, this potential depends on the height of the point of projection, the escape velocity, depends on it., , Q. What is escape velocity? Obtain the expression for the escape velocity., Ans: Escape Velocity: The minimum velocity with which a body may be projected such that it, escapes from the gravitational pull of the earth is called escape velocity., Expression for escape velocity: Consider a body at a certain height from the surface of the, earth., Let:, M = mass of the earth, m = mass of the body, R = radius of the earth, h = height of the body above the earth’s surface, r = R +h = distance of the body from the centre of the earth, 𝑣𝑒 = escape velocity of the body, g = acceleration due to gravity, The gravitational force of attraction between M and m separated by a distance r is:, 137

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Expression for orbital velocity of a satellite (𝒗𝒐 ):, Consider a satellite performing uniform circular, motion around the earth at a certain height from the, surface of the earth., Let: M=mass of the earth, m = mass of the satellite,, R = radius of the earth, h = height of the satellite, above the earth's surface,r=R+h= radius of the orbit of, the satellite= distance of the, Satellite from the centre of the earth and vo=orbital velocity of the satellite. As the satellite, performs uniform circular motion around the earth, it needs centripetal force. The required, centripetal force is provided by the gravitational force acting on the satellite by the earth., Centripetal force = Gravitational force, 𝑚𝑣𝑜2, 𝑟, , =, , 𝑣𝑜 2 =, , 𝐺𝑀𝑚, 𝑟2, , 𝐺𝑀, 𝑟, , 𝑣𝑜 = √, , 𝐺𝑀, , 𝑣𝑜 = √, , 𝐺𝑀, , 𝑟, , 𝑟, , 𝐺𝑀, , = √𝑅+ℎ∵GM = g𝑟 2, 𝐺𝑀, , = √𝑅+ℎ∵, , √𝑔𝑟 2, 𝑟, , =√𝑔𝑟, , ∵h<<r, , This is expression for orbital velocity 𝑣0 = √𝑔𝑅 and 𝑣𝑒 = 8 𝑘𝑚/𝑠, , Q. Name the force which governs the motion of satellites., Ans: Force of gravity., , Q. Name the natural satellite of earth., Ans: moon., , Q. What is the weight of the body at the Centre of the earth?, Ans: Zero., , Q. What is the height of the geostationary orbit from the surface of earth?, Ans: 36000km., 139

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Q. What is the distance of geostationary satellite from the Centre of the earth?, Ans: 42000km., , Q. What are geostationary satellites?, Ans: The satellites which appear to be stationary when observed from the earth are called, geostationary satellites. A geostationary satellite is also called geosynchronous satellite., , Q. What is the time period of geostationary satellite?, Ans: ≅ 24hours (23 hours and 56 minutes)., , Q. State the conditions necessary for a satellite to appear stationary (geostationary satellites)., Ans: The conditions for geostationary satellites:, (i) The time-period should be 24 hours., (ii) Its orbit should be in the equatorial plane of the earth., (iii) Its direction of motion should be the same as that of the earth about its polar axis., , Q. Mention the uses of polar satellites., Ans: Uses of polar satellites:, (i)., , To record the land and sea temperatures., , (ii). To take pictures of cloud and detect ozone hole layer over Antarctica., (iii). To make forecasting of climatic changes., (iv). To survey and scan the earth’s surface., (v). For ground water survey and preparing wasteland maps., , Q. Obtain the expression for energy of an orbiting satellite., Ans: Energy of satellite: The satellite revolving around the earth has kinetic energy due to its, motion and potential energy due to the gravitational force acting on it by the earth., Therefore, the total energy of the satellite is the sum of its kinetic energy and potential energy., Expression for energy satellite: Consider a satellite performing uniform circular motion, around the earth at a certain height from the surface of the earth., Let:, M = mass of the earth., 140

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m = mass of the satellite, R = radius of the earth, h = height of the satellite above the earth’s surface., r = R + h = radius of the orbit of the satellite or distance of the satellite from the Centre of the, earth, 𝑣𝑜 = orbital velocity of the satellite, As the satellite performs uniform circular motion around the earth, it needs centripetal force., The required centripetal force is provided by the gravitational force acting on the satellite by, the earth., Centripetal force = Gravitational force, 𝑚𝑣𝑜2, 𝑟, , 𝐺𝑀𝑚, , =, , 𝑟2, 𝐺𝑀𝑚, , 𝑚𝑣𝑜2 =, 1, 2, , 𝑟, , 𝑚𝑣𝑜2 =, , 𝐺𝑀𝑚, 2𝑟, , Kinetic energy of the satellite is:, 𝐾=, , 𝐺𝑀𝑚, 2𝑟, , ………………… (1), , Potential energy of the satellite is:, 𝑈= −, , 𝐺𝑀𝑚, 𝑟, , ……………… (2), , Total energy of the satellite = Kinetic energy + Potential energy, 𝐸=, , 𝐺𝑀𝑚, 2𝑟, , 𝐸=−, , −, , 𝐺𝑀𝑚, 2𝑟, , 𝐺𝑀𝑚, 𝑟, , ………………………… (3), , This gives the expression for the total energy of the satellite., Note: The negative total energy of the satellite shows that the satellite is bound to, the earth due to the gravitational force acting on it by the earth., , Q. State three Kepler’s laws of planetary motion., , 141

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Ans: 1) Law of orbit: It states that each planet moves around the sun in an elliptical path with, the sun at one of the focus., 2) Law of area: It states that the line joining the sun and the planet sweeps out equal area in, equal intervals of time., 3) Law of period: It states that the square of time period of the planet is directly proportional to, the cube of semi major axis of the elliptical path., , 𝑇 2 ∝ 𝑟3, , Q. State and explain Kepler’s Laws of planetary motion., Ans: Kepler’s Law planetary motion:, I. Law of orbit: It states that each planet moves around the sun in an elliptical path with the, sun at one of the focus., According to the first Law, the planet follows an elliptical path around the sun at one of the, focus of the ellipse as shown in the following figure., When a planet revolving around the sun in an elliptical orbit comes closer to the sun, its, angular velocity increases as its distance decreases . This results in decrease in the moment of, inertia. This means that a planet moves faster, when it is at its point of nearest approach to the, sun slower, when it is at its point of farthest approach to the sun. Hence, the motion of the, planet is in accordance with aw of conservation of angular momentum., II. Law of Areas: It states that the line joining the sun and the planet sweeps out equal areas, in equal intervals of time., According to this law, If a planet moves from A to B in a given time interval and from C, to D in the same time interval, then the areas ASB and CSD are equal., When a planet approaches the sun its velocity increases and when it recedes away from the, sun its velocity decreases., If a planet moves A to B in a given time interval and from C to D in the same time interval,, then Area of SAB = Area SCD., , 142

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III. Law of periods: It states that the square of time period of the planet is directly, proportional to the cube of semi major axis of the elliptical path., According to this law, if the time period of the planet is T and r is semi major axis , then, 𝑇 2 𝛼𝑟 3 . If 𝑇1 𝑎𝑛𝑑𝑇2 are the time periods of two planets in the elliptical paths of semi-major, axis 𝑟1 𝑎𝑛𝑑 𝑟2 respectively, then:, 𝑇, , 2, , 𝑟, , (𝑇1 ) =(𝑟2 ), 2, , 𝑇12 𝛼𝑟13 𝑎𝑛𝑑 𝑇22 𝛼𝑟23, , 3, , 2, , Q. How does the speed of the earth changes when it is nearer to the sun?, Ans: The planet's orbital speed change when its near to the sun as the object tends to move, faster as the planet is closer to the sun and stronger to the sun gravitational pull and faster, the planet moves and if it's farther from the is the weather is the gravitational attraction., , Q. Distinguish between mass and weight of a body., Ans:, Mass, , Weight, , It is the amount of matter contained in a, , It is the force with which a body is, , body, , attracted towards the centre of the earth., , It is a scalar quantity, , It is a vector quantity, , Its SI unit is kg, , Its SI unit is newton(N), , 143

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Therefore, the two spheres of equal masses is 𝟐𝟒. 𝟓𝒌𝒈, Q. A body of mass 0.5kg falling freely under gravity takes 5 second to reach the ground., Find the kinetic energy and potential energy of the body when body has travelled for 4, second?, Solution:, Given,, m = 0.5kg, u=0, t = 4 sec, a = g, ∴ From v = u + a t, v = 0 + 9.8 x 4, v = 39.2ms−1, ∴ kinetic energy of the body, when the body has travelled for 4 seconds, is, 1, , KE = 2 mv 2, 1, , KE = 2 x 0.1 x (39.2)2, K.E = 76.832 J, To Find Potential Energy (PE):, Distance travelled by the body in 5sec, is, 1, , 𝑆5 = ut + 2 gt 2, 1, , 𝑆5 = 0 + 2 x 9.8 x 52, 𝑺𝟓 = 𝟏𝟐𝟐. 𝟓𝐦, Distance travelled by the body in 4 sec is, 1, , 𝑆4 = ut + 2 gt 2, 1, , 𝑆5 = 0 + 2 x 9.8 x 42, 𝑺𝟓 = 𝟕𝟖. 𝟒𝐦, ∴ℎ = 𝑺𝟓 − 𝑺𝟒, h = 122.5 − 78.4 = 44.1m, ∴P.E = m g h, PE = 0.1 x 9.8 x 44.1, 𝐏𝐄 = 𝟒𝟑𝟐. 𝟏𝟖J, , 145

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Q. A body weighs 63N on the surface of earth. What is the gravitational force on it due to the, earth at a height is equals to half the radius of the earth? Given that radius of the earth =, 6400km., Solution:, Given,, g, , g′ =, , h, R, , (1+ )2, , Substituting ℎ =, , 𝑅𝑒, 2, , g, , g′ =, , 𝑅𝑒 2, , [1+ 2 ], 𝑅𝑒, , g′ =, , g, 1, (1+ )2, 2, , g′ =, , g, 3, ( )2, 2, , g′ =, , 4g, 9, , Weight of body of mass m at height h is given as:, W1 = mg ′, W1 = 63 x, , 4, 9, , W1 = 28N, , Q. A planet has same mass as that of earth, but its radius is half that of earth. If the, acceleration due to gravity on earth is ‘g’, what is the acceleration due to gravity on that, planet?, Solution:, Given,, Mass of earth: 𝑚𝑒 = 𝑚, Mass of planet: 𝑚𝑝 = 𝑚, Radius of earth: 𝑅𝑒 = 𝑅, Radius of planet: 𝑅𝑝 =, , 𝑅, 2, , Acceleration due to gravity on earth is g =, , Gme, R2e, , ………. (1), , 146

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Q. What is the value of acceleration due to gravity 1400km above the surface of the earth,, whose radius is 6400 km. (Given g = 9.8 m/s2), Solution:, Given,, g= 9.8m/s2, R= 6400 km, h= 1400 km,, 𝑔𝑅2, , We know that gh =, , (𝑅+ℎ)2, , 9.8×(6400) 2, , gh= (6400+1400)2, gh=, , 9.8×642 ×104, 782 ×104, , gh= 6.59 m/s2, , Q. Find the difference in the values of acceleration due to gravity, (i) at a height 2500km above the surface of the earth and, (ii) at a depth 2500km below the surface of earth?, (Given 𝒈 = 𝟗. 𝟖𝒎𝒔−𝟐 and R = 𝟔𝟑𝟕𝟖𝒌𝒎 .), Solution:, Given,, g = 9.8ms −2 ,, h = d = 2500km = 2500 × 103 m, R = 6378km = 6378 × 103 m, gR2, , At height: g ′ = (R+h)2, 9.8 × (6378 × 103 )2, , g ′ = [(6378 + 2500) × 103 ]2, g ′ = 5.57ms −2, 𝐴𝑡 𝑑𝑒𝑝𝑡ℎ ∶, d, , g ′′ = g [1 − R], 2500×103, , g ′′ = 9.8 [1 − 6378×103 ], g ′′ = 9.8 × 0.608, 𝐠 ′′ = 𝟓. 𝟗𝟓𝐦𝐬 −𝟐, , 148

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Q. Calculate the acceleration due to gravity:, (a) At height 16 km above the earth’s surface and, (b) At depth 12.8 km below the surface of earth., Given: Radius of earth is 6400km and acceleration due to gravity on the surface of earth,, g = 9.8𝒎𝒔−𝟐 ., Solution:, Given,, (a) Height: h = 16 km,, Radius of earth: R = 6400km,, Acceleration due to gravity: g = 9.8𝑚𝑠 −2, The acceleration due to gravity above the surface of earth:, 𝑔ℎ = 𝑔 (1 −, , 2ℎ, 𝑅, , ), 2𝑥16, , 𝑔ℎ = (9.8) (1 − 6400 ), 32, , 𝑔ℎ = (9.8) (1 − 6400 ), 𝒈𝒉 = 𝟗. 𝟕𝟓𝟏 𝒎𝒔−𝟐, (b) Depth: d = 12.8 km, The acceleration due to gravity below the surface of earth:, 𝑑, , 𝑔𝑑 = 𝑔 (1 − 𝑅 ), 12.8, , 𝑔𝑑 = (9.8) (1 − 6400), 𝒈𝒅 = 𝟗. 𝟕𝟖𝟎𝟒𝒎𝒔−𝟐, Q. Find the value of acceleration due to gravity, (a) at height 3000km above the surface of earth and, (b) at a depth 8km below the surface of earth., (Given: RE=6400km & g=9.8ms-2), Solution:, Given,, radius of earth R=6400km,, height above the surface of earth h=3000km,, the depth below the earth surface d=8km,, g=9.8 m/s-2., Acceleration due to gravity at height above the surface of earth:, 149

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Q.A earth satellite in a circular orbit at a height 200km above the surface of earth has a, period of 80 minute. Calculate the mass of the earth from this earth., (Given: Radius of the earth = 6400km), Solution:, Given,, Time period T=80min=80x60=4800sec, 𝑇 2 = 4800 𝑥 4800 = 2304 𝑥104, Radius of the earth R= 6400km=6400 x 103 m=64 x 105 m, Orbit height h= 200km = 200 x 103m = 2 x 105m, r = (R + h) = (64 x 105 + 2 x 105 ) = 66 x 105 m, 𝐺𝑀, , 𝑉0 = √𝑅+ℎ, v02 =, T=, , GM, (R+H), , ...........(1), , 2π(R+h), , [T]2 =, , V0, 4[𝜋]2[R+h], v20, , ........ (2), , From equation (1) and (2):, T 2 = 4π2, , [R+h]3, GM, [r]3, , M = 4π2 GT2, [66 x 105 ], , 3, , M = 4(3.14)2 6.7 x 10−11 x 2304 𝑥104, 287496 x 1015, , M = 4 x 9.856 x 6.7, M=, M=, , x 2304 x 10−7, , 4 x 9.856 x 2.87496 x 1015 x 105, 154408.2 x 10−7, 39.424 x 2.87496 x 1015 x 105, 15.44082 x 10−7 x 104, 113.424 x 1020, , M = 15.44082 x 10−3, 𝐌 = 𝟕. 𝟑𝟒𝟓𝟕 𝐱 𝟏𝟎𝟐𝟑 𝐤𝐠, 155

