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98, , CBSE New Pattern ~ Physics 11th (Term-I), , 06, Work, Energy, and Power, Quick Revision, 1. Work Work is said to be done by a force, when, the body is displaced actually through some, distance in the direction of the applied force., Thus, work is done on a body only if the, following two conditions are satisfied, ● A force acts on the body., ● The point of application of the force moves, in the direction of the force., 2. Work Done by a Constant Force Work done, by the force (constant force) is the product of, component of force in the direction of the, displacement and the magnitude of the, displacement. Then, the work done on the body by, the force is given by, Work done,W = F × s, SI unit of work is joule( J )., Its dimensions are [M 1L2T -2]., 3. Work Done when Force and Displacement, are Inclined to Each Other, Fy, , Fy, F, θ, , F, , Fx, , θ, , Fx, , s, , Work done, W = F × s = ( F cos q ) × s = Fs cos q, Two cases can be considered as given below for, the maximum and minimum work, Case I When F and s are in the same, direction, i.e. q = 0° , then work done is, W = Fs cos 0° = Fs (1) = Fs, , i.e. maximum work done by the, force., Case II When F and s are perpendicular to, each other, i.e. then, W = F × s = Fs cos 90° = Fs (0 ) = 0,, i.e. no work done by the force, when a, body moves in a direction, perpendicular to the force acting., 4. Work Done by a Variable Force Work, done by variable force is given as,, Wxi, , ®x, , f, , =, , ò, , x, , xi, , f, , F ×d x =, , ò, , x, , f, , xi, , (F cos q ) dx, , = Area under force-displacement curve, When the magnitude and direction of a force, vary in three dimensions, then it can be, expressed in terms of rectangular, components., So, work done from x i to x f ,, W =, , ò, , x, , xi, , f, , F x dx +, , ò, , x, , xi, , f, , F y dy +, , ò, , x, , xi, , f, , F z dz, , where, F x , F y and F z are the rectangular, components of force in x, y and z-directions,, respectively., 5. Conservative Force If the work done by the, force in displacing an object depends only on, the initial and final positions of the object and, not on the nature of the path followed, between the initial and final positions, such a, force is known as conservative force. e.g., Gravitational force is a conservative force.

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6. Non-Conservative Force If the work done, by a force in displacing an object from one, position to another depends upon the path, between the two positions. Such a force is, known as non-conservative force. e.g. Friction, is a non-conservative force., 7. Energy The energy of a body is defined as its, capacity or ability for doing work., The dimensions of energy are the same as, the dimensions of work, i.e. [M 1 L2 T -2 ]., ● It is measured in the same unit as work, i.e., joule in SI system and erg in CGS system., 8. Kinetic Energy The energy possessed by a, body by virtue of its motion is called kinetic, energy. In other words, the amount of work, done, by a moving object before coming to rest, is equal to its kinetic energy., 1, \ Kinetic energy, KE = mv 2, 2, where, m is a mass and v is the velocity of a, body., ● Relation between Kinetic Energy and, Linear Momentum, ●, , p = 2m K, 9. Work Energy Theorem or Work Energy, Principle It states that, work done by the net, force acting on a body is equal to the change, produced in the kinetic energies of the body., f, , 12. Potential Energy of a Spring For a small, stretch or compression, spring obeys Hooke’s, law, i.e. restoring force µ stretch or compression, - Fs µ x Þ Fs = - k x, where, k is called spring constant. Its SI unit, is Nm –1.The negative sign shows F s acts in the, opposite direction of displacement x., If the block is moved from an initial, displacement x i to final displacement x f , then, work done by spring force is, 1, 1, W s = kx i2 - kx f 2, 2, 2, \ Change in potential energy of a spring, 1, DU = – W s = k ( x 2f – x i2 ), 2, 1, If, x i = 0, then DU = kx 2f, 2, 13. Conservation of Mechanical Energy This, principle states that, if only the conservative, forces are doing work on a body, then its, mechanical energy (KE + PE) remains constant., i.e. K + U = constant = E, \ K i +Ui = K f +U f, The quantity K + U , is called the total, mechanical energy of the system., 14. Motion in a Vertical Circle A particle of, mass m is attached to an inextensible string of, length L and is moving in a vertical circle, about fixed point O (as shown), , K f – K i = ò Fnet × dx, i, , \ K f - K i =W, where, K f and K i are the final and initial, kinetic energies of the body., 10. Potential Energy The potential energy of a, body is defined as the energy possessed by the, body by virtue of its position or configuration., So, if configuration of the system changes, then, its potential energy changes., Dimensions = [ML2T –2 ], SI unit = Joule, 11. Gravitational Potential Energy, Gravitational potential energy of a body is the, energy possessed by the body by virtue of its, position above the surface of the earth., Gravitational potential energy, U = mgh, , O, , h, , L, , θ, , A, ●, , ●, , ●, , T, , u, , v, , B, mg cos θ, mg sin θ, , Minimum velocity at highest point, so that, particle complete the circle, v min = gL ,, at this velocity, tension in the string is zero., Minimum velocity at lowest point, so that, particle complete the circle, v min = 5 gL ,, at this velocity, tension in the string is 6 mg., When string is horizontal, then minimum, velocity is 3Rg and tension in this, condition is 3 mg.

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15. Power Power of a person or machine is, defined as the rate at which work is done or, energy is transferred., Average power (Pav ) = rate of doing work, work done (W ), =, time taken (t ), Thus, the average power of a force is defined, as the ratio of the work (W) to the total time (t)., 16. The instantaneous power of an agent at any, instant is equal to the dot product of its force, and velocity vectors at that instant., P = F×v, 17. Power is a scalar quantity and its dimensional, formula is [ML2T –3 ]., The SI unit of power is watt (W)., 1 joule, 1 watt =, = 1 Js - 1, 1 second, Another popular units of power are kilowatt and, horse power., 1 kilowatt = 1000 watt or 1 kW = 10 3 W, 1 horse power = 746 watt or 1 HP = 746 W, This unit is used to describe the output of, automobiles, motorbikes, engines, etc., 18. Collision A collision is an isolated event in, which two or more colliding bodies exert, strong forces on each other for a relatively, short time. For a collision to take place, the, actual physical contact is not necessary., Collision between particles have been divided, into two types which can be differentiated as, Elastic Collision, , Inelastic Collision, , A collision in which, there is absolutely no, loss of kinetic energy., , A collision in which there, occurs some loss of, kinetic energy., , Forces involved during, elastic collision must be, conserved in nature., , Some or all forces, involved during collision, may be non- conservative, in nature., , The mechanical energy, is not converted into, heat, light, sound, etc., , A part of the mechanical, energy is converted into, heat, light, sound, etc., , e.g. Collision between, subatomic particles,, collision between glass, balls, etc., , e.g. Collision between, two vehicles, collision, between a ball and floor,, etc., , 19. Conservation of Linear Momentum in, Collision Total linear momentum is, conserved at each instant during collision., \ p 1 + p 2 = constant, 20. Elastic Collision in One Dimension, In one-dimensional elastic collision, relative, velocity of separation after collision is equal to, relative velocity of approach before collision., u1 - u2 = v2 - v1, Velocities of the Bodies After the Collision, Velocity of Ist body after collision,, æ 2m 2 ö, æm - m2 ö, v1 = ç 1, ÷ u2, ÷ u1 + ç, èm1 + m2 ø, èm1 + m2 ø, , …(i), , Velocity of IInd body after collision,, æ 2m 1 ö, æm - m1 ö, v2 = ç 2, ÷u 1, ÷ u2 + ç, èm1 + m2 ø, èm1 + m2 ø, , …(ii), , Eqs. (i) and (ii) give the final velocities of the, colliding bodies in terms of their initial, velocities., The two cases under the action of same and, different masses can be considered as given, below, Case I, , When two bodies of equal masses, collide., i.e., , m 1 = m 2 = m (say), , From Eq. (i), we get, 2mu 2, = u 2 = velocity of body of, v1 =, 2m, mass m 2 before collision, From Eq. (ii), we get, 2mu 1, = u 1 = velocity of body of, v2 =, 2m, mass m 1 before collision., Case II When a light body collides against a, massive stationary body., Here, m 1 << m 2 and u 2 = 0, Neglecting m 1 in Eq. (i), we get, m 2u 1, v1 = = - u1, m2, From Eq. (ii), we get, v2 •0

