Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Downloaded from www.studiestoday.com Unit - 6 Gravitational 147 Downloaded from www.studiestoday.com Downloaded from www.studiestoday.com SUMMARY 1. Gravitational force between two point masses is F 2. Gm1 m2 r2 Acceleration due to Gravity (I) (II) GM 9.81ms 2 R2 At a height h from surface of earth on the surface of earth = g g1 h 1 R (III) 2 g (1 2h ) if h << R R At a depth d form the surfce of earth d ) R g1 = g if d = R i. e. on the surface of earth (IV) Effect of rotation of earth at latitude g1 = g – R2 cos2 - at the equator = 0 g1 = g – R2 = minimum value - At the pole = 900 g1 = g – R2 = maximum value - At the equator effect of rotation of earth is maximum and value of g is minimum. - At the pole effect of rotation of earth is zero and value of g is maximum. Field Strength Gravitational field strength at a point in gravirtational field is defined as, g 1 g /(1 3. F = gravitational force per unit mass m Due to point mass E GM 1 (towards the mass) E 2 2 r r Due to solid sphere E GM r R3 At r = 0, E = 0 at the center inside points Ei At r = R, E GM i.e. on the surface R2 out side points Eo = GM 1 or E o 2 2 r r 148 Downloaded from www.studiestoday.com Downloaded from www.studiestoday.com At r , E 0 Due to a sphericell shell inside points E=0 GM r2 outside points E0 = GM R2 on the surface E – r graph is discontinuous on the axis of a ring just outside surface E Er GMr (R2 r 2 ) 3 2 At r = 0, E = 0 i.e. at the center If r >> R, E 4. GM , i.e. ring behaves as a points mass r2 As r E 0 Gravitational potential :(i) Gravitational potential at a point in a gravitational field is defined as the negative of work done in moving a unit mass from infinity to that point per unit mass, thus Vp w wp m (ii) Due to point mass V Gm rm v as r 0 and v 0 as r (iii) Due to solid sphere inside points Vi – GM (1.5 R2 0.5r 2 ) R GM i.e. on the surface R V - r graph is parabola for inside points as r = R V GM r Due to sphercal shell out side points v (iv) inside points Vi GM R outside points Vi GM R 149 Downloaded from www.studiestoday.com Downloaded from www.studiestoday.com (v) on the axis of a ring Vr GM R2 r 2 GM i.e. at center R Gravitational potential Energy (i) This is the work done by gravitational forces in arranging the system from infinite sepration in the present position (ii) Gravitational potential energy of two point massess is at r = 0, V 5. Gm1m2 r (iii) To find the gravitational points energy of more than two points masses we have to make pairs of masses. Neighter of the pair should be repeated. For example in case of four point masses. U m4 m3 m4 m2 m4 m1 m3 m2 m3 m1 m2 m1 U G r23 r41 r32 r31 r21 r43 for n point masses total number of pairs will be (iv) n( n 1) 2 If a point mass m is plaled on the surface of earth the potential energy here is Uo Uo GMm R and potential energy at height h is Uh GMm (R h) the difference in potential energy would be mgh U = Uh – Uo = 1 + h/r If h << R, U = mgh 6. Relation between field strength E and potential V (i) if V is a function of only one variable (Say r) then dV E slope of U r graph dr (ii) If V is funtion at three coordinates variable; e x, y, and z then v v v E iˆ ˆj kˆ x y z 150 Downloaded from www.studiestoday.com Downloaded from www.studiestoday.com 7. Escape velocity (i) From the surface of earth Ve 2 gR 8. 2GM GM as g 2 R R =11.2 km / sec (ii) Escape velocity does not depend upon the angle which particle is projected form the surface and the mass of body Motion of satellites GM r (i) orbital speed Vo (ii) time period T 2 r 3/ 2 T 2 r 3 GM (iii) Kinetic energy K GMm 2r (iv) Potential energy U GMm ... r (v) Total Mechanical energy. E 9. GMm r Kepler’s laws - First law : Each planet moves in an elliptical orbit with the sun at one focus of ellipse - Second law : The radius vectors drawn form the sun to a planet, sweeps out equal area in equal time interval i.e. areal Velocity is constant. this law is derived from the law of conservation of angular momentum dA L dt 2m = constant here L is the angular momentum and m is mass of planet - Third law - - T 2 r3 where r is semi-major axis of elliptical path The gravitational force acting between two bodies is always attractive. It is independent of medium between bodies. It holds good over a wide range of distance. It is an action and reaction pair. It is conservative force. It is a central force and obey inverse square law as F 1/ r 2 The value of G is never zero any where but the value of g is zero at the center of earth. the acceleration due to gravity is independent of mass, shape, size etc of falling body. the rate of decrease of the acceteration due to gravity with height is twice as compared to that with depth. It the rate of rotation of earth increases the value of acceleration due to gravity decreases at all points on the surface of earth except at poles. 