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Q. The planet mars has two moons, Phobos and delmos., (i) Phobos has a period of 7 hours, 39 minutes and an orbital radius of 𝟗. 𝟒 𝐱𝟏𝟎𝟑 𝐤𝐦., Calculate the mass of mars., (ii) Assume that earth and mars moves in circular orbits around the sun, with martain, orbit beging 1.52 times the orbital radius of earth. What is the length of martain, year in days?, Solution:, Given,, orbital radius of R = 9.4 x103 km =9.4 x106 m, period of 7 hours, 39 minutes: T =459min=459 x 60 sec=27540sec=2.754 x104 s, (i) The mass of mars:, 𝐺𝑀, , 𝑉0 = √𝑅+ℎ, GM, , v02 = (R+H) ...........(1), T=, , 2π(R+h), V0, , [ T] 2 =, , 4[𝜋]2 [R+h], v20, , ........ (2), , From equation (1) and (2):, T 2 = 4π2, , [R+h]3, GM, [r]3, , Mm = 4π2 GT2, [9.4 x 106 ], , 3, , M = 4(3.14)2 6.7 x 10−11 x[2.754 x 104 ]2, M=, M=, , 4 x 9.856 x 830.584 x 1018, 6.7 x 10−11 x7.5845 x 108, 327.4494 x 1020, 50.5515 x 10−3, , 𝐌 = 𝟔. 𝟒𝟕𝟕𝟓 𝐱 𝟏𝟎𝟐𝟑 𝐤𝐠, (ii) martain year in days by using kepler’s third law:, 3, For mars: 𝑇𝑚2 = 𝑅𝑚, ……. (i), , 156

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PART-2, , 158

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9. MECHANICAL PROPERTIES OF SOLIDS, , Q. What is elasticity of a body?, Ans: The property of a body by virtue of which it tends to restore its original shape and size,, when the applied force is removed., , Q. Define stress ., Ans: The internal restoring force acting per unit area of cross – section of the deformed body is, called stress., Stress =, , 𝐅𝐨𝐫𝐜𝐞, 𝐀𝐫𝐞𝐚, , Q. What is elastic limit?, Ans: The maximum stress up to which the body shows the property of elasticity is called, elastic limit., Q. Define Young’s modulus and obtain the expression for Young’s modulus., Ans :Young's modulus (Y): Within proportionality limit, the ratio between longitudinal stress, to longitudinal strain is called Young's modulus., Consider an elastic wire of length L , radius r and cross sectional area A . Let ∆𝑳 or e be the, change in the length of the wire due to the stretching force F applied along the length of the, wire ., Longitudinal stress =, , Stretching force, , Longitudinal strain =, , Area, Change in length, original length, , F, , σ=A, ϵ=, , ∆L, L, , Young’s modulus of material of the wire can be written as :, Longitudinalstress, , Young’s modulus= Longitudinalstrain, 𝐹𝐿, , 𝐹𝐿, , Young’s modulus 𝑌 = 𝐴𝑒 or 𝑌 = 𝐴∆𝐿, Where Y is young’s modulus, F is Applied force, L is length of the wire, A is Area of the wire and ∆𝐿 or e is elongation of the wire, 159

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Q. Name the SI unit of stress ., Ans:𝑃𝑎 (𝑜𝑟) 𝑁𝑚−2, , Q. State Hooke’s law., Ans: For small deformation, stress is directly proportional to strain OR Within the elastic, limit, stress is directly proportional to strain., , Q. Why steel is more elastic than that of rubber?, Ans: The change in length (∆𝐿) for steel is less than that for rubber. Therefore young’s, modulus (Y) of steel is more and less for rubber., , Q. Which of the following has highest value of shear modulus?, (1) Aluminium (2) Tungsten and (3) Wood, Ans: (2)Tungsten, , Q. What is elasticity?, Ans: The property of a body by virtue of which it tends to restore its original shape and size,, when the applied force is removed is known as elasticity., , Q. How does strain depend on stress?, Ans: The Stress is directly proportional to the strain., , Q. Define Poisson’s ratio and write the formula for it., Ans: Poisson’s ratio is the ratio of lateral strain to the longitudinal strain. It is represented by σ., Poisson’s ratio mathematically written as,, 𝜎=, , −∆𝐷⁄𝐷, ∆𝐿⁄𝐿, , = (−, , ∆𝐷, , 𝐿, , ) (∆𝐿), 𝐷, , 160

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Q. Draw stress- strain graph of a metal wire. Explain yield point., Ans:, , Yield Point: Body returns to its original dimension after removal of the force., Q. Draw a typical stress – strain curve for a metal. Mark elastic limit (yield point in it)., Ans:, , Q. Define modulus of elasticity and obtain the expression for Young’s modulus of elasticity., Ans: Modulus of Elasticity: The ratio of stress to strain within the elastic limit is called, modulus of elasticity., Consider an elastic wire of length L, radius r and cross section area A. Let ∆𝒍 b the change in, the length of the wire due to stretching force Fapplied along the length of the wire., Tensile stress =, , stretching force, area, , Longitudinal strain =, , F, , → σ = A., , change in length, Original length, , ∆l, , = ε = L., , If F=Mg. Where M is mass of the load and A = πr 2 , then the Young’s modulus of material, wire can be written as: =, , σ, ε, , =, , F, A, ∆l, L, , MgL, , → Y = πr2 ∆l ., , This is the required expression for Young’s modulus in case of stretched string., , Q. How does the fractional change in length of a rod vary with its change in temperature?, Ans: Fractional change in length of a rod is directly proportional to the change in temperature., 161

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Q. Mention three types of moduli of elasticity., Ans: Types of moduli of elasticity are:, 1. Young’s modulus (Y), 2. Bulk modulus (B), 3. Shear modulus (G), Q. Define shear modulus. Write its expression and explain terms., Ans: Shear modulus: The ratio of shearing stress to the corresponding shearing strain is, defined as shear modulus of material and is denoted by G., 𝑆ℎ𝑒𝑎𝑟 𝑚𝑜𝑑𝑢𝑙𝑢𝑠 =, , 𝑆ℎ𝑒𝑎𝑟𝑖𝑛𝑔 𝑠𝑡𝑟𝑒𝑠𝑠, 𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑎𝑖𝑛, , 𝐺=, , 𝐹/𝐴, 𝐹/𝐴, 𝐹𝐿, =, ⇒𝐺=, 𝜃, ∆𝑥/𝐿, 𝐴∆𝑥, , Where G is shear modulus, F is force; A is area of cross section of material, ∆𝑥 is relative, displacement of the faces and 𝜃is small angular displacement., , Q. Explain different types of strains., Ans:, i. Longitudinal strain: The ratio of change in length, , L to the original length L of the, , body is called longitudinal strain., Longitudinal strain , , Change in length, L,  , Original length, L, , ii. Shearing strain or Tangential strain: Shearing or Tangential strain is defined as the ratio, of relative displacement of the faces x  to the length L  of the body., , Shearing Strain , , Re lative displaceme nt of the faces of the body, Length of the body, ,  tan   x, L, , Where:   For small angular displacement tan    ,, Shearing strain can be written as: angular displacement of the, 162

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body from the vertical position., ,  , , x, L, , iii. Volume strain or Hydraulic strain: Volume strain or hydraulic strain is defined as the, ratio of change in volume V  to the original volume V  ., Volume, , strain , , Change in volume, Original volume, , , , V, V, , In this case, there is a decrease in the volume of the body without any change its geometrical, shape as a result of hydraulic compression., , Q. Write stress-strain curve for a metal. What is proportional limit and yield point?, Ans : To study the behaviour of a metal wire under increasing load , a metal wire is suspended, from a rigid support and loaded at the other end . The load is increased gradually until it, breaks. A graph is plotted between the stress on Y – axis and the strain on the X -axis ., Proportionality limit : The maximum stress applied on the wire up to which stress is directly, proportional to the strain i.e., Hooke’s law is obeyed , is called proportionality limit ., Yield point (Elastic limit ): The maximum stress below which Hooke’s law is applicable is, called elastic limit . Or The elastic limit is the maximum stress on removal of which the body, regain its original dimension., , Q. Write any three applications of elastic behaviour of materials., Ans: Applications of Elastic behaviour of material:, 1. For designing a building, the structural design of the columns beams and supports require, the knowledge of the strength of materials., 2. for estimating maximum height of a mountain., 3. In overcoming of the problem of bending of beam under a load., , 163

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10. MECHANICAL PROPERTIES OF FLUIDS, Q. Discuss the variation of pressure in case of liquids at rest., Ans: Variation of pressure with depth: Consider a fluid of density 𝜌 at rest in vessel. Let 1, and 2 be the two points in the fluid. The point 1 is at a height h above the point 2. The, pressures at points 1 and 2 are P1 and P2 respectively. To find the pressure difference between, these points, a cylindrical element of fluid of base area A and height h is imagined., Weight of the fluid in the imaginary cylinder, = mass of the fluid in the cylinder × acceleration due to gravity, = Volume of the cylinder × Density of the fluid × acceleration, due to gravity, → 𝐹 = 𝐴ℎ𝜌………… (1), As the cylinder is in equilibrium, the net force due to the pressure, difference must be equal to the weight of the fluid in the cylinder., ∴ Force due to the pressure difference = Weight of the fluid in the cylinder, 𝑃2 𝐴 − 𝑃1 𝐴 = 𝜌𝑔ℎ𝐴 → 𝑃2 − 𝑃1 = 𝜌𝑔ℎ.…………… (2), It follows from this equation that the pressure difference depends on the vertical distance h, between the points 1 and 2, density of the fluid 𝜌 and acceleration due to gravity g . If the point, 1 is shifted to the top of the fluid (water) then it is opened to the atmosphere. The pressures, 𝑃1 = 𝑃𝑎 = atmospheric pressure and 𝑃2 = 𝑃 = 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒𝑎𝑡𝑑𝑒𝑝𝑡ℎbelow the free surface of the, fluid are replaced then, equation (2) becomes: 𝑃 − 𝑃𝑎 = 𝜌𝑔ℎ……………… (3), This excess of pressure 𝑃 − 𝑃𝑎 at a depth h is called gauge pressure., Equation (3) can be written as: 𝑃 = 𝑃𝑎 + 𝜌𝑔ℎ……36………...... (4), Thus the pressure P at a depth below the free surface of a liquid open to the atmosphere is, greater than the atmospheric pressure by an amount 𝜌𝑔ℎ, Q. Mention the expression for average pressure., F, , Ans: Pavg = A, , Q. What is the SI unit of pressure?, 168

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Ans: Pascal or N/m2, , Q. Write the dimensional formula of pressure., Ans: [M1L-1T-2], Q. Mention the device used to measure atmospheric pressure., Ans: The instrument used to measure atmospheric pressure is ‘Barometer’, , Q. State Pascal’s law., Ans: It states that whenever an external pressure is applied on any part of a fluid contained in a, vessel, then it is transmitted undiminished and equally in all directions., Q. Mention any one application of pascal’s law., Ans: 1) Hydraulic lift, 2) Hydraulic brake, 3) Hydraulic press, Q. Explain how Pascal’s law is applied in a hydraulic lift ., Ans: Hydraulic lift : It consist of two vertical cylinders joined together with a horizontal pipe., Each cylinder is provided with frictionless piston of different areas of cross section 𝐴1 and 𝐴2, such that 𝐴2 > 𝐴1 . The cylinders are filled with, incompressible liquids. The weight to be lifted is, placed on the piston having larger cross section area, and a force 𝐹1 is applied on the piston with a smaller, cross section area. Due to it,, 𝐹, , 𝐴, , The pressure 𝑃 = (𝐴1 ) 𝐴2 = (𝐴2 ) 𝐹1., 1, , 1, , Thus, the applied force 𝐹1 increases by a factor, As a result, the heavy load such as car, , 𝐴2, 𝐴1, , ., , gets lifted., , Q. What is buoyancy?, Ans:-An upward force exerted by the displaced fluid on a body immersed completely or, partially in a fluid is called buoyant force or buoyancy or upward thrust., , 169

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Q. Compare streamline and turbulent flow of a liquid., Ans:, Streamline flow, , Turbulent flow, , 1. The velocity of the fluid at a point, , 1. The velocity of the fluid at a, , remains constant both in magnitude, , point changes randomly both in, , and direction., , magnitude and direction., , 2. It is an orderly flow of fluid., , 2. It is disorderly flow of fluid, , 3., , 3., , No eddies and vertices are, , Eddies and vertices are, , formed., , formed, , 4. Velocity of the fluid is less than, , 4. Velocity of the fluid is greater, , the critical velocity., , than the critical velocity, , Q. Write the equation of continuity for streamline flow of fluid., Ans: A1V1 = A2V2., , Q. Write the mathematical form of Bernoulli’s theorem and explain the terms., 1, , Ans: P+2v2+gh = constant., Where,, P - pressure of the flow of fluid,  - density of the flow of fluid, v – velocity of the flow of fluid, g - acceleration due to gravity, h – height of the flow of liquid from the ground, , Q. State and explain Bernoulli’s theorem. Mention any two applications of Bernoulli’s, theorem., Ans: Bernoulli’s theorem: It states that along a, streamline flow, the sum of pressure; the kinetic, energy per unit volume and the potential energy per, unit volume remain constant., 170

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1, , P + ρv 2 +ρgh=constant, 2, , Consider an ideal fluid flowing streamline through a pipe AB or varying cross sectional area as, shown in the following figure., Let 𝐏𝟏 &𝐏𝟐 be the pressure and 𝐀𝟏 &𝐀𝟐 be the areas of cross section and 𝐯𝟏 &𝐯𝟐 be the speeds of, fluid at A and B respectively . In infinitesimally small interval of time ∆𝐭 , the fluid at A, moves through a distance 𝐯𝟏 ∆𝐭 to C and at the same time , the fluid at D moves through a, distance 𝐯𝟐 ∆𝐭 to B ., Applications of Bernoulli’s theorem:, 1. Filter pump, 2. Sprayer, , Q. On what principle venturimeter work?, Ans: Venturimeter works on Bernoulli’s principle., , Q. Define viscosity of fluid., Ans: The property of fluid by virtue of which the liquid opposes the relative motion between, layers is called viscosity of fluid., , Q. How the viscosity of the liquids varies with temperature?, Ans: The viscosity of liquid decreases with increase in temperature., , Q. How the viscosity of the gases varies with temperature?, Ans: The viscosity of gases increases with increase in temperature., , Q. What is terminal velocity?, Ans: The maximum constant velocity acquired by a body while falling freely through a, viscous medium is called terminal velocity., , 171