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21. Perfectly Inelastic Collision in One, Dimension, When the two colliding bodies together move, as a single body with a common velocity after, the collision, then the collision is perfectly, inelastic., In perfectly inelastic collision between two, bodies of masses m 1 and m 2, the body of mass, m 2 happens to be initially at rest (u 2 = 0 ). After, the collision, the two bodies move together, with common velocity v. The change in their, m 1 m 2u 12, kinetic energies is KE =, 2 (m 1 + m 2 ), \ DKE is a positive quantity., Therefore, kinetic energy is lost mainly in the, form of light, sound and heat., 22. Elastic Collision in Two Dimensions When, the collision between two bodies is not, head-on (the force during the collision is not, along the initial velocity). The bodies move, along different lines, then the collision is called, elastic collision in two dimensions., The three cases can be considered as given, below, Case I Glancing Collision In a glancing, collision, the incident particle does not lose, any kinetic energy and is scattered almost, undeflected. Thus, for such collision, when, q = 0°, f = 90°, u 1 = v 1 and v 2 = 0., 1, KE of the target particle = m 2v 22 = 0, 2, , Case II Head-on Collision In this type of, collision, the target particle moves in the, direction of the incident particle, i.e. f = 0°., m 1u 1 = m 1v 1 cos q + m 2v 2 and 0 = m 1v 1 sin q, So, the kinetic energy remains unchanged., Case III Elastic Collision of Two Identical, Particles When two particles of same mass, undergo perfectly elastic collision in two, dimensions, i.e. m 1 = m 2., \, q + f = 90°, Thus, after collision the two particles will move, at right angle to each other., 23. Inelastic Collision in Two Dimensions, When two bodies travelling initially along the, same straight line collide involving some loss of, kinetic energy and move after collision along, different directions in a plane, then it is called, inelastic collision in two dimensions., 24. Coefficient of Restitution or, Coefficient of Resilience, It is defined as the ratio of relative velocity of, separation after collision to the relative velocity, of approach before collision. It is denoted by e ., e =, , Relative velocity of separation (after collision), Relative velocity of approach (before collision), e =, , |v2 - v1 |, |u 2 - u 1 |, , where, u 1 & u 2 are velocities of two bodies, before collision and v 1 & v 2 are their respective, velocities after collision., , 25. Comparison between Different Types of Collisions, Collision, , Kinetic Energy, , Coefficient of Restitution, , Main Domain, , Elastic, , Conserved, , e =1, , Between atomic particles, , Inelastic, , Non-conserved, , 0 < e <1, , Between ordinary objects, , Perfectly inelastic, , Maximum loss of KE, , e =0, , During shooting, , Super elastic, , KE increases, , e >1, , In explosions

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102, , CBSE New Pattern ~ Physics 11th (Term-I), , Objective Questions, Multiple Choice Questions, 1. A bicyclist comes to a skidding stop in, 10 m. During this process, the force on, the bicycle due to the road is 200N and, is directly opposed to the motion. The, work done by the cycle on the road is, , 5. A body moves from point A to B under, the action of a force varying in, magnitude as shown in figure, then the, work done is (force is expressed in, newton and displacement in metre), 20, 15, 10, 5, F 0, , (NCERT Exemplar), , (a) + 2000 J, (c) zero, , (b) - 200 J, (d) - 20, 000 J, , 2. Force of 50 N acting on a body at an, , angle q with horizontal. If 150 J work is, done by displacing it 3 m, then q is, (a) 60°, (c) 0°, , –5, –10, –15, , (b) 30°, (d) 45°, , 3. A particle is pushed by forces, , 2$i + 3$j - 2k$ and 5$i - $j - 3k$, simultaneously and it is displaced from, point $i + $j + k$ to point 2$i - $j + 3k$ . The, work done is, (b) - 7 units, (d) -10 units, , (a) 7 units, (c) 10 units, , 4. Consider a force F = - x$i + y$j. The, work done by this force in moving a, particle from point A(1, 0 ) to B(0, 1), along the line segment is (all quantities, are in SI units), Y, B (0, 1), , 3, (a), 2, , (b) 2, , (c) 1, , (d), , 1, 2, , X, , 1, , 2, , R, 3, , 5, , 4, , s, , B, , (b) 22.5 J, (d) 27 J, , 6. A string of length L and force constant, k is stretched to obtain extension l. It is, further stretched to obtain extension l 1 ., The work done in second stretching is, 1, kl 1 (2l + l 1 ), 2, 1, (c) k (l 2 + l 12 ), 2, (a), , 1 2, kl 1, 2, 1, (d) k (l 12 - l 2 ), 2, (b), , 7. A uniform chain of length l and mass m, is lying on a smooth table and one-third, of its length is hanging vertically down, over the edge of the table. If g is, acceleration due to gravity, work, required to pull the hanging part on to, the table is, , (c), A (1, 0), , P, , (a) 30 J, (c) 25 J, , (a) mgl, , (0, 0), , Q, A, , mgl, 9, , mgl, 3, mgl, (d), 18, (b), , 8. IfW 1 , W 2 andW 3 are the work done in, moving a particle from A and B along, three different paths 1, 2 and 3, respectively (as shown) in the

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gravitational field of a point mass m, the, relation betweenW 1 , W 2 andW 3 is, B, , 1, , m, , 14. A mass of 5 kg is moving along a, , 2, , circular path of radius 1 m. If the mass, moves with 300 rev/min, its kinetic, energy (in J) would be, , 3, A, , (a) W1 > W2 > W3, (c) W1 < W2 < W3, , (b) W1 = W2 = W3, (d) W2 > W1 > W3, , 9. Amongst the given graphs which one, correctly represents the variation of the, kinetic energy (K ) of a body with, velocity (v )?, K, , K, , (b), , (a), , v, , v, , K, , K, , (d), , (c), v, , v, , 10. The kinetic energy of a body of mass, 4 kg and momentum 6 N-s will be, (a) 3.5 J, (c) 2.5 J, , (b) 5.5 J, (d) 4.5 J, , having a momentum p, which one of, the following correctly describes the, kinetic energy of the particle?, p2, 2m, , (b), , p, 2m, , (c), , v2, 2m, , (d), , (b) 100p 2, , (c) 5 p 2, , (d) zero, , 15. Two moving objects (m 1 > m 2 ) having, same kinetic energy are stopped by, application of equal retarding force., Which object will come to rest at, shorter distance?, (a), (b), (c), (d), , Bigger, Smaller, Both at same distance, Cannot say, , 16. A particle which is experiencing a, , force, is given by F = 3$i - 12$j,, undergoes a displacement of d = 4 $i., If the particle had a kinetic energy of, 3 J at the beginning of the, displacement, what is its kinetic energy, at the end of the displacement ?, (a) 9 J, , (b) 15 J, , (c) 12 J, , (d) 10 J, , from rest under the influence of a force, that varies with the distance travelled by, the particle as shown in the figure., The kinetic energy of the particle after, it has travelled 3 m is, , v, 2m, , 12. Two bodies of masses 4 kg and 5 kg are, moving with equal momentum. Then,, the ratio of their respective kinetic, energies is, (a) 4 : 5, (c) 1 : 3, , (a) 250p 2, , 17. A particle moves in one dimension, , 11. For a moving particle (mass m, velocity v), , (a), , (a) Heavy body, (b) Light body, (c) Both have same linear momenta, (d) None of the above, , Force, (in N), , same kinetic energy. Which will have, larger linear momentum?, , 2, , A, , B, , D, , 1, F, O, , (b) 2 : 1, (d) 5 : 4, , 13. A heavy body and a light body have, , C, , 3, , (a) 4 J, (c) 6.5 J, , 1, , 2, Distance, (in m), , E, , 3, , (b) 2.5 J, (d) 5 J