151 Downloaded from www.studiestoday.com Downloaded from www.studiestoday.com - If the radius of planet decreasees by n% keeping its mas unchanged, the accelreotion due to gravity on its surface increases by 2n%. If the mass of a planet increases by n% keeping its radius unchanged the acceleration due to gravity on its surface increases by n%. The value of g at a location gives the value of intensity of gravitational field at the location. The orbital velocity of a satellite is independent of mass of the satellite but depends upon the mass and radius of planet around which the rotation is taking place The value of orbital velocity for a satellite near the surface of earth is 7.92 kms–1. The direction of orbital velocity of satellite at an instant is along the tangent to the orbital path at that instant. The work done by a satellite in a complete orbit is zero. For a satellite orbiting close to the surface of earth (h << R), the time period of revolution 2 g / R 0.001237rad / sec T A geostationary satellite revolves around the earth from west to east. Its period of revoluton is one day i.e. 24 hours. The orbital velocity of geostationary satellite is 3.08 kms–1. Its height above the surface of earth is about 36000 km. The relative angular velocity of geostationary satellite w.r.t earth is zero. When a satelilte is orbiting in its orbit, no energy is required to keep it in its orbit. When the total energy of a satellite is negative, it will be moving in either a circular or an elliptical orbit. When the total energy of a satellite is zero, it will escape away from its orbit and its path becomes parabolic. When the height of satellites is increased its potential energy will increase and K.E. will decrease. When the velocity of satellite is increased, its total energy will increase and it will start orbiting in a circular path of larger radius. For a satellite orbiting in a circular orbit, the value of potenial energy is always greater than its K.E. If the velocity of a satellite orbiting the earth is increased by 41.4% or its K.E. is doubled, then it will escape away from the gravitational field of earth is 2 R / g 84.6 minutes and angular velocity. - - 2 orbital velocity If the gravitational force is inversely proportional to the nth power of distance r, then the orbital velocity of a satellite Escape velocity =. V0 r - n/ 2 and time period is T r ( n 2) 2 When a body is projected horizontally with velocity v, from any height from the surface of earth, then the following possibilities are there. (i) If v < v0, the body fails to revolve around the earth and finally falls to the surface of earth. (ii) If v = v0, the body will revolve around the earth in circular orbit. (iii) If v < ve the body will revolve around the earth in elliptical orbit. (iv) If v = ve, the body will escape from the gravitational field of earth. (v) If v > ve the body will escape, following a hyperbolic path. 152 Downloaded from www.studiestoday.com Downloaded from www.studiestoday.com - 1 th of its present size without any change in mass, the duration n of the day will be nearly 24/n2 hours. Force function F(r) is related with potential energy function U(r) by a relaion - dU dr A given planet will have atmosphere if the root mean square velocity of molecules in its atmosphere - If earth suddently contracts to F (i.e. Vrms = - In the weightlessness state, the bodies donot have weight but they do possess inertia on account of their mass. the bodies floating inside the space craft may collide with each other and crash. If a body is released from a height aqual to n times the radius of earth, then its striking velocity on the surface of earth is - 3RT ) is smaller than escape velocity for that planet. M 2ngR n 1 If polar ice caps melt then moment of inertia, Angular velocity will decrease and period of rotaiton of earth increase. The, line joining the places on earth having same value of g are called isogams. Gravity meter and Etvos gravity balance are used to measure changes in accelaration due to gravity. 153 Downloaded from www.studiestoday.com Downloaded from www.studiestoday.com Newton’s Law of Gravitation For the answer of the following questions choose the correct alternative from among the given ones. 1. 