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Q. What is meant by angle of contact?, Ans: The angle between the tangent to the liquid surface at the point of contact and the solid, surface inside the liquid is called the angle of contact., , Q. What is capillarity?, Ans: The tendency of a liquid in a capillary tube or absorbent to rise or fall as a result of, surface tension., , 172

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NUMERICALS, Q. What is the average pressure on human two thigh bones, each of cross-sectional area, 10cm2 support the upper part of human body of mass 50kg? (Taking g =10ms-2), Solution:, Given,, m = 50kg, g = 10 m/s2, F = mg = 50 x 10N, A = 2 x 10 x 10-4 m2, A = 20 x 10-4 m2, 𝐹, , 𝑃=𝐴, 500, , 𝑃 = 20 𝑋 10−4 = 2.5 x 105 N/m2, , Q. At a depth of 1000m in an ocean (a) What is the absolute pressure? (b) What is the gauge, pressure? (c) Find force acting on the window of area 20cm20cm of a submarine at this, depth, the interior of which is maintained at sea level atmospheric pressure. (Given: The, Density of sea water = 1.03103 kgm-3 and g=10ms-2), Solution: Given h= 1000m, , Patm = 1.013 x 105 pa, , ρ = 1.03 x 103 kg/m3, g= 10m/s2, (a) The absolute pressure: P = Patm + ρgh, P = 1.013 x 105 + 1.03 x 103 x 10 x 1000, P = 1.013 x 105 + 103 x 105, P = 104.013 x 105 pa, (b) What is the gauge pressure (P1 - Patm ) = ρgh, (P1 - Patm ) = 1.03 x 103 x 103 x 10 x 1000, (P1 - Patm ) = 103 x 105 pa, 𝐹, , (c) Force acting on the window: P= 𝐴, A = 20 x 20 cm = 400 x 10-4 m2, F =PA, F = 104.013 x 105 x 400 x 10-4, F = 4.16 X 105N., 173

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11. THERMAL PROPERTIES OF MATTER, Q. Name the instrument used to measure temperature of the substance., Ans: Thermometer., , Q. Write the relationship between degree Celsius ( 0C ) and degree Fahrenheit ( 0F ) of, temperature., Ans:, , 𝐶, , =, 5, , 𝐹−32, 9, , Q. Mention the value of steam point of water in Fahrenheit scale., Ans: 212𝑜 F, , Q. Write the relation between Celsius scale and Kelvin scale of temperature., Ans: C = K - 273 (or) T = t+273 (or) t =T - 273, , Q. Write any two differences between heat and temperature., Ans:, Heat, , Temperature, , 1. Heat is the energy that is transferred from, , 1. The degree of hotness or coldness of a, , one body to another body due to, , body is called its temperature., , temperature difference between the two bodies, 2. Calorimeter is used to measure the heat., , 2. Thermometer is used to measure, temperature., , Q. State Boyle’s and Charles Laws., Ans: Boyle’s Law: It states that At constant temperature, the volume of a given mass of a gas, is inversely proportional to its pressure., , 174

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Charles Laws at constant pressure: It states that At constant pressure, the volume of a given, mass of a gas is directly proportional to its temperature., Charles Laws at constant volume: It states that At constant volume, the pressure of a given, mass of a gas is directly proportional to its temperature., Q. According to Which law the pressure of given mass of gas varies inversely with volume at, constant temperature?, Ans: Boyle’s Law, Q. State and explain (i) Boyle’s law and (ii) Charles law at constant pressure ., Ans: (i) Boyle’s law: It states that at constant temperature, the volume of a given mass of a gas, is inversely proportional to its pressure., If V is the volume of a given mass of a gas and P is its pressure at constant temperature,, 1, , then 𝑉𝛼 𝑃, 𝑃𝑉 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡, If 𝑉1 , 𝑉2 , 𝑉3 , …………𝑉𝑛 are the volumes of a given mass of a gas at pressures, 𝑃1 , 𝑃2 , 𝑃3 , …………𝑃𝑛 respectively at constant temperature , then, 𝑃1 𝑉1 = 𝑃2 𝑉2 = 𝑃3 𝑉3 = ⋯ … … … … … = 𝑃𝑛 𝑉𝑛 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡, (ii) Charles law at constant pressure: It states that at constant pressure, the volume of a, given mass of a gas is directly proportional to its temperature., If V is the volume of a given mass of a gas and T is its temperature at constant pressure,, 𝑉𝛼𝑇, , then, 𝑉, 𝑇, , = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡, , If 𝑉1 , 𝑉2 , 𝑉3 , …………𝑉𝑛 are the volumes of a given mass of a gas at temperatures, 𝑇1 , 𝑇2 , 𝑇3 , …………𝑇𝑛 respectively at constant pressure , then, 𝑉1, 𝑇1, , 𝑉, , 𝑉, , 𝑉, , = 𝑇2 = 𝑇3 = ……………..=𝑇𝑛 = Constant, 2, , 3, , 𝑛, , 175

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Q. Write a note on absolute scale/Kelvin scale of temperature., Ans: Kelvin designed the scale of temperature with its zero at -273.15o C and size of each, degree same on Celsius scale. The lower fixed point is 273.15K and the upper fixed point is, 373.15K. The interval between lower and upper fixed points is divided into 100 equal parts., This scale of temperature is called absolute scale of temperature or Kelvin scale of, temperature. The unit of temperature on this scale is written as K., Q. What is the lowest possible temperature?, Ans: The lowest possible temperature is -273.150C., , Q. What is linear expansion?, Ans: The thermal expansion of a solid body in length is called linear Expansion., , Q. What are linear and volume expansion coefficients? show that 𝜶𝒗 = 𝟑𝜶𝟏, Ans: Linear Expansion: The thermal expansion of a solid body in length is called linear, Expansion., Volume Expansion: The thermal expansion of a solid body in volume is called volume, expansion., Relation between coefficient of linear expansion and coefficient of volume expansion:, Consider a cube of side L and Volume V., LT = L(1 + α∆T) --------------- (1), Cubing both sides:, , 𝐿3𝑇 = 𝐿3 (1 + 𝛼∆𝑇)3, , VT = V(1 + α3 ∆T 3 + 3α∆T + 3α2 ∆T 2 ) ------------- (2), ∵ (𝑎 + 𝑏)3 = 𝑎3 + 𝑏3 + 3𝑎2 𝑏 + 3𝑎𝑏2, Where 𝐿3𝑇 =VT = volume of the block at 𝑇 0 C, Since α is small, 𝛼 2 𝑎𝑛𝑑 𝛼 3 can be neglected., Equation (2) becomes:, VT = V(1 + 3α∆T)------------- (3), But: VT = V(1 + γ∆T)-------------- (4), 176

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Comparing Equations (3) and (4), V(1+ γ∆T) = V (1 + 3α∆T), 1+ γ∆T = 1 + 3α∆T, γ∆T = 3α∆T, 𝛄 = 𝟑𝛂 → 𝜶𝒗 = 𝟑𝜶𝟏 --------------(5), , Q. Draw a graph showing variation of volume of water with increasing in temperature., Ans:, , Q. What is an ideal gas? Write the ideal gas equation .At what condition real gas behaves like, ideal gas., Ans: A gas which obeys gas laws strictly at all the temperatures is called ideal gas, PV= nRT, For 1mol of gas PV=RT, Generally a gas behaves like an ideal gas at high temperature and low pressure, , Q. What is an ideal gas? Write the ideal gas equation and give the value of constant in the, equation with SI unit., Ans: The gas which obeys the gas laws at all values of temperatures and pressures is called an, ideal gas., Equation of an ideal gas:, 𝑷𝑽 = 𝑹𝑻, Where: P=pressure of the gas, V=volume of the gas, T= temperature of the gas and R=, universal gas constant., Value of constant in the equation: R=8.31Jmol-1K-1., 177

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(b) Specific heatcapacity of a gas at constant pressure (𝐂𝐩 ): Specific heat capacity of a gas, at constant pressure is defined as the amount of heat required to increase the temperature of, one kg of a gas by one kelvin or one degree Celsius ., 𝐂𝐩 of mass m to change its temperature by ∆𝐓 by supplying a heat ∆𝐐 is given by :, ∆Q, , Cp = m∆T, If m=1kg and ∆𝐓=1K, then above equation becomes, , 𝐂𝐩 = ∆𝐐 ., , (c)Latent heat of vaporization: The latent heat for a liquid –gas state change is called Latent, heat of vaporization (Lv )., The Latent heat of vaporization Lv of the substance of mass m which requires a heat Q is, given by:, Q, , Lv = m, If m=1kg then 𝐋𝐯 = 𝐐, Q. Define (a) Specific heat capacity of a gas at constant volume, (b) Specific heat capacity of gas at constant pressure, (c) Latent heat of fusion., Ans: (a) Specific heatcapacityof a gas at constant volume(𝐂𝐯 ): The amount of heat required, to increase the temperature of one kg of a gas by one kelvin or one degree Celsius., ∆𝐐, , 𝐂𝐯 = 𝐦∆𝐓, If m=1kg and ∆𝐓=1K , then above equation becomes, 𝐂𝐯 = ∆𝐐 ., (b) Specific heatcapacityof gas at constant pressure (𝐂𝐩 ): Specific heat capacity of a, gas at constant pressure is defined as the amount of heat required to increase the, temperature of one kg of a gas by one kelvin or one degree Celsius ., 𝐂𝐩 of mass m to change its temperature by ∆𝐓 by supplying a heat ∆𝐐 is given by :, 𝐂𝐩 =, , ∆𝐐, 𝐦∆𝐓, , If m=1kg and ∆𝐓=1K , then above equation becomes, 𝐂𝐩 = ∆𝐐 ., 179

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(c) Latent heat of fusion: It is defined as the amount of heat required to convert unit, mass of substance (1kg) from its solid state to liquid state as its melting point, without any changes in temperature., Q. What is latent heat? On what factors does it depend? Give its S.I unit., Ans: Latent heat is the amount of heat required to change the state of a substance at constant, temperature., Latent heat depends on,, Nature of the material of the substance, Pressure of the substance, S. I unit of the latent heat is Jkg -1, , Q. Define latent heat of fusion and latent heat of vaporization. Explain the variation of, temperature with heat energy for water at 1 atmosphere with graph., Ans: latent heat of fusion:The latent heat for a solid-liquid state change is called latent heat of fusion [𝐿𝑓 ]. (or) The, amount of heat required to convert unit mass of a substance (1Kg) from its solid state to liquid, state at its melting point without any change in temperature is known as latent heat of, fusion[𝐿𝑓 ]., Latent heat of vaporization:The latent heat for a liquid-gas state change is called latent heat of vaporization [𝐿𝑣 ]., (or) The amount of heat required to convert unit mass of a substance (1Kg) from its, liquid state to gaseous state at its boiling point without any change in temperature is, known as latent heat of vaporization[𝐿𝑣 ]., Graph to show the variation of temperature with heat energy for water at one atmosphere:, , 180

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Here we take Ice at -100C. As we start heating, the temperature of ice increases until it reaches, its melting point (00C). At this temperature, the addition of more heat does not increase the, temperature but causes the ice to melt, or changes its state. As is obvious from the graph, the, latent of fusion of ice is 𝐿𝑓 = 3.3 × 105 𝐽𝑘𝑔−1 i.e 3.3 × 105 𝐽 of heat is needed to melt 1 kg of, ice into water at 00C., Once the entire ice melts, the addition of more heat causes the temperature of water to rise., The temperature keeps on increasing till it reaches 1000C. The water starts boiling. The, addition of more heat to the boiling water causes vapourisation, without increase in, temperature. Obviously, the latent heat of vaporisation of water is, 𝐿𝑉 = 22.6 × 105 𝐽𝐾𝑔−1 ., i.e 22.6 × 105 𝐽 of heat is needed to convert 1 kg of water to steam at 100 0C. Additional heat, causes the temperature of the steam to rise., , Q. What is regelation? Give one of its practical applications., Ans: The phenomenon of refreezing the ice is called regelation, Ex: Skating is possible due to formation of water layer below the skates. Water is formed due, to the increase of pressure and it acts as a lubricant., , Q. How does the melting point of ice changes with increase of pressure?, Ans: The melting point of ice decreases with increase of pressure., , Q. How does the aquatic animals survive during winter even when lake or pond is covered by, ice?, Ans:, Anomalous expansion of water plays very important role in the life of animals in lakes,, ponds, seas ….. etc., This anomalous behaviour of water causes ice to form at the surface of a lake in cold weather., In cold countries (in Polar Regions), the temperature of atmosphere reaches as below as -20o C, in winter. In winter a lake cools and reaches up to 4°C, then water sinks to bottom because of, its increased density. Consequently, the surface reaches 0°C first and the lake becomes, covered with ice. Aquatic life is able to survive the cold winter as the lake bottom remains, 181

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unfrozen at a temperature of about 4°C due to density of water is maximum so that animals, can survive at bottom of lakes, ponds, seas…. etc., Q. Give the principle of calorimetry., Ans: If two bodies at two different temperatures are brought into thermal contact , then heat is, lost by the hot body and heat is gained by the cold body until they attain thermal, equilibrium (common temperature)., , Q. Distinguish between heat capacity and Specific heat capacity., Ans:, Heat capacity, , Specific heat capacity, , 1) Thermal capacity of a substance is, , 1) Specific heat capacity of a substance, , defined as the amount of heat required, , is defined as the amount of heat, , to increase its temperature by one, , required to increase the temperature of, , Kelvin or one degree Celsius., , the substance of mass one kg by one, Kelvin or one degree Celsius., , S=, , 2), , ∆Q, ∆, , SI unit is 𝐽𝐾𝑔 −1 𝐾 −1, , 3), , ∆Q, , 2), , C = m∆T, , 3), , S I unit is 𝐽𝐾 −1, , Q. Define 𝑪𝑷 𝒂𝒏𝒅 𝑪𝑽 and explain why 𝑪𝑷 > 𝑪𝑽, Ans : The specific heat capacity of a gas at constant pressure ( 𝒄𝒑 ) : specific heat capacity, of, , a gas at constant pressure is defined as the amount of heat required to increase the, , temperature of one kg of a gas by one Kelvin or one degree Celsius . 𝑐𝑝 =, , ∆𝑄, 𝑚∆𝑇, , The specific heat capacity of a gas at constant volume (𝒄𝒗 ):The amount of heat required, to increase the temperature of 1kg of a gas by one Kelvin or one degree Celsius.cv =, , ∆Q, m∆T, , 𝒄𝒑 𝐢𝐬 𝐠𝐫𝐞𝐚𝐭𝐞𝐫 𝐭𝐡𝐚𝐧 𝒄𝒗 ∶ When a gas is heated at constant volume, the entire heat energy is, utilized to increase the temperature of the gas (or to increase the internal energy). But when, it is heated at constant pressure, a part of heat energy is used to increase its temperature (or, internal energy) and the remaining part is used to do the work in expanding the gas against, the external pressure. Therefore, more heat is required to raise the temperature of unit mass, of gas (one mole of gas) by one Kelvin at constant pressure than that at constant volume., Hence, specific heat capacity of a gas at constant pressure is greater than that at constant, volume., 182