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surface of the earth, then its potential, energy, , The kinetic friction force is 15 N and, spring constant is 10000 N/m. The, spring compresses by, , (a) increases, (c) remains same, , (a) 5.5 cm, (c) 11.0 cm, , 18. When a person lifts a brick above the, (b) decreases, (d) None of these, , 19. A massless spring of spring constant k,, has extension y and potential energy E., It is now stretched from y to 2y. The, increase in its potential energy is, (a) 3E, (c) E, , (b) 2E, (d) 4E, , 20. A bread gives 5 kcal of energy to a boy., How much height he can climbs by, using this energy, if his efficiency is, 28% and mass is 60 kg?, (a) 15m, (c) 2.5 m, , (b) 5m, (d) 10 m, , 25. 300 J of work is done in sliding a 2 kg, block up an inclined plane of height, 10 m (taking, g = 10 ms -2 ). Work done, against friction is, (a) 200 J, (c) zero, , potential energy U as a function of, position r for a particle of mass m . If, the particle is released from rest at, position r0 , what will its speed be at, position 3r0 ?, U(r), 3 U0, , of gravity alone in vaccum. Which of, the following quantities remain constant, during the fall?, (NCERT Exemplar), , 2 U0, U0, , Kinetic energy, Potential energy, Total mechanical energy, Total linear momentum, , 22. A stone is projected vertically up to, reach maximum height h. The ratio of, its kinetic energy to its potential energy, 4, at a height h, will be, 5, (a) 5 : 4, (c) 1 : 4, , (b) 4 : 5, (d) 4 : 1, , 23. A spring of force constant 800 N/m has, an extension of 5 cm. The work done in, extending it from 5 cm to 15 cm is, (a) 16 J, (c) 32 J, , (b) 100 J, (d) 1000 J, , 26. The graph below represents the, , 21. A body is falling freely under the action, , (a), (b), (c), (d), , (b) 2.5 cm, (d) 8.5 cm, , O, , (a), (c), , 1r0, , 2r0, , 3U0, m, 2 U0, m, , 3r0, , 4r0, , r, , 4 U0, m, 6 U0, m, , (b), (d), , 27. A pebble is attached to one end of a, string and rotated in a vertical circle. If, string breaks at the position of, maximum tension, so from the figure, shown below, it will break at, A, , (b) 8 J, (d) 24 J, , C, , D, , 24. A 2 kg block slides on a horizontal floor, with a speed of 4 m/s. It strikes a, uncompressed spring and compresses it, till the block is motionless., , B, , (a) A, , (b) B, , (c) C, , (d) D

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28. What is the ratio of kinetic energy of a, particle at the bottom to the kinetic, energy at the top, when it just loops a, vertical loop of radius r?, (a) 5 :1, (c) 5 :2, , (b) 2 : 3, (d) 7 :2, , 29. A man can do work of 600 J in 2 min,, then man’s power is, (a) 7.5 W, (c) 5 W, , (b) 10 W, (d) 15 W, , 30. A particle is acted by a constant power., Then, which of the following physical, quantity remains constant?, (a), (b), (c), (d), , Speed, Rate of change of acceleration, Kinetic energy, Rate of change of kinetic energy, , 31. An object of mass m moves, horizontally, increasing in speed from, 0 to v in a time t . The power necessary, to accelerate the object during this time, period is, (a), , mv 2 t, 2, , (c) 2 mv 2, , mv 2, 2, mv 2, (d), 2t, , (b), , 32. A 60 HP electric motor lifts an elevator, having a maximum total load capacity, of 2000 kg. If the frictional force on the, elevator is 4000 N, the speed of the, elevator at full load is close to (take,, 1 HP = 746 W and g = 10 ms -2 ), (a) 2.0 ms -1, (c) 1.9 ms -1, , (b) 1.5 ms -1, (d) 1.7 ms -1, , 33. A car of mass m starts from rest and, accelerates, so that the instantaneous, power delivered to the car has a, constant magnitude P 0 . The, instantaneous velocity of this car is, proportional to, (a) t 2 P0, (c) t -1 / 2, , (b) t 1 / 2, (d) t / m, , 34. For a system to follow the law of, conservation of linear momentum, during a collision, the condition is, I. total external force acting on the, system is zero, II. total external force acting on the, system is finite and time of collision, is negligible, III. total internal force acting on the, system is zero, (a) Only I, , (b) Only II, , (c) Only III (d) I or II, , 35. Two identical balls A and B having, , velocities of 0.5 ms -1 and -0.3 ms -1, respectively, collide elastically in one, dimension. The velocities of B and A, after the collision respectively will be, (a), (b), (c), (d), , -05, . ms -1 and 0.3 ms -1, 0.5 m/s -1 and -0.3 ms -1, -0.3 ms -1 and 0.5 ms -1, 0.3 ms -1 and 0.5 ms -1, , 36. A particle of mass 1g moving with a, , velocity v 1 = ( 3$i - 2$j) ms -1 experiences, a perfectly elastic collision with another, particle of mass 2 g and velocity, v 2 = ( 4 $j - 6 k$ ) ms -1 . The velocity of the, particle is, (a) 2.3 ms-1, (c) 9.2 ms-1, , (b) 4.6 ms-1, (d) 6 ms-1, , 37. A particle of mass m 1 moves with, velocity v 1 collides with another particle, at rest of equal mass. The velocity of, second particle after the elastic collision, is, (a) 2v 1, , (b) v 1, , (c) - v 1, , (d) zero, , 38. During inelastic collision between two, bodies, which of the following, quantities always remain conserved?, (NCERT Exemplar), , (a), (b), (c), (d), , Total kinetic energy, Total mechanical energy, Total linear momentum, Speed of each body