2. 3. 4. 5. 6. 7. 8. 9. Two identical solid copper spheres of radius R are placed in contact with each other. The gravitational force between them is proportional to (A) R2 (B) R-2 (C) R-4 (D) R4 The gravitational force Fg between two objects does not depend on (A) sum of the masses (B) product of masses (C) Gravitational constant (D) Distance between the masses The atmosphere is held to the earth by (A) clouds (B) Gravity (C) Winds (D) None of the above Two sphere of mass m1 and m2 are situated in air and the gravitational force between them is F. The space around the masses is now filled with liquid of specific gravity 3. The gravitational force will now be (A) F (B) 3F (C) F/3 (D) F/9. A satellite of the earth is revolveing in a circular orbit with a uniform speed v. If the gravitational force suddennly disappears, the satellite will (A) Contineue to move with velocity v along the original orbit. (B) Move with a Velocity v, tangentially to the original orbit. (C) Fall down with increasing velocity. (D) Ultimately come to rest somewhere on the original orbit. Correct form of gravitational law is Gm1m 2 Gm1m 2 Gm1m 2 Gm1m 2 ˆ F = – F = – r F =– r (A) F = – (B) (C) (D) 2 2 3 r r r r3 Mass M is divided into two parts xM and (1 - x) M. For a given separation, the value of x for which the gravitational force between the two pieces becomes maximum is (A) 1 (B) 2 (D) 1/2 (D) 4/5 24 The earth (mass = 6 x 10 kg) revolves around the sun with angular velocity 2 10–7 rad/ sec in a circular orbit of radius 1.5 x 108 km. The force exerted by the sun on the earth is = ................... N (A) 18 1025 (b) zero (C) 27 1039 (D) 36 1021 Two particle of equal mass go round a circle of radius r. Under the action of their mutual gravitational force. The speed of each particle is =.................. (A) 10. 11. 1 1 2r Gm (B) Gm 2r (C) 1 Gm 2 r (D) 4Gm r The distance of the moon and earth is D the mass of earth is 81 times the mass of moon. At what distance from the center of the earth, the gravitational force will be zero (A) D/2 (B) 12D/3 (C) 4D/3 (D) 9D/10 One can easily “ Weight the earth ” by calculating the mass of earth using the formula (in usual notation) (A) g Re G (B) g/G Re2 (C) G 2 Re g (D) 154 Downloaded from www.studiestoday.com G 3 Re g Downloaded from www.studiestoday.com 12. Three equal masses of m kg each are plced the vertices of an equilateral triangle PQR and a mass of 2m kg is placed at the centroid 0 of the triangle which is at a distance of 2 m from each of vertices of triangle. The force in newton. acting on the mass 2m is = ............ 13. 14. 15. (A) 2 (B) 1 (C) 2 (D) zero Which of the follwing statement about the gravitational constant is true (A) It is a force (B) It has no unit (C) It has same value in all system of unit (D) It depends on the value of the masses. Two point masses A and B having masses in the ratio 4 : 3 are seprated by a distance of lm. When another point mass c of mass M is placed in between A and B the forces A and C is 1/3rd of the force between band C, Then the distance C form A is = ............... m (A) 23 (B) 1/3 (C) 1/4 (D) 2/7 The gravitational force between two point masses m1 ans m2 at separation r is given by mm F G 1 2 2 The constant k ............... r (A) Depends on system of units only. (B) Depends on medium between masses only. (C) Depends on both (a) and (b) (D) is independent of both (a) and (b) Acceleration Due to Gravity 16. As we go from the equator to the poles, the value of g ............... (A ) Remains constant (B) Decreases (C) I ncreases (D) Decreases upro latitude of 45 17. If R is the radius of the earth and g the acceleration due to gravity on the earth’s surface, the mean density of the earth is = ............... (A) 4 G / 3gR 18. 19. 20. 22. (B) 3 R / 4 gG (C) 3g / 4 RG (D) RG / 12 g g=10ms-2. The radius of the earth is 6400 km and In order that a body of 5 kg weights zero at the equator, the angular speed of the earth is = ............... rad/sec (A) 1/80 (B) 1/400 (C) 1/800 (D) 1/600 The time period of a simple pendulum on a freely moving artificial satellite is ............... sec (A) 0 (B) 2 (C) 3 (D) Infinite A spherical planet far out in space has mas Mo and diameter Do. A particle of m falling near the surface of this planet will experience an acceleration due to gravity which is equal to (A) GM 0 / Do 2 21. o (B) 4mGMo / Do 2 (C) 4GMo / Do 2 (D) GmMo / Do 2 A body weights 700 g wt on the surface of earth How much it weight on the surface of planet whose mass is 1/7 and radius is half that of the earth (A) 200g wt (B) 1400g wt (C) 50 g wt (D) 300g wt. 2 The value of g on he earth surface is 980 cm/sec . Its value at a height of 64 km from the earth surface is ............... cm5–2 (a) 960.40 (B) 984.90 (C) 982.45 (D) 977.55 155 Downloaded from www.studiestoday.com Downloaded from www.studiestoday.com 23. 