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Q. What is conduction of heat? State and explain the law of thermal conductivity and hence, define the coefficient of thermal conductivity and write its SI unit., Ans: Conduction of heat: The process of transmission of heat from one part of the body at a, higher temperature to another at a lower temperature, without the actual movement of the particles of the, medium is known as conduction of heat., Laws of thermal conductivity: the amount of heat, transferred across the ends of the rod (Q) is:, Directly proportional to cross sectional area of the rod (A), Directly proportional to temperature difference (𝑇1 − 𝑇2 ) between the ends,, Directly proportional to time for which the heat flows (t) and, Inversely proportional to length of the rod (L)., According to the laws of thermal conductivity:, Q  (𝑇1 − 𝑇2 ),, Q  t,, 1, , Q ,, 𝐿, , QA, 𝑄=, , 𝐾𝐴(𝑇1−𝑇2)𝑡, 𝐿, , -------------- (1), , Where K= constant called co-efficient of thermal conductivity or thermal conductivity of the, material of the rod., If A = 1,, , (𝑇1−𝑇2 ), 𝐿, , = 1 and t = 1, then equation (1) becomes: K = Q, , The thermal conductivity of a material (co-efficient of thermal conductivity) is defined, as the, , amount of heat transferred per second across a rod of that material in steady, , state per unit area of cross section , when its opposite ends are maintained at unit, temperature gradient ., S I unit of thermal conductivity is: W𝑚−1 𝐾 −1 OR J𝑠 −1 𝑚−1 𝐾 −1, , 183

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Q. Name modes of transfer of heat., Ans: (i) Conduction,, (ii) Convection,, (iii) Radiation., Q. Why cooking pots have copper coating on the bottom?, Ans: Being a good thermal conductor copper promotes (or) The fast and uniform distribution, of heat over the bottom of the pot for uniform cooking. (or) Copper conducts heat five, times better than iron and even twenty times better than stainless steel., , Q. What is convection of heat?, Ans: It is the process by which heat flows from the region of higher temperature to the region, of lower temperature by the actual movement of the particles of the medium., , Q. Mention the factors on which heat flow by conduction in a metal bar depends., Ans: The heat flow by conduction in a metal depends on:, (i) cross sectional area of the metal bar,, (ii) temperature difference (T1-T2) between the ends of the metal bar,, (iii) time for which the heat flows (t) and, (iv) length of the bar(L)., , Q. Mention any three factors on which amount of heat added to increase the temperature of, the substance depends., Ans: Three factors on which amount of heat added to increase the temperature of the, substance depends:, 1. Cross sectional area of the rod., 2. Time for which the heat flows., 3. length of the rod., , 184

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Q. Explain the three distinct modes of heat transfer. State the laws of thermal conductivity., Ans: (i) Conduction: The process of transmission of heat from one part of the body at a, higher, , temperature to another at a lower temperature without the actual movement of the, , particles of the medium is known as conduction., Explanation: When one end of the solid is heated, the atoms or molecules at the heated end, absorb thermal energy and begin to vibrate with larger amplitude. They collide with the, neighbouring atoms and transfer a part of energy to them. These atoms in turn begin to vibrate, more vigorously collide with the neighbours and transfer a part of thermal energy to them. The, process continues. Thus, heat is transferred from the hot end to the cold end. The medium takes, an active part in the process. The particles of medium do not leave their positions but they, transfer energy through their vibrations. Therefore, the heat energy is transferred without the, actual movement of the particles of the medium in conduction., (ii) Convection: The process of transmission of heat from a place of higher temperature to a, place of lower temperature by the actual movement of the particles of the medium is known as, convection., Explanation: When a glass vessel containing water is heated, the water molecules at the, bottom of the vessel heated up. These heated molecules become light and rise to the surface of, the water. The places vacated by the heated molecules are occupied by colder molecules. On, receiving the heat energy from the source, the colder molecules become light and move, towards the surface. This process repeats and the currents are produced in water. These currents, are known convection currents. Thus, in convection, heat is transferred by the actual movement, of the particles of the medium., (iii) Radiation: The process of transmission heat from one place to another without affecting, the medium is known as radiation., Explanation: When thermal radiation falls on other bodies, it is partly reflected and partly, absorbed. The amount of heat that a body can absorb by radiation depends on the colour of the, body., Law of thermal conductivity: The amount of heat transferred across the ends of the rod (𝑄) is, directly proportional to cross sectional area of the rod (𝐴), directly proportional to temperature, difference (𝑇1 − 𝑇2 ) between the ends, directly proportional to time for which the heat flows (𝑡), and inversely proportional to length of the rod (𝐿), Mathematically: 𝑸 ∝, , 𝑨(𝑻𝟏 −𝑻𝟐 ) 𝐭, 𝑳, , 185

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Q. Mention the properties of Thermal radiations., Ans: Thermal radiations are, 1. Travel through vacuum with speed of light., 2. Do not affect the medium through which they pass., 3. Are electromagnetic in nature., 4. Obey the law of rectilinear propagation., 5. Obey inverse square law of distance., 6. Exhibit the phenomenon of interference, diffraction and polarization., Q. What is a black body?, Ans: A body which completely absorbs the entire heat radiation incident on it is called a perfect, black body. ( or) A body which emit the radiations of all wavelengths when at higher, temperature is called perfect black body., , Q. State Newton’s law of cooling., Ans: The rate of cooling of a hot body is directly proportional to the difference in average, temperature of the body to the temperature of the surroundings provided, temperature, difference is small., , Q. State and explain Newton’s law of cooling., Ans: Newton’s law of cooling: It states that the rate of loss of heat of a hot body is directly, proportional to the temperature difference between the body and the surrounding provided the, temperature difference is small., 𝑑𝑄, , If − 𝑑𝑡 is the rate of heat lost and T1, and T2 are the temperatures of the, surrounding and the body respectively,, then according to Newton’s laws of, cooling:, 𝑑𝑄, , − 𝑑𝑡 𝛼 (𝑇2 − 𝑇1 ), 𝑑𝑄, , − 𝑑𝑡 = 𝑘 (𝑇2 − 𝑇1 ), 𝑑𝑄, 𝑑𝑡, , = −𝑘 (𝑇2 − 𝑇1 ) ............. (1), 186

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𝑑𝑄, 𝑑𝑡, , = −𝑘(𝑇2 − 𝑇1 ), , Where: k= a positive constant depending on area and nature of the surface of the body., Wien's displacement law: The wavelength (𝜆𝑚 ) corresponding to maximum energy emitted, by a black body is inversely proportional to its absolute temperature(T), 1, , 𝜆𝑚 𝛼 𝑇, 𝜆𝑚 𝑇 = 𝑏, Where b= Wien’s constant= 2.9 × 10−3 𝑚 𝐾., , Q. Write a note on Greenhouse effect., Ans: The phenomenon which keeps the earth’s surface warm at night is called greenhouse effect., The radiations from the sun heat up the earth. Due to its lower temperature, the visible light is, absorbed by the earth’s surface and re-radiated as infrared radiations. These radiations cannot pass, through the lower atmosphere. They are trapped by low lying clouds and gases such as carbon, dioxide and water vapour (green house gases). Hence the radiations are reflected back to the, earth’s surface. As a result, the object on the earth got heated up and keeps the earth’s surface, warm at night. Thus, the infrared radiations play an important role in keeping the earth’s warmth, through green house effect., , 188

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Q. Convert 1000C into (i) Fahrenheit (0F) and (ii) Kelvin (K) scale of temperature., Solution:, Given,, (i). Relation between Centigrade scale and Fahrenheit scale:, 𝐹=, , 9𝐶, , 𝐹=, , 9(100), , 5, , + 32, , 5, , + 32, , F = 180+32, F = 2120F, (ii). Relation between Centigrade scale and Kelvin scale:, K = C+273, K= 100+273, K = 373K, , Q. A brass rod of length 50 cm and diameter 3 mm is joined to a steel rod of the same length, and diameter. What is the combined change in length of the combined rod at 250 o C, if, the original lengths are at 40 o C? Is there a thermal stress developed across the junction?, The ends of the rod are free to expand. (α brass = 2 x 10-5 K-1 and α steel = 1.20 x 10-5 K-1)., Solution:, Given,, ∆𝑇 = 𝑇2 − 𝑇1 = 250℃ − 40℃ => ∆𝑇 = 210℃, At 𝑇1 length of brass rod = 50cm, At 𝑇1 diameter of brass rod 𝑑1 =3cm., At 𝑇2 length of steel rod = 50cm, At 𝑇2 diameter of steel rod 𝑑2 =3mm., α brass = 2 x 10-5 K-1 and α steel = 1.20 x 10-5 K-1, Change in length for brass rod ∆𝑙𝑏𝑟𝑎𝑠𝑠 = 𝑙𝑏𝑟𝑎𝑠𝑠 𝛼𝑏𝑟𝑎𝑠𝑠 ∆𝑡, ∆𝑙𝑏𝑟𝑎𝑠𝑠 = 50x2 x 10−5 x 210, 190

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From equations (1) and (2), then we get, KA(T1 −T2 )t, L, , K=, , = mice x Lice, , mice x Lice x L, 𝐴(𝑇1−𝑇2 )𝑡, 0.15x3.35x105 x0.1, , K = (0.1)2 (373−0)(2400), K=, , 0.15x3.35x105 x10−2, 0.1(373)(24), , K=, , 0.5025x103, 895.2, 502.5, , K = 895.2, , K = 0.561Wm-1K-1, , Q. A copper block of mass 2.5 kg is heated in furnace to a temperature of 500°C and then, placed on a large ice block. What is the maximum amount of ice that can melt? Specific, heat of copper = 0.39Jg-1 oC-1 Latent heat of fusion of water = 335 Jg-1, Solution:, Given;, A copper block of mass m=2.5Kg=2500g, Temperature 𝑇 = 500oc, Specific heat of copper C=0.39Jg-1 oc-1, Latent heat of fusion of water L=335jg-1, According to the principle of calorie metry, mc𝜃 = ML, M=, , mc θ, L, , M=, , (2500)(0.39)(500), 335, , M=, , 487500, 335, , M = 1455.22g, 𝑴 = 𝟏. 𝟒𝟓𝒌𝒈, 192

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and K  390Wm 1 K 1, To find:, , Q, t, , Area of the tube is given by:, , A  2 r l  2  3.14  0.1 2  1.256m2, A  1.256m 2, Heat loss per second is given by:, Q KAT1  T2 , , t, d, , Q 390  1.256  373  273, , t, 5  10 3, Q,  9.7968  10 6 Js 1  9796.8  103 js 1  9796.8kJs1, t, , Q. A cylindrical copper rod of diameter 0.04m and length of 0.5m is heated at one end by, steam. The other end is kept at 𝟎𝟎 C .Find the quantity of heat conducted across any, section per minute. (Given: thermal conductivity of copper s=385W𝒎−𝟏 𝑲−𝟏 ), Solution:, Given,, d=0.04m,, r=0.02m,, h=l=0.5m,, t=1min=60sec,, k= 385W𝑚−1 𝐾 −1, 𝑄, 𝑡, , =, , 𝐾𝐴(𝜃1 −𝜃2 ), 𝑙, , A = 2𝜋𝑟ℎ + 2𝜋𝑟 2, A = 2 × 3.142 × 0.02 × 0.5 + 2 × 3.142 × (0.02)2, A = 0.06284 + 2 × 3.142 × 0.0004, A = 0.0653𝑚2 ., , 200

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12. THERMODYNAMICS, , Q. State Zeroth law of thermodynamics., Ans: If two systems are separately in thermal equilibrium with a third system, then they must, be in thermal equilibrium with each other., , Q. What is the significance zeroth law of thermodynamics?, Ans: Concept of temperature, , Q. What are extensive thermos dynamical variables? Give an example., Ans: The thermodynamic variable which indicates the size of the system is known as an, extensive thermodynamic variable., , Q. State and explain the first law of thermodynamics., Ans: First law of thermodynamics: It states that the heat given to a system is equal to the sum, of the increase in internal energy and external work done by the system., If dQ is the amount of heat given to a system, dU is the increase in the internal energy of a, system and dW is the external work done by the system , then according to the first law of, thermodynamics, dQ = dU + dW, Q. Give the mathematical form of first law of thermodynamics., Ans:- dQ = dU + dW, dQ is the amount of heat given to a system ,dU is the increase in the internal energy of a, system and dW is the external work done by the system., Q. What is the significance of first law of thermodynamics?, Ans: (1) This law introduces the concept of internal energy., (2) This law is applicable to all the three phases of matter (i.e., solid, liquid and gas), Q. Write the physical significance of first law of thermodynamics., 207

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Ans: The first law of thermodynamics is an illustration of law of conservation of energy., Q. What is Isothermal and adiabatic process., Ans: Isothermal process: It is a thermodynamic process in which the pressure and volume of, the system change at constant temperature., Adiabatic process : It is a thermodynamic process in which the pressure, volume and, temperature of the system change but there is no exchange of heat between the system, and its surroundings., Q. Draw P – V diagram for adiabatic process., Ans:, , Q. In which process, pressure of the system remains unchanged?, Ans: Isobaric., (When the pressure of a system remains constant during a thermodynamic process, the process, is called isobaric.), Q. What is isothermal process? Derive an expression for work done in an isothermal process., Ans: A thermodynamic process in which the pressure and volume of the system change at, constant temperature is called isothermal process., Derivation for an expression for work done in isothermal process: Consider 𝜇 moles of an, ideal gas of volume 𝑉, and pressure 𝑃 enclosed in a perfectly non conducting cylinder of conducting base provided, with a, movable frictionless piston of cross sectional area A., When the gas is allowed to expand, the piston moves through a small distance dx., The small work done by the gas is given by, 𝑑𝑊 = 𝑃𝐴𝑑𝑥 𝑑𝑊 = 𝑃𝑑𝑉-------(1), , ∵ 𝐴𝑑𝑥 = 𝑑𝑉 = increase in the volume of gas, , For an isothermal process of an ideal gas of 𝜇 moles:, 𝑃𝑉 = 𝜇𝑅𝑇, 208