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39. Two objects of mass m each moving, , with speed u ms -1 collide at 90°, then, final momentum is (assume collision is, inelastic), (a) mu, (c) 2 mu, , (b) 2 mu, (d) 2 2 mu, , 40. A body of mass 5 ´ 10 kg moving with, 3, , speed 2 ms -1 collides with a body of, mass 15 ´ 10 3 kg inelastically and sticks, to it. Then, loss in kinetic energy of the, system will be, (a) 7.5 kJ (b) 15 kJ, , (c) 10 kJ (d) 5 kJ, , 41. If the linear momentum of a body is, increased by 50%, then the kinetic, energy of that body increases by, ……… ., (a) 100%, , (b) 125%, , (c) 225% (d) 25%, , 42. A ball of mass m moves with speed v, and strikes a wall having infinite mass, and it returns with same speed, then, the work done by the ball on the wall, is ……… ., (a) zero, (c) m / v J, , (b) mv J, (d) v/mJ, , 43. A body of mass 5 kg is thrown, vertically up with a kinetic energy of, 490 J. The height at which the kinetic, energy of the body becomes half of the, original value is ……… ., (a) 12.5 m, (c) 2.5 m, , (b) 10 m, (d) 5 m, , 44. If two persons A and B take 2 s and 4 s,, respectively to lift an object to the same, height h, then the ratio of their powers, is ……… ., (a) 1 :2, (c) 2 :1, , (b) 1 :1, (d) 1 : 3, , 45. At time t = 0, particle starts moving, along the x-axis. If its kinetic energy, increases uniformly with time t , the, net force acting on it must be, , proportional to t n , where the value of, n is ……… ., (a) 1, (c) 2, , (b) - 1/2, (d) 1/2, , 46. A man of mass m, standing at the, bottom of the staircase, of height L, climbs it and stands at its top. Which, amongst the following statement is, (NCERT Exemplar), correct?, (a) Work done by all forces on man is equal to, the rise in potential energy mgL., (b) Work done by all forces on man is zero., (c) Work done by the gravitational force on, man is mgL., (d) The reaction force from a step does some, work because the point of application of, the force does not move while the force, exists., , 47. Which of the following statement is, correct about non-conservative force?, (a) It depends on velocity of the object., (b) It depends on the particular path taken by, the object., (c) It depend on the initial and final positions of, the object., (d) Both (a) and (b), , 48. Which of the following statement is, correct?, (a) Conservation of mechanical energy does, not consider only conservative force., (b) Conservation of energy consider both, conservative and non-conservative forces., (c) Conservation of energy consider only, conservative force., (d) Mass converted into energy in nuclear, reaction is called mass-defect., , 49. Which of the following statement does, not specify an example of perfectly, inelastic collision?, (a) A bullet fired into a block if bullet gets, embedded into block., (b) Capture of electrons by an atom., (c) A man jumping on to moving boat., (d) A ball bearing striking another ball bearing.

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50. Match the Column I (angle) with, Column II (work done) and select the, correct option from the codes given, below., Column I, , Column II, , A., , q < 90°, , p., , Friction, , B., , q = 90°, , q., , Satellite rotating, around the earth, , C., , q > 90°, , r., , Coolie is lifting a, luggage, , Codes, A, , B, , C, , (a) p, , q, , r, , (b) r, , q, , p, , (c) p, , r, , q, , (d) r, , p, , q, , Column I, , Column II, , A., , Velocity is, proportional to, , p., , t, , B., , Displacement is, proportional to, , q., , t2, , C., , Work done is, proportional to, , r., , t3, , B, , C, , (a) p, , q, , r, , (b) r, , q, , p, , (c) p, , q, , q, , (d) r, , p, , p, , (a) Both A and R are true and R is the, correct explanation of A., (b) Both A and R are true but R is not the, correct explanation of A., (c) A is true but R is false., (d) A is false and R is also false., , =, , by a machine delivering a power, proportional to time (P µ t ). Then,, match the Column I with Column II, and select the correct option from the, codes given below., , A, , For question numbers 52 to 60, two, statements are given-one labelled, Assertion (A) and the other labelled, Reason (R). Select the correct answer to, these questions from the codes (a), (b), (c), and (d) are as given below, , 52. Assertion Stopping distance, , 51. A body is moved along a straight line, , Codes, , Assertion-Reasoning MCQs, , Kinetic energy, , Stopping force, Reason Work done in stopping a body, is equal to change in kinetic energy of, the body., , 53. Assertion A spring of force constant k, is cut into two pieces having lengths in, the ratio 1 : 2. The force constant of, series combination of the two parts is, 2k / 3., Reason The spring connected in series, are represented by k = k1 + k 2 ., , 54. Assertion According to the law of, conservation of mechanical energy,, change in potential energy is equal and, opposite to the change in kinetic, energy., Reason Mechanical energy is not, conserved., , 55. Assertion Decrease in mechanical, energy is more in case of an object, sliding up a relatively less inclined, plane due to friction.

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Reason The coefficient of friction, between the block and the surface, decreases with the increase in the angle, of inclination., , 56. Assertion For looping a vertical loop of, , the force and the displacement over which it, acts. Consider a constant force F acting on, an object of mass m. The object undergoes a, displacement d in the positive x-direction as, shown in figure., , radius r, the minimum velocity at lowest, point should be 5gr ., , F, , θ, , Reason In this event, the velocity at the, highest point will be zero., , 57. Assertion Kilowatt-hour is the unit of, energy., Reason One kilowatt hour is equal to, . ´ 10 6 J., 36, , 58. Assertion There is no loss in energy in, elastic collision., Reason Linear momentum is conserved, in elastic collision., , 59. Assertion Quick collision between two, bodies is more violent than a slow, collision; even when the initial and final, velocities are identical., Reason The momentum is greater in, first case., , 60. Assertion Two particles are moving in, the same direction do not lose all their, energy in completely inelastic collision., Reason Principle of conservation of, momentum does not holds true for all, kinds of collisions., , Case Based MCQs, Direction Answer the questions from, 61-65 on the following case., Work, A farmer ploughing the field, a construction, worker carrying bricks, a student studying for a, competitive examination, an artist painting a, beautiful landscape, all are said to be working., In physics, however, the word ‘Work’ covers a, definite and precise meaning. Work refers to, , x, , d, , The work done by the force is defined to be, the product of component of the force in the, direction of the displacement and the, magnitude of this displacement, thus, W = ( F cos q ) d = F × d., , 61. The earth is moving around the sun in, a circular orbit, is acted upon by a, force and hence work done on the, earth by the force is, (a), (b), (c), (d), , zero, positive, negative, None of the above, , 62. In which case, work done will be, zero?, (a) A weight-lifter while holding a weight of, 100 kg on his shoulders for 1 min, (b) A locomotive against gravity is running on, a level plane with a speed of 60 kmh - 1, (c) A person holding a suitcase on his head, and standing at a bus terminal, (d) All of the above, , 63. Find the angle between force, , F = ( 3$i + 4 $j - 5k$ ) unit and, displacement d = (5$i + 4 $j + 3k$ ) unit., (a) cos-1 (0.49), (c) cos-1 (0.60), , (b) cos-1 (0.32), (d) cos-1 (0.90), , 64. Which of the following statement(s), is/are correct for work done to be, zero?, I. If the displacement is zero., II. If force applied is zero.

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(a) Only I, (c) Only II, , (b), , (a), , (b) I and II, (d) I, II and III, , t, , 65. A proton is kept at rest. A positively, charged particle is released from rest at, a distance d in its field. Consider two, experiments; one in which the charged, particle is also a proton and in another,, a positron. In same time t , the work, done on the two moving charged, particles is, (a) same as the same force law is involved in, the two experiments, (b) less for the case of a positron, as the, positron moves away more rapidly and the, force on it weakens, (c) more for the case of a positron, as the, positron moves away a larger distance, (d) same as the work is done by charged, particle on the stationary proton, , Direction Answer the questions from, 66-70 on the following case., Kinetic Energy, The energy possessed by a body by virtue of, its motion is called kinetic energy. In other, words, the amount of work done, a moving, object can do before coming to rest is equal to, its kinetic energy., 1, \ Kinetic energy, KE = mv 2, 2, where, m is a mass and v is the velocity of a, body., The units and dimensions of KE are Joule (in, SI) and [ML 2 T -2 ], respectively., Kinetic energy of a body is always positive. It, can never be negative., , 66. Which of the diagrams shown in figure, most closely shows the variation in, kinetic energy of the earth as it moves, once around the sun in its elliptical, orbit?, , KE, , KE, , III. If force and displacement are, mutually perpendicular to each other., , KE, , (c), , KE, , (d), , 67. A force which is inversely proportional, to the speed is acting on a body. The, kinetic energy of the body starting from, rest is, (a) a constant, (b) inversely proportional to time, (c) directly proportional to time, (d) directly proportional to square of time, , 68. The kinetic energy of an air molecule, (10 -21 J) in eV is, (a) 6.2 meV, (c) 10.4 meV, , (b) 4.2 meV, (d) 9.7 meV, , 69. Two masses of 1 g and 4 g are moving, with equal kinetic energy. The ratio of, the magnitudes of their momentum is, (a) 4 : 1, (c) 1 : 2, , (b) 2 :1, (d) 1 : 16, , 70. An object of mass 10 kg is moving with, velocity of 10 ms -1 . Due to a force, its, velocity become 20 ms -1 . Percentage, increase in its KE is, (a) 25%, (c) 75%, , (b) 50%, (d) 300%, , Direction Answer the questions from, 71-75 on the following case., PE of Spring, There are many types of spring. Important, among these are helical and spiral springs as, shown in figure., (a), , (b)