24. 25. If earth rotates faster than its present speed the weight of an object will. (A) increases at the equator but remain unchanged of the poles. (B) Decreases at the equator but remain unchanged at poles. (C) Remain unchanged at the equator but decreases at poles. (D) Remain unchanged at the equator but increases at the poles. The moon’s radius is 1/4 that of earth and its mass is 1/80 times that of the earth. If g represents the acceleration due to gravity on the surface of earth, that on the surface of the moon is .............. (A) g/4 (B) g/5 (c) g/6 (D) g/8 The depth of at which the value of acceleration due to gravity becomes 1/n the time the value of at the surface is (R = radius of earth) 26. 27. 28. 29. 30. 31. 32. 33. 34. ( n 1) n n (D) R n 1 If the density of smalll planet is that of the same as that of the earth while the radius of the planet is 0.2 times that of lthe earth, the gravitational acceleration on the surface of the planet is ............... (A) 0.2g (B) 0.4g (c) 2g (D) 4g If mass of a body is M on the earth surface, than the mass of the same body on the moon surfae is (A) M/6 (B) 56 (C) M (D) None of these An object weights 72 N on the earth. Its weight at a height R/2 from earth is = .............. N (A) 32 (B) 56 (C) 72 (D) zero If the radius of earth is R then height ‘h’ at which value of ‘g’ becomes one - fourth is (A) R/4 (B) 3R/4 (C) R (D) R/8 If the mass of earth is 80 times of that of a planet and diameter is double that of planet and ‘g’ on the earth is 9.8 ms-2 , then the value of ‘g’ on that planet is = ............... ms-2 (A) 4.9 (B) 0.98 (c) 0.49 (D) 49 Assuming earth to be a sphere of a uniform density, what is value of gravitational acceleration in mine 100 km below the earth surface = ............... ms-2 (A) 9.66 (B) 7.64 (C) 5.00 (D) 3.1 Let g be the acceleration due to gravity at earth’s surface and k be the rotational K.E. of earth suppose the earth’s radius decreases by 2% keeping alt other quantities same then (A) g decreases by 2% and K decreases by 4 % (B) g decreases by 4% and K increases by 2% (C) g increases by 4% and K increases by 4% (D) g decreases by 4% and K increases by 4% A body weight 500 N on the surface of the earth. How much would it weight half way below the surface of earth (A) 125N (B) 1250N (C) 500N (D) 1000N The radii of two planets are respectively R1 and R2 and their densities are respectively 1 and 2 the ratio of the accelerations due to gravity at their surface is ............... (A) R/n (B) R (A) g1 : g 2 1 2 . 2 2 R1 R2 (C) g1 : g2 R1 2 : R21 (C) R/n2 (B) g1 : g2 R1 R2 : 12 (D) g1 : g2 R1 1 : R22 156 Downloaded from www.studiestoday.com Downloaded from www.studiestoday.com 35. 36. 37. 38. 39. 40. 41. 42. 43. 44. 45. 46. 47. At what height over the earth’s pole, the free fall acceleration decreases by one percent = .................. km (Re = 6400 km). (A) 32 (B) 80 (C) 1.253 (D) 64 Weight of a body is maximum at (A) moon (B) poles of earth (C) Equator of earth (D) Center of earth At what distance from the center of earth, the value of aceeleration due to gravity g will be half that of the surfaces (R = Radius of earth) (A) 2R (B) R (C) 1.414 R (D) o.414 R The acceleration due to gravity near the surface of a planet of radius R and density d is proportional to (A) d/R2 (B) dR2 (C) dR (D) d/R The acceleration due to gravity is g at a point distance r from the center of earth R. if r < R then (A) g r (B) g r2 (c) g r-2 (D) g r-1 The density of a newly discoverd planet is twice that of earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of earth. If the radius of the earth is R, the radius of planet would be ..................... (A) 2R (B) 4R (C) 1/4 R (D) ....... Density of the earth is doubled keeping its radius constant then acceleration, due to gravity will be .................. ms-2(g = 9.8 ms2) (A) 19.6 (B) 9.8 (C) 4.9 (D) 2.45 Weight of body of mass m decreases by 1% when it is raised to height h above the earth’s surface. If the body is taken to a depth h in a mine. change in its weight is (A) 2% decreases (B) 0.5% decreases (C) 