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Step II: When heat is supplied to the gas at constant pressure , the gas expands and pushes the, piston through a distance dx .The work done by the gas to increase its volume by dV against the, external pressure P is:, dW=P dV ------------ (4), The amount of heat supplied d𝑄2 is used to do the work dW and to increase the internal energy, by dU. Due to increase in internal energy , the temperature of the gas increases by dT., d𝑄1 = 𝑑𝑊 + 𝑑𝑈, d𝑄1 = 𝑃𝑑𝑉 + 𝐶𝑉 𝑑𝑇 ------- (5), Step III: By the definition of molar specific heat capacity of a gas constant pressure (𝐶𝑃 ),, d𝑄1 = 𝐶𝑃 𝑑𝑇--------- (6), Step IV: Ideal gas equation is:, PV=RT, , Where: R = Universal gas constant., , Differentiating the equation,, PdV = RdT, , ------------------- (7), , From equation (5) , (6) and (7):, 𝐶𝑃 𝑑𝑇 = 𝑃𝑑𝑉 + 𝐶𝑉 𝑑𝑇, 𝐶𝑃 𝑑𝑇 = 𝑅𝑑𝑉 + 𝐶𝑉 𝑑𝑇, 𝐶𝑃 = 𝑅 + 𝐶𝑉, 𝑪𝑷 − 𝑪𝑽 = 𝑹, Q. What is refrigeration?, Ans: The term refrigeration means cooling a space, substance or system to lower and maintain, its temperature below the ambient one., , Q. State second law of Thermodynamics as given by Clausius., Ans: It state that heat cannot flow from a cold body to a hot body without the performance of, work by some external agency., , Q. Give Clausius statement for second law of thermodynamics., , 210

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Ans: It state that heat cannot flow from a cold body to a hot body without the performance of, work by some external agency., , Q. What is the efficiency of Carnot’s engine ,when T1 = T2 ?, 𝑻, , Ans: 𝜼 = 𝟏 − 𝑻𝟐, 𝟏, , ∵T1 = T2, 𝜼=𝟏, The efficiency of Carnot engine is 100%, , Q. Explain the working principle of Carnot’s cycle with necessary P-V diagram., Ans:, , Working principle of Carnot’s cycle: The cyclic change of the working substance, when the, engine starts working is known as Carnot cycle ., First stroke (Isothermal expansion): The cylinder containing working substance (ideal gas), at pressure 𝑃1 and volume 𝑉1 is placed on the surface of temperature 𝑇1 . The working, substance attains the temperature 𝑇1 . This initial state is represented by a point A in the PV, diagram .The pressure on the piston is allowed to decrease and the volume of the system, increases. Since the base of the cylinder is perfectly conducting, the gas absorbs heat. The, temperature increased due to absorption of heat is decreased due to expansion. Therefore, temperature of the system 𝑇1 remains constant as the gas is allowed to expand isothermally., Thus the process is an isothermal expansion. The process continues till the point B so that the, pressure and volume of the system become 𝑃2 and 𝑉2 . This isothermal expansion is indicated, by the curve AB in the diagram., Heat absorbed by the gas from the reservoir at temperature 𝑇1 (work done by the gas), 211

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is given by :, 𝑾𝟏 = 𝑸𝟏 = 𝝁𝑹𝑻𝟏 𝐥𝐨𝐠 𝒆, , 𝑽𝟐, 𝑽𝟏, , ---------------------(1), , This work is done by the gas (system), so it is positive., Second stroke (Adiabatic expansion): The cylinder is removed from the source and is placed, on the insulating stand so that it is completely insulated from the surroundings. The pressure on, the piston decreases and the volume of the system increases. Since there is no exchange of heat, between the system and the surroundings, the temperature of the system decreases during, expansion. The process is allowed to expand adiabatically till its temperature decreases from, 𝑇1 to 𝑇2 . This process is an adiabatic expansion. The process continues till the point C so that, the pressure and the volume of the system becomes 𝑃3 and 𝑉3 . This adiabatic expansion is, indicated by the curve BC in the diagram., Work done in the adiabatic expansion is given by:, 𝑾𝟐 =, , 𝝁𝑹, [𝑻 − 𝑻𝟐 ], 𝜸−𝟏 𝟏, , --------------- (2), , This work is done by the gas (system) , so it is positive., Third stroke (Isothermal compression): The cylinder is placed on the sink of temperature 𝑇2 ., The pressure on the piston increases and the volume decreases. Due to this, the gas is, compressed and the temperature increases. Since base of the cylinder is conducting, there is a, transfer of heat from the gas to the sink. The heat rejected by the gas due to compression, flows, to the sink to keep its temperature 𝑇2 constant therefore, temperature of thegas remains, constant. Thus the process is an isothermal compression. The process continues till the point D, so that the pressure and volume of the system becomes 𝑃4 and 𝑉4 . This isothermal comression is, indicated by the curve CD in the diagram., Heat released by the gas from the reservoir at temperature 𝑇2 (work done by the gas), is given by :, 𝑾𝟑 = 𝑸𝟐 = 𝝁𝑹𝑻𝟐 𝐥𝐨𝐠 𝒆, , 𝑽𝟑, 𝑽𝟒, , ------------- (3), , This work is done by the gas (system) , so it is negative., Fourth stroke (Adiabatic compression): Now the cylinder is placed on the insulating stand., The pressure on the piston is further increased and the gas is compressed till it attains the, initial temperature 𝑇1 to return to its initial state A. thus the process is an adiabatic, compression. This adiabatic compression is represented by the curve DA in the diagram., 212

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13. KINETIC THEORY OF GASES, Q. State any three assumptions of Kinetic theory of gases., Ans: The kinetic theory of gases is based on the following basic assumptions., 1. A gas consists of a large number of molecules. These molecules are identical in all respects, and perfectly elastic hard spheres., 2. The molecules are so small that the volume of a molecule is negligible in comparison with, the volume of the gas., 3. The gaseous molecules are in a state of continuous random motion with all possible, velocities., , Q. What is an ideal gas?, Ans: A gas which obeys all gas laws at all temperatures and pressures is called an ideal gas., , Q. Write equation of state of perfect gas., Ans:𝑃𝑉 = RT, Q. Deduce an expression for ideal gas of ‘n’ number of molecules., 1, , Ans: PV = 3 mv2rms, For a given mass of a gas PV αv2rms, According to kinetic interpretation of temperature, v2rmsα T, It follows from above discussion that., PV αT, It can be written as, PV = RT (∵ for 1 mole of an ideal gas equation), Where R=Universal gas constant., For n number of moles, above equation becomes, PV=nRT, , 221

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Q. What is law of Equiportition of energy? Write the expression for it., Ans: Equiportition of energy: It states that for any system in thermal equilibrium, the total, energy of the system is equally distributed among its various degrees of freedom and energy, associated with each degree of freedom of a molecule is 𝐸 =, , 1, 2, , 𝑘𝐵 𝑇., , Q. State the law of Equiportition of energy. Give the expression for the average energy of a, Mono- atomic and diatomic molecule., Ans: It states that in thermal equilibrium, the total energy of a particle is equally distributing, possible modes of energy, with each mode having an average energy equal to 𝐸 =, , 1, 2, , 𝑘𝐵 𝑇., , Where 𝐾𝐵 a constant is called Boltzmann’s constant and T is the absolute temperature of, the gas., Total internal energy of one mole of a gas having f degrees of freedom is:, 𝑈 = 𝑁𝐴 𝑓 × 1⁄2 𝐾𝐵 𝑇, 1, , 𝑈 = 𝑓 × 2 (𝑁𝐴 𝐾𝐵 )𝑇, 1, , 𝑈 = 𝑓 × 2 𝑅𝑇, , ∵ 𝑁𝐴 𝐾𝐵 = 𝑅, , Mono- atomic gas molecule: A mono – atomic gas molecule contains only one atom. As the, mono- atomic gas molecule behaves like a point mass, its moment of inertia about its axis of, 1, , 1, , 1, , rotation is zero. 𝑈 = 2 𝑚𝑉𝑥2 + 2 𝑚𝑉𝑦2 + 2 𝑚𝑉𝑍2 ……….(1), Therefore, a mono- atomic gas molecule can acquire energy in three ways: along x-axis,, along y-axis and z- axis. Hence, a mono-atomic gas molecule has three degrees of freedom., Diatomic gas molecule: A diatomic atomic gas molecule contains two atoms. The molecule, is capable of translational as well as rotational motion. If the line joining the two atoms is, taken as X-axis , then the molecules can either rotate about Y and Z -axis and not about X𝟏, , 𝟏, , 𝟏, , 𝟏, , 𝟏, , 𝟐, , 𝟐, , 𝟐, , 𝟐, , 𝟐, , axis . 𝑼 = 𝒎𝒗𝟐𝒙 + 𝒎𝒗𝟐𝒚 + 𝒎𝒗𝟐𝒛 + 𝑰𝒚 𝝎𝟐𝒚 + 𝑰𝒛 𝝎𝟐𝒛 ………..(2), Therefore, a diatomic gas molecule can acquire energy in three ways due to translational, motion and two ways due to rotational motion. Hence, the molecule has three degrees of, freedom due to translational motion and two degrees of freedom due to rotational motion., Thus, a diatomic gas molecule has in all five degrees of freedom., , Q. How many degrees of freedom for a diatomic gas molecule?, , 223

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Ans: The molecule of a diatomic gas has only five degrees of freedom (f = 5), Q. How many degrees of freedom are there for a monatomic gas?, Ans: Three [3], , Q. Mention the degrees of freedom for a linear tri-atomic gas molecule., Ans: 7 (Seven), , 𝑪, , Q. Calculate: 𝑪𝒑 for monatomic gas., 𝒗, , 3, , Ans: The molar specific heat capacity of gas at constant volume is: 𝐶𝑉 = 2 𝑅, 5, , The molar specific heat capacity of gas at constant pressure is: 𝐶𝑃 = 2 𝑅, 𝐶, , The ratio of specific heat capacities is given by: 𝛾 = 𝐶𝑃, , 𝑉, , 5, , 𝛾 = 23, 2, , 𝑅, 𝑅, , 5, , 𝛾 = 3., , Q. Define mean free path., Ans: The average distance covered by a molecule between two successive collisions, is called, mean free path., Mean free path=, , 𝒌𝑩 𝑻, √𝟐𝝅𝒅𝟐 𝑷, , Q. Define mean free path of molecule of gas. Write its expression. Explain the terms., Ans: The average distance covered by a molecule between two successive collisions, is called, mean free path., Mean free path=, , 𝒌𝑩 𝑻, √𝟐𝝅𝒅𝟐 𝑷, , Where: 𝑘𝐵 is Boltzmann’s constant, 𝑇is absolute temperature, 𝑃is pressure of the gas, 224

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𝑑 2 is square of the diameter of the gas molecule, , Q. Obtain the expression for the Mean free path of a gas molecule., Ans: The assumptions made to derive the expression for mean free path are:, 1. The molecules of the gas are considered as hard spheres, each of diameter d. The collisions, between two molecules are perfectly elastic., 2. All the other molecules except the molecule under consideration are at rest., 3. A molecule of the gas under consideration collides with all those molecules whose center’s, are at distance d from the Centre of the molecule under consideration., 4. Consider a gas in a container, having n molecules per unit volume. Let d be the diameter of, each molecule. Consider molecule A moving and all other molecules at rest. The molecule, A collides with all other molecules like B and C whose Centers are at distance d from its, Centre. If the molecule A moves a distance L, then it collides with all the molecules within, the cylinder of volume 𝜋 𝑑 2 𝐿., The number of molecules in a cylinder of volume:, 𝜋𝑑 2 𝐿 = 𝑛𝜋𝑑 2 𝐿, Number of collisions suffered by the molecule A= number of molecules in a cylinder., 𝑁𝐴 = 𝑛𝜋𝑑 2 𝐿, The mean free path of the moving molecule A is:, 𝐿, , 𝛌 = 𝑛𝜋𝑑2 𝐿, 1, , 𝛌 = 𝑛𝜋𝑑2 ........... (1), If the motion of all the molecules is taken into, account, then number of collisions is more, The expression (1) for mean free path can be written, 𝛌=, , 1, √2𝑛𝜋𝑑2, , ........... (2), , Number of molecules per unit volume =, 𝑛=, , as:, , Avogadro’s number, 𝑣𝑜𝑙𝑢𝑚𝑒𝑜𝑓𝑡ℎ𝑒𝑔𝑎𝑠, , 𝑁𝐴, 𝑉, , Expression (2) becomes, λ=, λ=, λ=, , 1, 𝑁, √2𝜋𝑑2 ( 𝑉𝐴 ), , 𝑉, √2𝜋𝑑2 𝑁𝐴, 𝑃𝑉, √2𝜋𝑑2 𝑁𝐴 𝑃, , 225

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λ=, λ=, , RT, √2𝜋𝑑2 𝑁𝐴 𝑃, , (since PV = RT), , 𝑅, )𝑇, 𝑁𝐴, √2𝜋𝑑2 𝑃, , (, , (𝒌 )𝑻, , 𝑅, , 𝑩, ′, 𝛌 = √𝟐 𝝅𝒅, 𝟐 𝑷 (𝑠𝑖𝑛𝑐𝑒 𝑁 = 𝑘𝐵 = 𝑏𝑜𝑙𝑡𝑧𝑚𝑎𝑛𝑛 𝑠𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡) .......... (3), 𝐴, , From this expression it follows that, the mean free path (λ)is :, 1. Directly proportional to absolute temperature of the gas (T), 2. Inversely proportional to pressure of the gas (P), 3. Inversely proportional to square of the diameter of the gas molecule., , Q. Define rms speed of gas molecules. Obtain the expression for rms speed of gas molecule, on the basis of kinetic interpretation of temperature., Ans: Root mean square speed or velocity is defined as the square root of the mean of the, squares of the velocities of the individual molecules of the gas., 2, Expression: 𝑣𝑟𝑚𝑠, =, , 𝑣𝑟𝑚𝑠 = √, , 2, 𝑣12+𝑣22+𝑣32 +⋯+𝑣𝑛, , 𝑛, , 2, 𝑣12+𝑣22 +𝑣32+⋯+𝑣𝑛, , 𝑛, , Where: 𝑣1 , 𝑣2 , 𝑣3 … … . 𝑣𝑛 are velocities of 𝑛 molecules, , Q. State the Dalton’s law of partial pressure., Ans: The total pressure exerted by a mixture of non-reacting gases occupying a vessel is equal, to the sum of the individual pressures which each gas would exert if it alone occupied the, whole vessel., i.e. 𝑃 = 𝑃1 + 𝑃2 +……………+𝑃𝑛, , Q. Mention the factor on which internal energy of an ideal gas depends., Ans: Internal energy of an ideal gas depends only on temperature., , Q. What is an Ideal gas? Write the Ideal gas equation and give the value constant in the, equation with S.I. unit., , 226