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Usually, we assume that the springs are, massless. Therefore, work done is stored in, the spring in the form of elastic potential, energy of the spring. Thus, potential energy, of a spring is the energy associated with the, state of compression or expansion of an, elastic spring., , 71. The potential energy of a body is, increases in which of the following, cases?, (a), (b), (c), (d), , If work is done by conservative force, If work is done against conservative force, If work is done by non-conservative force, If work is done against non- conservative, force, , 72. The potential energy, i.e. U (x ) can be, assumed zero when, (a) x = 0, (b) gravitational force is constant, (c) infinite distance from the gravitational, source, (d) All of the above, , 73. The ratio of spring constants of two, springs is 2 : 3. What is the ratio of their, potential energy, if they are stretched, by the same force?, (a) 2 : 3, (b) 3 : 2, (c) 4 : 9, (d) 9 : 4, , Direction Answer the questions from, 76-80 on the following case., Principle of Conservation of Energy, Total energy of an isolated system always, remains constant. Since, the universe as a, whole may be viewed as an isolated system,, total energy of the universe is constant. If one, part of the universe loses energy, then other, part must gain an equal amount of energy., The principle of conservation of energy, cannot be proved as such. However, no, violation of this principle has been observed., , 76. When we rub two flint stones together,, got them to heat up and to ignite a heap, of dry leaves in the form of, (a) chemical energy, (c) heat energy, , 77. Which graph represents conservation of, total mechanical energy?, Energy, , –xm, , U, xm, , E=K+U, , X, , Energy, U, , (b), –xm, , increases by 15 J when stretched by, 3 cm. If it is stretched by 4 cm, the, increase in potential energy is, (b) 30 J, (d) 36 J, , stretched through a distance x is 10 J., What is the amount of work done on, the same spring to stretch it through an, additional distance x?, (b) 20 J, (d) 40 J, , K, xm, Energy, U, , (c), –xm, , 75. The potential energy of a spring when, , (a) 10 J, (c) 30 J, , K, , (a), , 74. The potential energy of a spring, , (a) 27 J, (c) 33 J, , (b) sound energy, (d) electrical energy, , K, xm, , K=E, , X, , E=K+U, , X, , Energy, U, , (d), –xm, , K, xm, , E=K+U, , X

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78. In the given curved road, if particle is, released from A, then, , 79. U is the potential energy, K is the, kinetic energy and E is the mechanical, energy. Which of the following is not, possible for a stable system?, , m A, , (a) U > E, h, , (b) U < E, , (c) E > K, , (d) K > E, , 80. A body of mass 5 kg is thrown, , B, , (a) kinetic energy at B must be mgh, (b) kinetic energy at B must be zero, (c) kinetic energy at B must be less than mgh, (d) kinetic energy at B must not be equal to, potential energy, , vertically up with a kinetic energy of, 490 J. The height at which the kinetic, energy of the body becomes half of the, original value is, (a) 12.5 m, (c) 2.5 m, , (b) 10, (d) 5 m, , ANSWERS, Multiple Choice Questions, 1. (c), 11. (a), 21. (c), , 2. (c), 12. (d), 22. (c), , 3. (b), 13. (a), 23. (b), , 4. (c), 14. (a), 24. (a), , 5. (b), 15. (c), 25. (b), , 6. (d), 16. (b), 26. (c), , 7. (d), 17. (c), 27. (b), , 8. (b), 18. (a), 28. (a), , 9. (c), 19. (a), 29. (c), , 10. (d), 20. (d), 30. (d), , 31. (d), 41. (b), , 32. (c), 42. (a), , 33. (b), 43. (d), , 34. (a), 44. (c), , 35. (b), 45. (b), , 36. (b), 46. (b), , 37. (b), 47. (d), , 38. (c), 48. (b), , 39. (c), 49. (d), , 40. (a), 50. (b), , 54. (d), , 55. (c), , 56. (c), , 57. (b), , 58. (b), , 59. (a), , 60. (c), , 63. (b), 73. (b), , 64. (d), 74. (a), , 65. (c), 75. (c), , 66. (d), 76. (a), , 67. (c), 77. (c), , 68. (a), 78. (a), , 69. (c), 79. (a), , 51. (c), , Assertion-Reasoning MCQs, 52. (a), , 53. (c), , Case Based MCQs, 61. (a), 71. (b), , 62. (d), 72. (d), , 70. (d), 80. (d), , SOLUTIONS, 1. Here, work is done by the frictional force on, the cycle = - 200 ´ 10 = - 2000 J., As the road is not moving, hence work done, by the cycle on the road is zero., , 2. Given, F = 50 N, W = 150 J, and s = 3 m, Work done, W = Fs cos q, 150 = 50 ´ 3 ´ cos q, 150, cos q =, =1, 150, Þ, q = 0°, , 3. Net force, F = 2$i + 3$j - 2k$ + 5$i - $j - 3k$, = 7 $i + 2$j - 5k$, Displacement, d = 2i$ - $j + 3k$ - $i - $j - k$, = $i - 2$j + 2k$, Work done = F × d = ( 7 i$ + 2$j - 5k$ ) × ( $i - 2$j + 2k$ ), = 7 - 4 - 10 = - 7 units, , 4. Work done by a variable force on the, particle,, W = ò F × dr = ò F × ( dxi$ + dy$j)

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\ In two dimension, dr = dx$i + dy$j, and it is given F = - xi$ + y$j, W = ò ( - x$i + y$j) × ( dxi$ + dy$j), , \, , = ò - x dx + y dy, = ò - x dx +, , ò y dy, , As particle is displaced from A(1, 0 ) to B( 0, 1),, so x varies from 1 to 0 and y varies from 0 to, 1., So, with limits, work will be, 0, , W = ò - x dx +, 1, , 0, , ò, , 1, , 0, , y dy, 1, , é - x2 ù, é y2 ù, =ê, +, ú, ê ú, ë 2 û1 ë 2 û0, 1, = [ 0 - ( - 1) 2 + (1) 2 - 0 ] = 1 J, 2, , 5. Work done = Area under F -s curve, WAB = W12 + W23 + W34 + W45, = Area under AP + Area under PQ, + Area under QR - Area above RB, 1, = 10 ´ 1 + (10 + 15), 2, 1, 1, ´ 1 + ´ 1 ´ 15 - ´ 1 ´ 15, 2, 2, = 10 + 12.5 = 22.5 J, , 6. Work done in stretching a string to obtain an, extension l,, 1 2, kl, 2, Similarly, work done in stretching a string to, obtain an extension l 1 is, 1, W2 = kl 12, 2, \ Work done in second case,, 1, W = W2 - W1 = k ( l 12 - l 2 ), 2, l, 7. The weight of hanging part æç ö÷ of chain is, è 3ø, æ1 ö, ç mg ÷ . This weight acts at the centre of, è3 ø, gravity of the hanging part, which is at a, æl ö, distance of ç ÷ from the table., è6ø, W1 =, , Hence, work required to pull hanging part,, , \, , W = force ´ displacement, mg l mgl, W =, ´ =, 3 6 18, , 8. Gravitational force is a conservative force, and work done by or against the force in, moving a body depends only on the initial, and final positions of the body and not on, the nature of path followed by it., So, W1 = W2 = W3, 1, 9. As we know that, KE = mv 2, 2, So, kinetic energy is directly proportional to, the square of velocity., K µv 2, As this equation resembles equation of, parabola as m is constant, hence option (c), represents a parabola., , 10. The kinetic energy K and momentum p of a, body are related as, p2, , where m is the mass of the body, K =, 2m, Here, p = 6 N-s and m = 4 kg, (6 )2, K =, = 4.5 J, 2´4, , 11. The kinetic energy of the particle is, 1, K = mv 2, 2, As, momentum, p = mv, or, p 2 = m 2v 2, p2, or, v2 = 2, m, 1 æ p2 ö p2, \, K = mç 2÷ =, 2 èm ø 2m, p2, 2m, where, p is the momentum and m is the mass, of the body., As,, (given), p1 = p2, K1 m2 5, \, =, =, K2 m1 4, , 12. Kinetic energy of a body, K =, , 13. Kinetic energy of a body, K =, or, , p = 2mK, , p2, 2m