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Ans: A gas, which obeys gas laws (Boyle’s law and Charles’s law) at all temperatures and, pressure, is called a perfect gas or an ideal gas., Ideal gas or perfect gas equation:, 𝑃𝑉 = 𝑛𝑅𝑇 or 𝑃𝑉 = µ𝑅𝑇, Where : 𝑃 = 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 , 𝑉 = 𝑣𝑜𝑙𝑢𝑚𝑒 , 𝑇 = 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 , 𝑅 = 𝑢𝑛𝑖𝑣𝑒𝑟𝑠𝑎𝑙 𝑔𝑎𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡, and 𝑛 𝑜𝑟 µ = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠, 𝑹 = 𝟖. 𝟑𝟏 𝑱 𝒎𝒐𝒍𝒆−𝟏 𝑲−𝟏, , Q. What is an ideal gas? Write ideal gas equation. At what condition real gases behaves as, ideal gas?, Ans: Ideal gas: A gas, which obeys gas laws ( Boyle’s law and Charles’s law ) at all, temperatures and pressure is called a perfect gas or an ideal gas ., Ideal gas or perfect gas equation:, 𝑃𝑉 = 𝑛𝑅𝑇 (or) 𝑃𝑉 = µ𝑅𝑇, Where: P = pressure, V = volume , T = temperature , R = universal gas constant, And n or µ = number of moles., Condition for real gases behaves as ideal gas:, Real gases behave as ideal gases at low pressure and high temperature., , Q. Define rms speed of molecules of a gas. How it is related to the kinetic energy of the gas?, Ans: Root mean square speed: It is defined as the square root of the mean of the squares of, speeds of individual molecules of a gas., 1, , 2, ∵ KE = 2 mVrms, 2, Vrms, =, , 2 KE, , Vrms = √, , m, 2KE, m, , 227

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Q. Determine the molar specific heat at constant volume for a mono atomic gas molecule., Ans: The total internal energy of one mole of a gas having ‘f’ degrees of freedom is given by:, 1, , U = NA f × K B T, 2, , 1, , U = f × 2 ( NA K B ) T, 1, , U = f × 2 RT……..(i), , ∵ NA K B = R, , For mono atomic gas f=3, then for one mole of mono atomic gas:, 1, , U = f × 2 RT, 1, , U = 3 × RT, 2, , 3, , U = 2 RT, The molar specific heat capacity of gas at constant volume is:, CV =, , CV =, , du, dt, , 3, 2, , d( RT), dT, 𝟑, , 𝐂𝐕 = 𝟐 𝐑, Q. Using kinetic theory of gases derives an expression for finding the pressure of an ideal gas, in terms of mean squared speeds of the molecules., Ans: Expression for the pressure of an ideal gas: Consider an ideal gas enclosed in a cubical, box of length 𝑙 . Let there be 𝑛 number of identical gas molecules each of mass. Consider one, such molecule moving with a velocity 𝑣1 ., The component of 𝑣1 along the three mutually Perpendicular axes X, Y and Z are, 𝑣1𝑥 , 𝑣1𝑦 and 𝑣1𝑧 respectively., It can show that:, 2, 2, 2, 𝑣12 = 𝑣1𝑥, + 𝑣1𝑦, + 𝑣1𝑧, ………………………… (1), , The momentum of the molecule along X -axis is:, 𝑝1 = 𝑚𝑣1𝑥 …………………………………….. (2), When this molecule strikes the face , then it rebounds with the same speed towards the, face 𝐵 ., 228

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14. OSCILLATIONS, , Q. What is periodic motion? Give an example., Ans: A motion which repeats itself after equal intervals of time is called periodic motion., Ex:- Motion of earth around the sun., Q. Define periodic motion and Write expression for acceleration of particle in SHM., Ans: If the motion of a body is repeated in regular intervals of time, then the motion is called, periodic motion., Acceleration of particle in SHM is 𝑎 = −𝜔2 𝑦, , Q. Define frequency and period of oscillation., Ans: Frequency: The number of oscillations completed by the particle in one second., Time period: The time taken by the particle to complete one oscillation., , Q. Write the relation between time period and frequency of oscillation., 1, , Ans: Frequency (f) is equal to reciprocal of time period (T) i.e 𝑓 = 𝑇, , Q. A particle takes 32s to make 20 oscillations. Calculate time period and frequency., Ans:T =, , Total time taken, No.of oscillations, , T=, , 32, 20, , T = 1.6sec, Frequency: f =, , 1, T, , =, , 1, 1.6, , = 0.625 Hz, , Q. Find the period of sound wave whose frequency is 20 Hz., 1, , 1, , Ans: T = 𝑓 = 20 = 0.05𝑠𝑒𝑐, , 232

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Q. The function are y=𝐥𝐨𝐠(𝝎𝒕). The displacement y increases monotonically with time t, is it, periodic function or non periodic function?, Ans: Non periodic function., , Q. What is meant by phase of a oscillating particle?, Ans: The phase of a vibrating particle at any instant is the quantity that gives the position and, direction of motion of the particle at that instant., , Q. What is SHM? Write its characteristics and give its graphical representation., Ans: SHM: A particle is said to be in simple harmonic motion if it moves to and fro about to, and fro about its mean position under the action of restoring force which is directly, proportional to its displacement from the mean position and is always directed towards the, mean position., , Characteristics of SHM:, 1. SHM is periodic and oscillatory, 2. Force as well as acceleration are proportional to displacement., 3. Force as well as acceleration are always directed towards the mean position and opposite, to the direction of displacement., 4. Time period and frequency are independent of amplitude and they depend only on the, force constant and mass of the oscillating body., Q. Write the equation for SHM and explain the symbols used., Ans: 𝑦 = 𝐴 sin 𝜔𝑡, Where: y is displacement,, A is Amplitude,, 𝜔 is the angular velocity and, t is time interval., 233

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Q. What are the conditions of motion of a particle executing S.H.M?, Ans: Conditions for particle executing S.H.M: The restoring torque (or) Angular, acceleration acting on the particle should always be proportional to the angular displacement of, the particle and directed towards the equilibrium position., Τα–θ, , ( Or), , θ α– Τ, , Where: Τ – Torque,, , α – Angular acceleration,, , θ – Angular displacement, , Or,, Conditions for particle executing S.H.M: The restoring force acting on the particle, moving to, and fro about mean position, is directly proportional to the displacement and acts in a direction, opposite to displacement., , Mathematically,, 𝑭 ∝ −𝒚, 𝑭 = −𝒌𝒚, , Where, ‘F’ is the restoring force and ‘y’ is the displacement of the particle from the mean, position. Negative sign indicates that force in opposite direction to the displacement and ‘k’ is, the constant of proportionality called force constant or spring constant, Q. Draw a graph of displacement versus time for a particle executing SHM., Ans:, , Q. Draw displacement energy curve of a particle executing simple harmonic motion., Ans:, , 234

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Q. Obtain the force law for simple harmonic motion., Ans: It states that the restoring force is directly proportional to displacement of the particle, from its mean position., , Q. Write the expression for velocity and acceleration in case of simple harmonic motion with, usual notation., Ans: Velocity of a particle executing SHM: The rate of change of displacement of the, particle is called velocity of the particle., Differentiating both sides of equation w.r.t time:, , 𝑑𝑦, 𝑑𝑡, , =, , 𝑑, 𝑑𝑡, , (𝐴 sin 𝜔𝑡 ) → 𝑣 = 𝐴𝜔 cos 𝜔𝑡, , This equation gives velocity of the particle executing SHM., Acceleration of a particle executing SHM: The rate of change of velocity of particle is, called acceleration of the particle ., Differentiating both sides of equation wrt t:, , 𝑑𝑣, 𝑑𝑡, , =, , 𝑑, 𝑑𝑡, , (𝐴𝜔 cos 𝜔𝑡) → 𝑎 = −𝜔2 𝐴 sin 𝜔𝑡, , This equation gives acceleration of the particle executing SHM., Q. Mention the expression for velocity of a particle executing SHM. Explain the symbols., Ans: 𝑉 = 𝐴𝜔𝑐𝑜𝑠𝜔𝑡, Where: ‘V’ is the velocity at any instant, ‘A’ is the amplitude,, ‘ 𝜔′ is the angular frequency and, ‘t’ is the time., (or), 𝑽 = 𝝎√𝑨𝟐 − 𝒚𝟐, Where: ‘V’ is the velocity at any instant, ‘y’ is the displacement from the mean position, ‘A’ is the amplitude, ‘ 𝜔′ is the angular frequency, Q. Derive expressions for the velocity and acceleration of a particle executing simple, harmonic motion., Ans: Velocity of a particle executing SHM:, 235

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The rate of change of displacement of the particle is called Velocity of the particle., The displacement of the particle executing SHM is 𝑦 = 𝐴 sin(𝜔𝑡), Differentiating both sides of above equation w.r.t is:, dy, dt, , d, , = dt (Asinωt), , V = Aωcosωt, v = Aω√1 − sin2 ωt, v = Aω√1 −, v = Aω√, , y2, A2, , ∵, , y, A, , = sinωt, , A2 −y2, A2, , 𝐕 = 𝐀𝛚√𝐀𝟐 − 𝐲 𝟐, , Accelerating of a particle executing SHM:The rate of change of velocity of a particle is celebration., Since 𝑉 = 𝐴𝜔𝑐𝑜𝑠𝜔𝑡, Differentiating both sides of equation wrt ‘t’ is, dv, dt, , d, , = dt (Aωcosωt), , a = −ω2 A sinωt, 𝐚 = −𝛚𝟐 𝐲, Q. What are free oscillations and forced oscillations?, Ans: Free oscillations: The oscillations made by a body which are not subjected to any, external force are called free oscillations, Forced oscillations: When an external periodic force is applied, the body begins to oscillate, with the frequency equal to that of the external periodic force. These oscillations are called, forced oscillations., Q. What are damped oscillations? Give an example., Ans: The oscillations whose amplitude decreases with time are called damped oscillations., Example of damped oscillations: clock pendulum (or) guitar string, , Q. Derive the expressions for kinetic energy, potential energy and total energy of a simple, Harmonic oscillator., Ans: Energy of a executing SHM: A particle performing simple harmonic motion is acted, upon by a restoring force which is always directed towards the mean position. The restoring, 236

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force always opposes the displacement of the particle from the mean position. To displace the, particle work must be done against the restoring force. This work done is converted into, potential energy in the particle. The particle possesses kinetic energy due to its motion. The, sum of kinetic energy and potential energy gives the total energy of the particle., Consider a particle of mass m executing SHM with amplitude A & constant angular, , frequency 𝜔 . Let y be the displacement of the particle at any time t and k be the force, constant., The restoring force F acting on the particle is given by:, F = −ky ……………. (1), Negative sign implies that restoring force F and displacement Y are oppositely directed. To, displace the particle work must be done., The small work done dw in displacing the particle through a small displacement dy is given by:, dw = −F. dy, 𝑑𝑤 = −(−ky)dy, 𝑑𝑤 = 𝑘𝑦𝑑𝑦 …………….. (2), Total work done to displace the particle y from its mean position is given by:, y, , ∫ dw = ∫0 kydy, y, , y2, [, w=k ], 2 0, 1, , w = 2 k[y2 − 0], 1, , k, , w = 2 ky2 ∵ ω2 = m → k = mω2, 1, , w = 2 mω2 y2, This work is equal to potential energy., 1, , U = 2 mω2 y2 ………………… (3), This gives the expression for potential energy of the particle executing simple harmonic, motion., Displacement of the particle executing SHM is given by:, y = A sin ωt …………………..(4), Differentiating both sides of equation (4) w.r.t. t:, 237

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‘y’ is the displacement and, ‘ω’ is the angular frequency., , Q. Where is the velocity of the body maximum and minimum in the case of simple harmonic, motion?, Ans: (i) Maximum: Mean position or equilibrium position., (ii) Minimum: Extreme position, , Q. Where is the potential energy of a body maximum and minimum?, Ans: (i) Potential energy of a body maximum at extreme position., (ii) Potential energy of a body minimum at mean position., , Q. At what positions kinetic energy of simple harmonic vibrator becomes maximum and, minimum?, Ans: (i) Kinetic energy of simple harmonic vibrator is maximum at mean position., (ii) Kinetic energy of simple harmonic vibrator is minimum at extreme position., , Q. Derive an expression for time period of simple pendulum., Ans: Simple pendulum: A heavy point mass suspended by a weightless, inextensible and, perfectly flexible string fixed rigidly to a support is called simple pendulum., Consider a simple pendulum which consists of a small bob of mass m tied to a light string of, length 𝑙. The other end of the string is tied to a rigid support. When the bob is displaced, slightly through an angle 𝜃 from the mean position O and then released, it starts oscillating, simple harmonically about O. The forces acting on the bob, are: weight of the bob mg acting vertically downward and, tension (T) acting along the string towards the point of, suspension S. The weight mg is resolved into two, components mgcos 𝜃 and mgsin 𝜃 as shown in the figure., The component mgcos 𝜃 is balanced by the tension Ts. The, unbalanced component, , mgsin 𝜃 acts as restoring force, , 239

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which tends to bring the bob back to the mean position., The restoring force F acting on the bob is given by:, F=-mgsin 𝜃------------- (1), Negative sign shows that force and displacement are oppositely directed., Since 𝜃 is very small, sin 𝜃= 𝜃, Equation (1) becomes, F = -mg 𝜃 ------------------ (2), From Newton’s second law of motion:, F = ma, , ------------------- (3), , Where: a = acceleration of the bob, From equation (2) and (3), ma = − mg 𝜃, a = -g 𝜃 ----------------- (4), From the diagram:, 𝜃=, , 𝑦, 𝑙, 𝑦, , 𝑎 = −𝑔 𝑙 -------------- (5), As, , 𝑔, 𝑙, , is constant, a α – y, , This shows that acceleration of the bob of pendulum is directly proportional to its displacement, and the direction of acceleration is opposite to the direction of motion of the displacement., Thus motion of the bob is simple harmonic., The acceleration of the body executing SHM is given by:, a = −𝜔2 𝑦 --------------- (6), From equation (5) and (6), 𝜔2 =, , 𝑔, 𝑙, , 𝜔=√, , 𝑔, 𝑙, , 2𝜋, 𝑔, =√, 𝑇, 𝑙, 240