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(given), Since,, KH =KL, where, subscripts H and L represents for, heavy and light bodies., pH, mH, \, =, pL, mL, So,, and, , mH >mL, pH > pL, , 14. Given, mass, m = 5 kg, Radius, R = 1 m, Revolution per minute, w = 300 rev/min, = ( 300 ´ 2p ) rad/min, 300 ´ 2 ´ p, rad/s = 10 p rad/s, =, 60, Linear speed, v = w R = 10 p ´ 1, = 10p m/s, 1, 2, \ KE = mv, 2, 1, = ´ 5 ´ (10 p ) 2, 2, 1, = 100 p 2 ´ 5 ´, 2, = 250 p 2 J, , 15. Applying work-energy theorem on both, moving objects,, 1, m 1v 12 = Fx 1, 2, 1, and, m 2v 22 = Fx 2, 2, Since, both moving objects have same kinetic, energy,, 1, 1, i.e., m 1v 12 = m 2v 22 Þ Fx 1 = Fx 2, 2, 2, Þ, x1 = x2, Therefore, both the objects will come to rest, at the same distance., , 16. Work done in F is given by DW = F × d, By substituting given values, we get, Þ, DW = ( 3i$ - 12$j ) × ( 4 $i ), … (i), Þ, DW = 12 J, Now, using work-energy theorem, we get, work done, DW = change in kinetic, energy, DK, or, … (ii), DW = K 2 - K 1, , Comparing Eqs. (i) and (ii), we get, K 2 - K 1 = 12 J or K 2 = K 1 + 12 J, Given, initial kinetic energy, K 1 = 3 J, \ Final kinetic energy, K 2 = 3 J + 12 J = 15 J, , 17. \ Work done on the particle, = Area under the curve ABC, W = Area of square ABFO + Area of DBCD, + Area of rectangle BDEF, 1, = 2 ´ 2 + ´ 1 ´ 1 + 2 ´ 1 = 6.5 J, 2, Now, from work-energy theorem,, DW = K f - K i, (Q K i = 0), Þ K f = DW = 6.5 J, , 18. Potential energy of brick above the earth’s, surface is given by, U = mgh, i.e., U µh, Hence, when a brick is lifted above the, surface of the earth, then its potential energy, increases., , 19. The potential energy of the spring is, 1 2, …(i), ky, 2, Now, it is stretched from y to 2y, so its, potential energy becomes, 1, E ¢ = k ( 2y ) 2, 2, [using Eq. (i)], = 2ky 2 = 4 E, \ The increase in its potential energy is, DE = E ¢ - E = 4 E - E = 3E, E =, , 20. Energy received by the boys from bread, = 5000 cal = 5000 ´ 4.2, = 21 ´ 10 3 J, According to law of conservation of, mechanical energy,, 28, mgh =, ´ 21 ´ 10 3, 100, 28 ´ 21 ´ 10 3, \, h =, = 10 m, 100 ´ 9.8 ´ 60, , 21. As the body is falling freely under gravity,, the potential energy decreases and kinetic, energy increases but total mechanical energy, (PE + KE) of the body and earth system will, be constant as external force on the system is, zero.

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22. At a height, , 4, h,, 5, , 26. According to the law of conservation of, , 4, 4, Potential energy = mg ´ h = mgh, 5, 5, Total energy = mgh, \ Kinetic energy at that height, 4, 1, =mgh - mgh = mgh, 5, 5, 4, \ At a height h, the ratio of, 5, 1, mgh, KE 5, 1, =, =, 4, PE, mgh 4, 5, , 23. The work done on the spring is stored as the, PE of the body and is given by, U =ò, or, , x2, x1, , U =ò, , x2, x1, , F ext dx, kx dx, , 1, k ( x 22 - x 12 ), 2, 800, =, [( 015, . ) 2 - ( 0.05) 2 ], 2, = 400 (0.2 ´ 0.1) = 8 J, =, , 24. According to work-energy theorem,, loss in kinetic energy = work done against, friction + potential energy of spring, 1, 1, mv 2 = f × x + kx 2, 2, 2, 1, 1, 4, Þ, ´ 2 ( 4 ) = 15 x + ´ 10000 x 2, 2, 2, 2, Þ 5000 x + 15x - 16 = 0, \, x = 0.055 m = 5.5 cm, , 25. Net work done in sliding a body up to a, height h on an inclined plane, = Work done against the gravitational, force + Work done against the frictional force, …(i), Þ, W = Wg + W f, But, W = 300 J, Given, m = 2 kg and h = 10 m, Wg = mgh = 2 ´ 10 ´ 10 = 200 J, Putting these values in Eq. (i), we get, 300 = 200 + W f, Þ, W f = 300 - 200 = 100 J, , energy, U i + K i = U f + K f, So, by putting the values in Eq. (i),, 1, 3U 0 + 0 = 2U 0 + mv 2, 2, 2U 0, Þ, v =, m, , ...(i), , 27. FBD of pebble,, A, , C, , D Pebble, mg cosθ, B, mg sinθ, mg, 2, , mv, mv 2, Þ T = mg cos q +, l, l, Tension is maximum, when cos q = 1 and, velocity is maximum. Both conditions are, satisfied at q = 0°, i.e. at lowest point B., , \ T - mg cos q =, , 28. At top point, the tension (T H ) in string, becomes zero, so velocity of the particle is, v H = gr, At the bottom, the velocity of the particle is, v L = 5gr, Therefore, the ratio of kinetic energies at, bottom and top is, 1, mv L2 æ ö 2, KL, v, 2, =, =ç L÷, K H 1 mv 2 è v H ø, H, 2, 5 gr 5, =, = = 5 :1, 1, gr, Hence, the ratio of kinetic energies is 5 : 1., , 29. Given, W = 600 J, and t = 2 min = 2 ´ 60 = 120 s, W 600, \ Power, P =, =, =5W, t, 120, dW, 30. By definition, P =, dt, Q Work done = Kinetic energy, dW d (KE), P=, =, = constant, Þ, dt, dt

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W, t, Since, K i = initial KE = 0, 1, and, K f = final KE = mv 2, 2, From work-energy theorem,, work done = change in KE, 1 2, mv - 0, K f -K i, \, P =, = 2, t, t, mv 2, Þ, P =, 2t, , 36. From conservation of momentum,, , 31. Power, P =, , m 1 v1 + m 2 v2 = (m 1 + m 2 ) v, $, 1 ´ ( 3i - 2$j) + 2 ´ ( 4 $j - 6 k$ ) = (1 + 2) v, Þ 3$i + 6 $j - 12k$ = 3 v, \ Velocity,, , For elastic collision,, æm - m 1 ö, 2m 1v 1, v2 = ç 2, ÷ v2 +, èm 1 + m 2 ø, m1 + m2, , to pull the lift,, F = weight carried + friction = mg + f, = ( 2000 ´ 10 ) + 4000 = 24000 N, Power delivered by motor at speed v of load,, P = F ´v, P 60 ´ 746, Þ, v =, =, = 1865, ., = 1.9 ms -1, F, 24000, dv, dv, 33. As, P0 = Fv = æçm ö÷ v = mv, è dt ø, dt, Þ P0 × dt = mvdv, Integrating both sides, we get, 0, , Þ, Þ, Þ, , P0 dt = ò mv dv, mv 2, 2, 2P0 t, 2, v =, m, v µ t 1/ 2, , P0 t =, , 34. From Newton’s second law, F =, , After putting given values, we will get, 2m 1v 1, v2 =, Þ v2 =v1, 2m 1, , 38. When we are considering the two bodies as, system, the total external force on the system, will be zero., Hence, total linear momentum of the system, remain conserved., , 39. Speed of objects = u ms -1, Since, both objects collide with 90°., According to the law of conservation of, momentum,, Total moment before collision, = Total momentum after collision, $, $, |mui + muj| = p, f, , m 2u 2 + m 2u 2 = p f Þ p f = 2 mu, , 40. Given, mass of body, m 1 = 5 ´ 10 3 kg, dp, dt, , dp, =0, dt, Þ p = constant, Thus, if total external force acting on the, system is zero during collision, then the, linear momentum of the system remains, conserved., , If, , v = | v| = 1 + 4 + 16 = 4.6 ms -1, , 37. Given, mass, m 1= m 2 = m and velocity, v = v 1, , 32. At maximum load, force provided by motor, , ò, , v = $i + 2$j - 4 k$, , Þ, , F = 0, then, , and mass of another body, m 2 = 15 ´ 10 3 kg, Velocity, v 1 = 2 ms -1, For perfectly inelastic collision ( e = 0 ),, Loss in kinetic energy of system,, 1 m 1m 2, DE K =, ´ v 12, 2m1 + m2, =, , 35. As, we know in a elastical collision of two, identical bodies, i.e. m A = m B , the particles, mutually exchange their velocities., So, ( v i ) A = 0.5 ms -1 and ( v i ) B = -0.3 ms -1 ., After collision,, ( v f ) A = - 0.3 ms -1 and ( v f ) B = 0.5 ms -1 ., , 1 5 ´ 10 3 ´ 15 ´ 10 3, ´, ´ 22, 2 5 ´ 10 3 + 15 ´ 10 3, , = 7.5 ´ 10 3 J = 7.5 kJ, , 41. Kinetic energy of the body,, , p2, 2m, Since, the mass remains constant, so K µ p 2 ., K =

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2, , Hence, total work done = - mgL + mgL = 0., As the point of application of the contact, forces does not move, hence work done by, reaction forces will be zero., , K 2 p 22 é150 ù, 9, =, = 2 =ê, ú, K 1 p 1 ë100 û, 4, , \, , æK, ö, ö, æ9, Thus, ç 2 - 1÷ ´ 100 = ç - 1÷ ´ 100, ø, è, èK 1, ø, 4, = 125%, , 47. If the work done or the kinetic energy, , 42. As, work done = force ´ displacement, As, there is no displacement produced in the, wall, so work done by the ball on the wall is, zero., Alternative Method As, there is no change, in kinetic energy of the ball, so according to, work-energy theorem, work done should be, zero., , 43. According to the law of conservation of, energy,, 1, 1 æ1, ö, mv 2 = ç mv 2 ÷ + mgh, ø, 2, 2è2, Þ, , 490 = 245 + 5 ´ 9.8 ´ h ,, 245, h =, = 5m, 49, , ÞPA : PB =, , mgh 1 /t 1 æ h 1 ö, =ç ÷, mgh 2 /t 2 è h 1 ø, , …(i), , æ t2 ö t2 4 2, ç ÷= = =, è t1 ø t1 2 1, , PA : PB = 2 : 1, dk, 45. Given, k µ t Þ, = constant, dt, Þ, K µt, 1 2, mv µ t Þ v µ t, 2, dK, Also, P = Fv =, = constant, dt, 1, 1, F µ Þ, F µ, Þ, v, t, Þ, , Þ, , 48. In elastic collision, the conservation of, mechanical energy consider only, conservative force while conservation of, energy consider both conservative and, non-conservative force., Mass converted into energy in nuclear, reaction is called nuclear energy., Thus, the statement given in option (b) is, correct, rest are incorrect., , 49. Whenever there is a collision between two, , 44. Given, t 1 = 2 s, t 2 = 4 s, and h 1 = h 2 = h, mgh 1, mgh 2, and PB =, As, PA =, t1, t2, , depend on other factors such as the velocity, or the particular path taken by the object, the, force would be called non-conservative., Thus, the statements given in options (a) and, (b) are correct, rest is incorrect., , F µ t -1 / 2, , 46. When a man of mass m climbs up the, staircase of height L, work done by the, gravitational force on the man is mgl, work, done by internal muscular forces will be mgL, as the change in kinetic energy is almost, zero., , bodies, the total momentum of the bodies, remains conserved. If after the collision of, two bodies, the total kinetic energy of the, bodies remains the same as it was before, collision, then it a perfectly elastic collision., A ball bearing striking another ball bearing is, an example of elastic collision. If two bodies, strick together after the collision, then the, collision is said to be perfectly inelastic, collision., Options (a), (b) and (c) are examples of, perfectly inelastic collisions., , 50. Work done by an agent is given by, W = F × s = Fs cos q, where, F is the applied force, s is the, displacement and q is the smaller angle, between F and s ., A. If q < 90° , i.e. acute angle, then work done, is positive, as in case of coolie lifting, luggage., B. If q = 90° , i.e. right angle, then work done, is zero, as in case of satellite rotation, around the earth., C. If q > 90° , i.e. obtuse angle, work done is, negative, as in case of friction., Hence, A ® r, B ® q and C ® p.