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1, 1 𝑔, √, =, 𝑇 2𝜋 𝑙, 𝑙, , 𝑇 = 2𝜋√𝑔, This gives the expression for time period of oscillation of simple pendulum., , Q. What happens to the time period of a simple pendulum, if its length is doubled?, Ans: Increases (or) Time period becomes: T   2  T ., , Q. What happens to the time period of a simple pendulum, when it is taken to moon?, Ans: Time period increases., , Q. How does the period of oscillation of a pendulum depend mass of the bob and length of, the pendulum., Ans: Mass does not affect the pendulum's swing. The longer the length of string, the farther, the pendulum falls; and therefore, the longer the period, or back and forth swing of, the pendulum. The greater the amplitude, or angle, the farther the pendulum falls; and, therefore, the longer the period., , Q. Obtain an expression for period of oscillations of a load attached to a spring., Ans: Consider a spring of constant k. One end of the spring is attached to a rigid support and, the other to a block of mass m. The block is free to move on a frictionless horizontal surface., If the block is slightly displaced, the spring gets elongated and a restoring force is produced in, the spring. When the block is released, it executes SHM about the mean position., If the block is displaced towards right through a small, displacement y, the restoring force produced in the, spring is given by:, F = - k y ------------ (1), Negative sign shows that F and y are oppositely directed., From Newton’s second law:, F = m a ------------ (2), Where a = instantaneous acceleration of the block., 241

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NUMERICALS, Q. A particle takes 32 s to make 20 oscillations. Calculate time period and frequency., Solution:, Given,, 32, , Time period: (T) is 20 = 1.6 𝑠, Frequency: (f) = (1/T) =1/1.6 =0.625 Hz, Q. On a average a human heart is found to beat 75 times in a minute. Calculate its frequency, and time period., Solution:Time period (T) is, , 60, 72, , = 0.833 𝑠, , 72, , Frequency (f) is 60 = 1.2 𝐻𝑧, Q. A particle executes simple harmonic motion given by: y  0.24 sin 400t  0.5 in SI unit., Find: (i) amplitude, (ii) Frequency, (iii) time period of vibration and (iv) initial phase., Solution:, y  0.24 sin 400t  0.5 ………………………………... (1), , , , y  A sin  t  0, ,  ………………………………………... (2), , On comparing like terms in equations (1) and (2):, , A  0.24m, ,   400rad s 1, , 0  0.5rad, i. Amplitude:, , A  0.24m ., ii. Frequency:, Angular frequency is given by:   2 f, , , 2, 400, f , 2  3.14, 400, f , 6.28, , f , , f  63.69 Hz, , 243

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iii. Time period of vibration:, Frequency of vibration is given by: f , , T, , 1, f, , T, , 1, 63.69, , 1, T, , T  0.0157s, iv. Initial phase is given by:, , 0  0.5rad ., , Q. x = 0.02 sin (πt) describes the motion of a body executing SHM. Calculate the period and, amplitude of the motion. What is the displacement at t = 0.1s?, Solution:, Given,, x = 0.02 sin(πt), The standard wave equation is x = A sin 𝟂t, (i) Amplitude: A = 0.02m, (ii) Time period: 𝜔 = 𝜋, 2𝜋, 𝑇, , =𝜋, , T = 2sec, (iii) Displacement: x = 0.02 sin (π) (0.1), x = 0.02 sin (0.1 π), x = 0.02 sin (0.314), x = 0.02 (0.054), x = 1x10-4 m, , 244

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Q. A spring balance has a scale that read from 0 to 50kg. The length of the scale is 20 cm. A, body suspended from this balance, when displaced and released, oscillates with a period, of 0.60s. What is the weight of the body?, Solution:, Given,, Length of the scale = 20 cm = 0.2 m, Total scale reading = 50 kg, 50, , Scale reading per unit length of the scale = 0.2 = 250 𝑘𝑔𝑚−1, Period of oscillation T= 0.6 s, T2 g, , 𝑙, , We have 𝑇 = 2π√g or 𝑙 = 4π2, ∴𝑙=, , 0.62 ×9.8, 4π2, , = 0.089 m., , Reading on the scale corresponding to the length:, 𝑙 = 0.089m = 250 × 0.089 = 22.34kg, Weight of the body= 22.34 kg wt, W=22.34 x 9.8N, W =219 N., , Q. A particle executes SHM along X-axis. It’s displacement varies with time according to the, 𝝅, , equation 𝒙(𝒕) = 𝟐. 𝟓 𝐜𝐨𝐬(𝟒𝝅𝒕 + ), Where x(t) in metre and ‘t’ is in second. Determine, 𝟔, , the (i) amplitude, (ii) frequency, (iii) period and (iv) phase constant of the motion., Solution:, Given,, 𝝅, , 𝒙(𝒕) = 𝟐. 𝟓 𝐜𝐨𝐬(𝟒𝝅𝒕 + ),, 𝟔, , (i) Amplitude: (A)= 2.5 m, (ii) Frequency: 𝜔 = 2𝜋𝑓 = 4𝜋 = 𝑓 = 2 𝐻𝑧, (iii) Time period: (T) =(1/f) =1/2 = 0.5 s, (iv) Phase constant: ∅ =, , 𝜋, 6, , rad, , 245

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15. WAVES, Q. What are mechanical waves? Give an example for mechanical wave., Ans: Mechanical waves are which require a material medium for their propagation., Ex:- Sound waves, water waves, waves on the string etc., Q. Write any three characteristics of a progressive wave., Ans: Characteristics of a progressive wave: (any three), (i) A progressive wave is formed due to continuous vibrations of the particles of the, medium., (ii) A progressive wave travels with a certain velocity., (iii) Three is a flow of energy in the direction of the wave., (iv) No particle of the medium is at rest., (v) Amplitude of all the particles is same., (vi) Phase changes continuously from particle to particle., Q. Define (i) Wave amplitude, Ans : (i) Wave amplitude(A) : It is the maximum displacement of a particle of the medium, from its mean position ., , Q. At a node of stationary wave, what is the value of amplitude?, Ans: Amplitude is minimum., , Q. Define the terms wave amplitude, wave period and wave length of a wave., Ans: (i) Wave Amplitude (A): It is the maximum displacement of a particle of the medium, from its mean position., (ii) Wave period (T): It is the time taken by a particle of the medium to complete one, vibration., (iii) Wave length: It is the distance covered by the wave in one time period or It is the, shortest distance between the two particles of the medium in the same, phase. (or) It is the distance between two consecutive crests or troughs, (or) It is the distance between the centers of two consecutive, compressions or rarefactions., 246

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Q. The distance between two successive nodes is 2m. Calculate the wavelength of wave., 𝜆, , Ans: The distance between successive nodes or the distance between successive antinodes = 2., 𝜆, 2, , = 2 ⇒ 𝜆 = 4𝑚., , Q. Calculate the frequency of sound waves whose time period is 0.5s., 1, , Ans: Frequency = time period, 1, , f=T, 1, , f = 0.5, 𝐟 = 𝟐 𝐇𝐳, , Q. What is a stationary wave? Give an example., Ans: When two progressive waves of the same amplitude, frequency and speed travel in, opposite directions along a straight line superposed each other, the resultant wave does not, travel in either direction and is called stationary wave., Example: 1. Waves produced in air column of an organ pipe, 2. Waves produced in a string fixed at two ends., , Q. Give an example for transverse wave., Ans: Light wave, , Q. Derive the equation V= fλ, Ans: Consider a wave of wave length λ, time period T and frequency f travelling with a, velocity v., wave length, , Wave velocity = wave period, λ, , 𝑣=𝑇, ∴ 𝑣 = 𝑓λ ( since, , 1, T, , = f), , 247

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Q. Distinguish between Longitudinal wave and transverse wave., Ans:, Longitudinal Waves, , Transverse Waves, , 1. The particles of the medium vibrates, , 1. The particles of the medium vibrate, , along the direction of wave propagation., , perpendicular to the direction of wave, propagation ., , 2. Propagate in the form of compression, , 2. Propagate in the form of crest and, , and rarefaction., , troughs., , 3.cannot be polarized, , 3.can be polarized, , 4. Travel through solids, liquids and gases., , 4. Travel through solids and on Liquid, surfaces., , 5. There is pressure variation throughout, , 5.There is no pressure variations, , the medium., 6. Sound wave is an example of, , 6. Light wave is an example of transverse, , longitudinal wave., , wave., , Q. Give any three differences between progressive wave and stationary wave., Ans:, Progressive wave, 1.It is formed due to continues vibrations, of the particles of the medium, , Stationary wave, 1. It is formed due to super position of two, identical waves travelling in a straight line, in opposite direction., , 2.It travels in a particular direction, , 2. It does not travel in any direction., , 3. There is a transfer of energy., , 3.There is no transfer of energy, , 4.In a progressive wave no particle is at, , 4. In stationary wave, the particles at nodes are, , rest, 5. The amplitude of all the particles is, same., 6.The phase changes continuously from, Particle to particle., , at rest., 5. The amplitude is different for different, particles., 6. All the particle in the loop is in same phase, and they are in opposite phase in adjacent, loops., 248

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Q. Write any three Characteristics of standing waves., Ans: Characteristics of stationary wave:, 1. A stationary wave does not move in any direction., 2. In a stationary wave, there is no propagation of energy from one part of a medium to, another., 3. Stationary wave may be either longitudinal or transverse., 4. In a stationary wave, nodes and antinodes are formed alternately and they are equally, spaced., Q. Write Newton’s formula for speed of sound in air., 𝑃, , Ans: 𝑣 = √𝜌Where,, v = volume, P =pressure, 𝜌 =density of the gaseous medium., 11. Mention Newton’s formula for speed of sound in gases and explain the terms used., Ans: Newton’s formula for speed of sound in gases ., 𝐏, , 𝐯 = √𝛒, Where,, , v = volume, P =pressure, 𝜌 =density of the gaseous medium., Q. Give Newton’s formula for speed of sound in air and hence explain Laplace correction to, Newton’s formula., Ans: Newton’s formula for speed of sound in air:, v, , P, , , , .................. (1), , Newton’s – Laplace formula for speed of sound in a gas : According to Laplace , when the, sound passes through a gas in the form of compression and rarefaction, the pressure and, volume change adiabatically. During the arrival of compression, there is an increase in the, 249

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temperature and at the time of formation of rarefaction, there is a fall in the temperature. As air, is a bad conductor of heat and the compression and rarefaction arrive so rapidly that there is no, time for exchange of heat with the surrounding medium. Hence, propagation of sound in a gas, is an adiabatic but not an isothermal. The speed of sound in a gaseous medium is given by:, Equation (1) leads to the form, v=, , P, ……….................. (2), , , This is called Newton’s - Laplace formula for speed of sound in a gas., For air at NTP: P = 1.013, , , , 105 Nm-2,  = 1.4 and  = 1.293 Kgm-3, , Equation (2) becomes:, , v=, , 1.4  1.013  10 5, =331ms-1, 1.293, , This is in close agreement with the experimental value of 332 ms -1., , Q. Derive the Newton’s – Laplace formula for the velocity of sound in air., Ans: According to Laplace, when the sound pressure through a gas in the form of compression, and rarefaction, the pressure and volume change adiabatically. During the arrival of, compression, there is an increase in the temperature and, , at the time of formation of, , rarefaction, there is a fall in the temperature. As air is a bad conductor of heat and the, compression and rarefaction arrive so rapidly that there is no time for exchange of heat with the, surrounding medium., , Hence, propagation of sound in a gas is an adiabatic but not an, , isothermal .The speed of sound in a gaseous medium is given by:, 𝐾, 𝑉 = √ − − − − − − − − − − − (1), 𝜌, Where K = Bulk modulus of gas and 𝜌 = density of the gaseous medium, An ideal gas equation for adiabatic process is:, 𝑃𝑉 𝛾 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 − − − − − − − − − (2), , 250

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𝐶, , Where: P = Pressure, V = volume and 𝛾 = 𝐶𝑃 = ratio of specific heat capacity of the gas at, 𝑉, , constant pressure to that constant volume., Differentiating both sides of equation (2), 𝑃𝛾𝑉 𝛾−1 𝑑𝑣+𝑉 𝛾 dP = 0, 𝑉 𝛾 (𝛾𝑃𝑉 −1 𝑑𝑉 + 𝑑𝑃) = 0, 𝛾𝑃𝑉 −1 dv + dP=0, 𝛾𝑃, , 𝑑𝑉, + 𝑑𝑃 = 0, 𝑉, , 𝛾𝑃, , 𝑑𝑉, = −𝑑𝑃, 𝑉, , 𝛾𝑃 = −, , 𝑑𝑃, 𝑑𝑉⁄, 𝑉, , But, the bulk modulus of gas is given by:, 𝑑𝑃, 𝑑𝑉⁄, 𝑉, Equation (1) leads to the form of, 𝐾=, , 𝛾𝑃, , V=√, , 𝜌, , Q. Derive an expression for fundamental frequency of oscillation of a stretched string., Ans: A string of a given length is stretched between two rigid supports. The string is made to, vibrate in different modes by plucking it at different points. If the string is plucked at some, point, the particle of the string at that point starts vibrating in a direction perpendicular to the, length of the string. This gives rise to transverse waves which travel along the string in both, directions. These waves get reflected from two fixed ends. The reflected waves are identical in, all respects and travel in opposite directions .Their superposition gives rise to stationary waves., Since the two ends of the string are fixed, nodes are formed at the two ends of the string. There, may or may not be nodes between the two ends of the string., Modes of vibration of a stretched string: Consider a string of length l and linear density m, stretched under a tension T between two fixed ends. When the string is plucked at different, 251