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51. As, power, P µ t, So,, , W = ò Pdt = ò at dt, , or, W µt2, Since, work done is equal to change is KE., Hence, v 2 µ t 2 or v µ t, ds, Further, v =, dt, ds, \, µ t or ds µ t dt, dt, (by integration), or, s µt2, Hence, A ® p, B ® q and C ® q., , 52. According to work-energy theorem, work, done by a body is equal to change in kinetic, energy of the body., 1, …(i), Þ, W = DKE = mv 2, 2, But, W = stopping force ´ stopping distance, …(ii), W = F ×d, From Eqs. (i) and (ii), we have, Stopping distance (d), ö, æ1, Kinetic energy ç mv 2 ÷, ø, è2, =, Stopping force ( F ), Therefore, both A and R are true and R is, the correct explanation of A., F, 1, 53. As we know, k =, Þ k µ, l, l, k 2 l1 1, = =, Þ, k 1 l2 2, , In series,, , k 1 = 2k , k 2 = k, 1 1, 3, 1, 1, 1, =, +, =, + =, k ¢ k 1 k 2 2k k 2k, , 2k, 3, Therefore, A is true but R is false., , Q, , k¢=, , 54. According to the law of conservation of, mechanical energy, for conservative forces,, the sum of kinetic energy and potential, energy remains constant and throughout the, motion it is independent of time., This is the law of conservation of mechanical, energy, i.e. KE + PE = total energy =, constant., Therefore, A is false and R is also false., , 55. Mechanical energy consists of both PE and, KE. In the given cases, some of the, mechanical energy is converted into heat, energy and it is more in the case when, inclination is less due to increased (as q, decreases, value of cos q will increases), friction force on an inclined plane., f r = mmg cos q, The coefficient of friction does not depend, on the angle of inclination of the plane. It, depends only on the nature of surfaces in, contact., Therefore, A is true but R is false., , 56. At the lowest point of a vertical circle, the, minimum velocity at bottom,, v min = 5gr, Velocity at highest point, v = gr, Therefore, A is true but R is false., Work done (or energy), 57. Power =, Time, Þ Work done = Power ´ Time, W = P ´t, If, P = 1 kilowatt, t = 1 hour, then, W = 1 kilowatt ´ 1 hour, = 1 kilowatt-hour, = 10 3 watt ´ 60 ´ 60 s, . ´ 10 6 J, = 36, Therefore, both A and R are true but R is, not the correct explanation of A., , 58. In elastic collision, total energy, kinetic energy, and momentum remain conserved, therefore, no loss in energy occurs in elastic collision., Therefore, both A and R are true but R is, not the correct explanation of A., , 59. As momentum, p = mv or p µ v , i. e., momentum is directly proportional to its, velocity, so the momentum is greater in a, quicker collision between two bodies than in, slower one., Hence, due to greater momentum quicker, collision between two bodies will be more, violent even initial and final velocities are, identical., Therefore, both A and R are true and R is, the correct explanation of A.

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60. If two particles are initially moving in the, same direction, then their resultant, momentum will not be zero. Therefore, their, resultant momentum cannot be zero after a, completely inelastic collision., As, kinetic energy is directly proportional to, the square of the momentum, hence kinetic, energy cannot be zero. This implies, not all, the energy in inelastic collision is lost., Therefore, A is true but R is false., , 61. When earth is moving around the sun in a, circular orbit, then gravitational attraction on, earth due to the sun provides required, centripetal force, which is in radially inward, direction, i. e. in a direction perpendicular to, the direction of motion of the earth in its, circular orbit around the sun., As a result, the work done on the earth by, the force will be zero. i.e. W = Fd cos 90° = 0., , 62. Work done by weight-lifter is zero, because, there is no displacement., In a locomotive, work done is zero because, force due to gravity and displacement are, mutually perpendicular to each other., In case of a person holding a suitcase on his, head and standing at a bus terminal, work, done is zero because there is no, displacement., Hence, options (a), (b) and (c) are correct., 63. Given, F = ( 3$i + 4 $j - 5k$ ) unit, and, d = ( 5i$ + 4 $j + 3k$ ) unit, \, , F × d = F x d x + F yd y + F z d z, = 3 ( 5) + 4 ( 4 ) + ( -5) ( 3), = 16 units, , Now,, , F × F = F 2 = F x2 + F y2 + F z2, = 9 + 16 + 25, = 50 units, , Þ, and, , F = 50 units, d × d = d 2 = d x2 + d 2y + d z2, = 25 + 16 + 9, , Þ, \, , = 50 units, d = 50 units, 16, cos q =, 50 50, , =, , 16, = 0.32, 50, , F × dö, æ, çQ cos q =, ÷, è, Fd ø, , q = cos -1 ( 0.32), , Þ, , 64. The work done in displacing an object by, applying force F is given by, W = F × s = Fs cos q, So, work done will be zero, when, (i) either applied force F or displacement s is, zero., (ii) the force and displacement are mutually, perpendicular to each other. i.e. q = 90°., So, all statements are correct., , 65. Force between two protons is same as that of, between proton and a positron., As positron is much lighter than proton, it, moves away through much larger distance, compared to proton., We know that, work done = force ´ distance., As, forces are same in case of proton and, positron but distance moved by positron is, larger, hence work done will be more in case, of positron., , 66. When the earth is closest to the sun, speed, of the earth is maximum, hence KE is, maximum. When the earth is farthest from, the sun, speed is minimum, hence KE is, minimum but never zero and negative., This variation is correctly represented by, option (d)., K, 67. F =, (given), v, dv K, m, =, dt, v, Þ, Þ, Þ, , ò mv dv = ò K dt, m, , v2, = Kt, 2, KE µ t, , 68. The kinetic energy of an air molecule is, =, , 10 -21 J, ~, - 0.0062 eV, 16, . ´ 10 -19 J/eV, , This is the same as 6.2 meV., , 69. As we know that, linear momentum, p, = 2mK, , æ, p2 ö, çQ K =, ÷, è, 2m ø

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For same kinetic energy, p µ m, p1, m1, 1 1, =, =, = =1: 2, p2, m2, 4 2, , 70. Initial velocity = 10 ms -1, Final velocity = 20 ms -1, 1, Initial KE = ´ 10 ´ 10 ´ 10 = 5 ´ 10 2 J, 2, 1, Final KE = ´ 10 ´ 20 ´ 20 = 20 ´ 10 2 J, 2, ( 20 - 5) ´ 10 2, ´ 100 = 300%, % increase =, 5 ´ 10 2, , 71. Potential energy of a body increases when, work is done against a conservative force, e.g., if we raise the height of an object, its, potential energy increases because work is, done against gravitational force which is a, conservative force., , 72. The zero of the potential energy is arbitrary., It is set according to convenience. For the, spring force, we took U ( x ) = 0, at x = 0, i.e., the unstretched spring had zero potential, energy., For the constant gravitational force mg , we, took U = 0 on the earth’s surface., Also, for the force due to the universal law of, gravitation, the zero is best defined at an, infinite distance from the gravitational source., , 73. F = k 1x 1, F = k 2x 2, k, x, k 1x 1 = k 2x 2 Þ 1 = 2, k 2 x1, PE (1) k 1 x 12, =, PE (2) k 2 x 22, 2, , æk ö, 3, k, k, = 1 ´ ç 2÷ = 2 =, k2 èk1 ø, k1 2, 1, 2, , 74. PE of spring = kx 2 Þ PE µ x 2, \, , PE = 15 ´, , When x becomes 2x, the potential energy will, be, 1, 1, U ¢ = k ( 2x ) 2 = 4 ´ kx 2, 2, 2, = 4 ´ 10 = 40 J, \ Work done = U ¢ - U = 40 - 10 = 30 J, , 76. One of the greatest technical achievements of, human kind occurred when we discovered, how to ignite and control fire . We learnt to, rub two flint stones together (mechanical, energy), got them to heat up and to ignite a, heap of dry leaves (chemical energy), which, then provided sustained warmth., , 77. Parabolic plots of the potential energy U and, kinetic energy K of a block attached to a, spring obey in a Hooke’s law. The two plots, are complementary, one decreasing as the, other increases. The total mechanical energy, E = K + U remains constant., Energy, U, , –xm, , X, , KE depends only on initial and final point, and not on path covered, i.e. at B, KE = mgh ., , 79. We know that, PE + KE = Mechanical energy, U +K =E, Þ, U = E -K, Now, K can never be negative, so, U <E, K = E -U, Now, U can be negative, so K > E is possible., , 80. According to the law of conservation of, energy,, 1, 1 æ1, ö, mu 2 = ç mu 2 ÷ + mgh, ø, 2, 2è2, , 16 ~, ( 4), = 15 ´, - 27 J, 9, ( 3) 2, , through a distance x,, 1, U = kx 2 = 10 J, 2, , K, xm, , 78. In a conservative field loss of PE or gain of, , 2, , 75. Potential energy of the spring when stretched, , 0, , E=K+U, , Þ, , 490 = 245 + 5 ´ 9.8 ´ h, 245, h =, = 5m, 49