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points, it vibrates in different modes. The string may vibrate in one loop, two loops, three loops, etc. There are called modes of vibration., Fundamental mode of vibration: If the string is plucked in the middle, it vibrates in one loop, as shown in the figure., Let 𝑓1 be the frequency of vibration and 𝜆1 be the corresponding wavelength., Length of the string = Distance between two successive nodes., 𝑙=, , 𝜆1, 2, , 𝜆1 = 2l ---------- (1), Speed of wave is:, v =𝑓1 𝜆1 --------- (2), Speed of transverse wave in the string is:, 𝑇, , v =√𝑚 --------- (3), From the equation (2) and (3), 𝑇, , 𝑓1 𝜆1 =√𝑚, , 𝑓1 =, , 1 𝑇, √, 𝜆1 𝑚, 𝟏, , 𝑻, , 𝒇𝟏 = 𝟐𝒍 √𝒎 ------------ (4), This gives the fundamental frequency (first harmonic), Q. What harmonics are present in (a) an open pipe (b) an closed pipe?, Ans: (a) an open pipe: All harmonics (even and odd) are present in the vibrations of an air, column in an open pipe., 𝑓1 : 𝑓2 : 𝑓3 ………… = 1:2:3…………, (b) a closed pipe : Only odd harmonics are present in the vibrations of an air column in a, closed pipe ., 𝑓1 : 𝑓2 : 𝑓3 ………… = 1:3:5…………, 252

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Q. Discuss the modes of vibrations of air column in closed pipe., Ans: Harmonics in closed pipe:, First harmonic or fundamental:, The first harmonic in a closed pipe should have a node at the, closed end and an anti node at the open end., 𝜆, , The vibrating length 𝑙 = 4 , 𝜆 = 4𝑙, 𝑣, , Then 𝑓1 = 𝜆, 𝑣, , ∴ 𝑓1 = 4𝑙, This is known as fundamental frequency., Third harmonic or first overtone:, The next possible harmonic should have one more node and anti node than fundamental, frequency. (two nodes and two antinodes), The vibrating length 𝑙 =, Then, , 𝑣, , 𝑓3 = 𝜆 =, 𝑓3 =, , 3𝑣, 4𝑙, , 3𝜆, 4, , ,𝜆=, , 4𝑙, 3, , 3𝑣, 4𝑙, , = 3𝑓1, , This is known as third harmonic or first overtone., Fifth harmonic or second overtone:, It will have three nodes and three antinodes., The vibrating length=, Then,, 𝑓5 =, , 𝑣, , 𝑓5 = 𝜆 =, , 5𝜆, 4, , ,𝜆=, , 4𝑙, 5, , 5𝑣, 4𝑙, , 5𝑣, = 5𝑓1, 4𝑙, , This is known as fifth harmonic or second overtone., ∴Similarly, the frequencies higher harmonics can be derived in the same way, ∴ In closed pipe, the frequencies of overtones are in the ratio of odd numbers i.e 1: 3: 5…., It can be observed that in the case of an open organ pipe the entire harmonic can be formed., But in the case of closed organ pipe only odd harmonics are possible., , 253

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Q. Discuss the modes of vibration of air column in an open pipe., Ans: Consider a pipe of length l. Let v be the speed of sound in air., First or fundamental mode of vibration: In this, , mode, two antinodes are formed at the open, , ends with a node in between as shown in the figure., Let f1 be the frequency of vibration of the air column and  1 be the corresponding wavelength., Length of the air column = Distance between two successive antinodes., , 1, 2, , l=, , 1  2l, Speed of wave is: v = f1  1, f1 =, , v, 1, , f1=, , v, 2l, , ……………………...(1), , This gives the fundamental frequency of vibration or first harmonic., , Second mode of vibration : In this mode, three antinodes and two nodes are formed as shown, in the figure., Let f2 be the frequency of vibration of air column and  2 be the corresponding wavelength., Length of the air column = 2  Distance between two successive antinodes., l=2, , 2, 2, , 2 , , 2l, 2, , Speed of wave is:v = f2  2, f2 =, , v, 2, , f2 = 2., , v,  2. f 1 …………………(2), 2l, , This is called second harmonic or first overtone., Third mode of vibration: In this mode, four antinodes & three nodes are formed as shown, in the figure., Let f3 be the frequency of vibration of air column and  3 be the corresponding wavelength., Length of the air column = 3  Distance between two consecutive antinodes, 254

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l=3., 3 , , 3, 2, , 2l, 3, , Speed of wave is: v = f3  3, f3 =, , v, 3, , f3 = 3 ., , v,  3.f 1 …………….(3), 2l, , This is called the third harmonic or second overtone., , Conclusion: From equations (1), (2) and (3):, , f1 : f2 : f3 = 1 : 2 : 3, , ∴ All harmonics (even and odd) are present in the vibrations of an air column in an open pipe., , Q. What is Doppler Effect?, Ans: The apparent change in the frequency of sound due to the relative motion between the, source and the observer is called Doppler Effect., Q. What is Doppler Effect? Mention one of its applications., Ans: Doppler effect: The apparent change in the frequency of sound due to the relative, motion between the source and the observer is called Doppler effect., Applications: SONAR AND RADAR AND MEDICAL IMAGING., SONAR:- Sonar is Sound navigation and ranging., RADAR:- Radar is Radio detection and ranging., Q. What is Doppler Effect of sound? Derive expression for apparent frequency of sound, when source is moving away from stationary listener., Ans: Doppler Effect: The apparent change in the frequency of sound due to the relative, motion between the source and the observer is called Doppler effect., Apparent frequency of sound when source is moving away from stationary listener, (Observer):, Let S be source of sound moving with velocity 𝑉𝑠 away from stationary listener O., Distance travelled in time period: T = Vs . T, 255

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Therefore apparent wavelength: λ′ = λ + Vs T, If frequency is 𝑓 then,, 𝜆′ = 𝜆 +, λ′ =, , 𝑉𝑠, 𝑓, , fλ+Vs, f, , λ′ =, , ∵𝑉 = f𝛌, , V + Vs, f, , If apparent frequency heard by listener is f ′, f′ =, , V, λ′, , f′ =, , f′ =, , V, [, , V+Vs, f, , ], , fV, V + Vs, , f′ = f [, , V, ], V + Vs, , Thus apparent frequency is less than the actual frequency as source moves towards listener., , Q. Derive the expression for apparent frequency of sound, when an observer is moving, towards a stationary source., Ans: Derivation of general expression for apparent frequency: Consider a source S emitting, sound of frequency f. Let v be the velocity of sound. Let the source move towards the observer, with velocity vs and the observer move, away from the source with velocity v0., , In one second, the source travels a distance SS = vs and wave travels a distance SP = v. In one, second, source emits f waves such that these waves will be contained in a length S 1P = v-vs., The apparent wavelength of these waves is:, 𝜆1 =, , 𝑣−𝑣𝑠, 𝑓, , 256

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These waves approach the observer O with a relative velocity (v-v0). The number of waves, received by the observer in one second (apparent frequency as heard by the observer) is:, 𝑓1 =, , 𝑣−𝑣0, 1, , 𝑓1 =, , 𝑣−𝑣0, 𝑣−𝑣𝑠, ), 𝑓, , (, , 𝑣−𝑣, , f1 = [𝑣−𝑣0]f, 𝑠, , When an observer in moving towards a stationary source, Vs = 0 and v0 = -v0, f1 = f [, , 𝑣−(−𝑣0 ), , f1 = f [, , 𝑣+𝑣0, , 𝑣−0, 𝑣, , ], , ], , Q. What are beats? Give one application of beats., Ans: The periodic rise and fall (waxing and waning) in the intensity of sound due to the, superposition of two sound waves of slightly different frequencies travelling in the same, direction is called beats., Applications: (i) to determine the unknown frequency of tuning fork, (ii) In tuning musical instruments., , Q. Give the theory of beats., Ans: Phenomenon of Beats: The phenomenon of regular rise and fall in the intensity of sound,, when two waves of nearly equal frequencies travelling along the same line and in the same, direction superpose each other is called beats., One rise and one fall in the intensity of sound constitute one beat and the number of beats per, second is called beat frequency., Mathematical treatment of beats:, Consider two waves of same amplitude ‘A’ but of slightly different frequencies ‘f1 ’ and ‘f2 ’., These waves are represented by the following equations, y1 = a sin 1t = A sin 2f1t ------------ (1), y2 = a sin 2t = A sin 2f2t ----------- (2), according to the principle of superposition, the resultant displacement is given by, y = y1 + y2 = Asin2f1t + A sin2f2t, y = A [sin 2 f1t + sin 2 f2t], 257

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f =100 Hz, 𝜔, , (d)Wave velocity: (v) = 𝐾, v=, v=, , 200𝜋, 0.5𝜋, 200 ×10, 5, , V= 400m/s, , Q. A transverse harmonic wave on a string is described by, 𝛑, , 𝐲(𝐱, 𝐭) = 𝟑. 𝟎 𝐬𝐢𝐧(𝟑𝟔𝐭 + 𝟎. 𝟎𝟏𝟖𝐱 + 𝟒 ) Where x and y are in centimetre and‘t’ is in, seconds. The positive direction of x is from right to left., (a) Is this a travelling wave or stationary wave? If it is travelling what are the speed a, direction of its propagation., (b) What are its amplitude and frequency?, (c) What is the initial phase at the origin?, (d) What is the least distance between the two successive crests in the wave?, Solution:, (a)The equation of progressive wave travelling from left to right is given by the displacement, function:, y(x, t) = a sin(ωt + kx + ɸ) ────────────────────(1), The given equation is, π, y(x, t) = 3.0 sin(36 + 0.018x + ) ────────────────(2), 4, On comparing both the equations, we find that equation (1) represents a travelling wave,, propagating from right to left., Now using equations (1) and (2) , we can write:, 𝜔 = 36 𝑟𝑎𝑑/𝑠and𝑘 = 0.018𝑐𝑚−1, We know that:, ω, , f = 2𝜋 and 𝜆 =, , 2𝜋, 𝑘, , 261

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𝑓 ′ =Apparent frequency, (a)When the train approaches him, 𝑓 ′ =(, , 𝑣, 𝑣−𝑣𝑠, , ) 𝑓0, , 340, , =(340−20) 600, 340, , =(320)600, =637∙5Hz, (b)When the train recedes him, 𝑓 ′ =(, , 𝑣, 𝑣+𝑣𝑠, , ) 𝑓0, , 340, , =(340+20) 600, 340, , =(360)600, =566∙67Hz, , Q. A train standing at the outer signal of railway station blows a whistle of frequency 380Hz, in still air. What is the frequency of the whistle for a platform observer, when the train, (a) approaches the platform with a speed of 20 m/s?, (b) recedes from the platform with a speed of 20 m/s? Given the velocity of sound = 340, m/s., Solution: Given v = 340m/s, Velocity of source(Train)vs=20m/s, Velocity of observer Vo = 0m/s, 𝑣−𝑣, , Apparent frequency f l = f[𝑣−𝑣0 ], 𝑠, , (a) f l = f[, , 𝑣, 𝑣−𝑣𝑠, , ], , 340, , fl = 380[340−20], , 266

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340, , fl = 380[320 ], fl = 403.75Hz, 𝑣, , (b) f l = f[𝑣+𝑣 ], 𝑠, , 340, , fl = 380[340+20], 340, , fl = 380[, , 360, , ], , fl = 358.88 Hz, , Q.A train is moving at a speed of 54kmph. As it is receding from the observer, it whistles., The frequency of the note being 220Hz. Calculate the frequency of sound as heard by the, observer. Given : velocity of sound = 345 ms-1, v, , Solution:, , vo= 0 ms-1, -vs, , S, , O, , 5, , Source Velocity, vs=54kmph = 54x 18 = 15m/s, Observer Velocity vo= 0m/s, v=345m/s, Frequency of note: f = 220Hz, Frequency of sound heard by man f l= ?, 𝑣−𝑣0, , Apparent frequency f l = f[, , 𝑣−𝑣𝑠, , ], , 𝑣, , fl = f[𝑣+𝑣 ], 𝑠, , 345, , fl = 220 [345+15], 345, , fl = 220 [360], fl = 210.83Hz, 267

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Q. Two car approaches each other with a common speed of 20ms-1, the first car sounds of, horn and a passenger in the other car estimates it to be 700Hz, the speed of sound is, 332ms-1., (a) Calculate the actual frequency of the horn?, (b) When the cars move away from each other, what is the estimated frequency of the, same passenger?, Solution: vs=vo= 20ms-1, v = 332 ms-1, f1 =700Hz,, (a), 𝑓1 = 𝑓 (, , 𝑣 + 𝑣𝑜, 332 + 20, ) ⇒ 700 = 𝑓 (, ) ⇒ 700 = 𝑓(1.13), 𝑣 − 𝑣𝑠, 332 − 20, , f = 619Hz, Therefore the actual frequency of the horn,𝐟 = 𝟔𝟏𝟗𝐇𝐳, , (b)For f = 619Hz,When the cars move away from each other, the frequency is,, , 𝑓1 = 𝑓 (, , 𝑣 − 𝑣𝑜, 332 − 20, ) = 619 (, ) = 619x(0.887), 𝑣 + 𝑣𝑠, 332 + 20, , 𝒇′ = 𝟓𝟒𝟗𝑯𝒛, , Q. A train is moving at a speed of 72kmh-1 towards a station sounding a whistle of frequency, 640Hz. What is the apparent frequency of the whistle as heard by a man standing on the, platform when train approaches him? Given; Speed of sound = 340 ms -1, Solution: vs = 72kmh-1 = 20ms-1, f = 640Hz, v = 340 ms-1, 𝑓 ′ =?, 𝑣, , Apparent frequency of the sound, 𝑓 ′ = 𝑓 (𝑣−𝑣 ), 𝑠, , 𝑓 ′ = 640 (, , 340, ), 340 − 20, , 𝑓 ′ = 680 𝐻𝑧, , Q. A Train moving at a speed of 72kmph towards a station sounding a whistle of frequency, 600H𝒛 . What are the apparent frequencies of the whistle as heard by a man on the, platform?, 270

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(b) Frequency, (c) Period, Solution:Given, 𝒙 = 𝟓 𝐜𝐨𝐬(𝟐𝟎𝒕 + 𝟐𝝅), We have,, Y=A cos (𝜔𝑡 + 𝜙𝑜 ), (a) Amplitude; A=5m, (b) To Calculate Frequency, 𝜔 = 20, 2πf = 20, 𝑓=, , 10, 𝐻𝑧, 𝜋, 1, , (c) Time period is ; 𝑇 = 𝑓, T=, , 𝜋, 𝑠𝑒𝑐, 10, , Q. A stone dropped from the top of a tower of height 300m splashes into water of a pond, near the base of the tower. When is the splash heard at the top? Given that the speed of, sound in air is 340 ms-1(g = 9.8 ms-2)., Solution:, Given, s=300m, u=0ms-1, 1, s = ut + 𝑎𝑡 2, 2, s=, , 1 2, at, 2 1, , 1, 300 = (9.8)𝑡12, 2, 𝑡12 =, , 600, 9.8, , 𝑡12 = 61.22, , 275