1% increases (D) 0.5% increases If density of earth increased 4 times and its radius becomes half of then out weight will be... (A) Four times it present calue (B) doubled (C) Remain same (D) halved A man can jump to a height of 1.5 m on a planet A what is the height ne may be able to jump on another planet whose density and radius are respectively one- quater and one- third that of planet A (A) 1.5 m (B) 15 m (C) 18 m (D) 28 m If the value of ‘g’ acceleration due to gravity, at earth surface fis 10ms–2. its value in ms–2 at the center of earth, which is assumed to be a sphere of Radius ‘R’ meter and uniform density is (A) 5 (B) 10/R (C) 10/2R (D) zero A research satellite of mass 200 kg. circles the earth in an orbit of avrage radius 3R/2 where R is radius of earth. Assuming the gravitational pull 10 N, the pull on the satellite will be =..........N (A) 880 (B) 889 (C) 890 (D) 892 Acceleration due to gravity on moon is 1/6 of the acceleration due to gravity on earth. If the e ratio of densities of earth e and moon m is 5 3 then radius of moon Rm in terms m of Re will be ............... (A) 5 Re 18 (B) 1 Re 6 (C) 3 Re 16 (D) 1 2 3 157 Downloaded from www.studiestoday.com Re Downloaded from www.studiestoday.com 48. 49. 50. The acceleration of a body due to the attraction of the earth (radius R) at a distance 2R from the surface of the earth is = ............... (g = acceleration due to gravity at the surface of earth) (A) g/9 (B) g/3 (C) g/4 (D) 9 The height at which the weight of a body becomes 1/16 th its weight on the surface of (radius R) is (A) 3R (B) 4R (C) 5R (D) 15R A spherical planet has a mass Mp and diameter Dp A particle of mass m falling freely near the surface of this planet will experience an acceleration due to gravity, equal to (A) 51. 4 GMp Dp 2 (B) GMpm Dp 2 (C) GMp Dp 2 (D) 4GMpm Dp 2 Assuming the earth to have a constant density, point out which of following curves show the variation acceleration due to gravity from center of earth to points far away from the surface of earth ............... (D) None of these Gravitational potential, Energy and Escape Velocity 52. 53. 54. In a gravitational field, at a point where the gravitational potential is zero (A) The gravitational field is necessarily zero (B) The gravitational field is not necessarily zero (C) Nothing can be said definetely, about the gravitational field (D) None of these The mass of the earth is 6.00 1024 kg and that of the moon is 7.40 1022 kg. The constant of gravitation G = 6.67 10-11 Nm2 kg–2. The potential energy of the system is -7.79 1028 Joules. the mean distance between the earth and moon is = ............. meter. (A) 3.80 108 (B) 3.37 108 (C) 7.60 108 (D) 1.90 102 The masses and radii of earth and moon are M1, R1 and M2, R2 respectively. Their centres are d distance of apart. The minimum velocity with which a particle of mass m should be projected from a point midway between their centres so that it esacapes to infinity is............... (A) 2 G (M1 M 2 ) d (B) 2 (C) 2 Gm ( M1 M 2 ) d (D) 2 2G (M1 M 2 ) d Gm (M 1 M 2 ) d ( R1 R2 ) 158 Downloaded from www.studiestoday.com Downloaded from www.studiestoday.com 55. A rocket is launched with velocity 10 kms-1. If radius of earth is R then maximum height attained 56. by it will be = .............. (A) 2R (B) 3R (C) 4R (D) 5R What is the intensity of gravitational field at the center of spherical shell Gm (B) g (C) zero (D) None of these r2 Escape velocity of a body of 1 kg. on a planet is 100 ms-1. Gravitational potential energy of the body at the planet is =............... J (A) -5000 (B) -1000 (C) -2400 (D) 5000 A body of mass m kg starts falling from a point 2R above the earth’s surface. Its K.E. when it has fallen to a point ‘R’ above the Earth’s surface = ..................... [R - Radius of Earth, M-mass of Earth G-Gravitational constant] (A) 57. 58. 1 GMm 1 GMm 2 GMm 1 GMm (B) (C) (D) 2 R 6 R 3 R 3 R The Gravitational P.E. of a body of mas m at the earth’s surface is -mgRe. Its gravitational potential energy at a height Re from the earth’s surface will be = ............ here (Re is the radius of the earth) (A) 59. 61. 1 1 mg Re (D) mg Re 2 2 A body is projected vertically upwards from the surface of a planet of radius R with a velocity equal to half the escape velocity for that planet. The maximum height attained by the body is .......... (A) R/3 (B) R/2 (C) R/4 (D) R/5 Energy required to move a body of mass m from an orbit of radius 2R to 3R is ................ 62. GMm G m G m G m (B) (C) (D) 2 2 12R 3R 8R 6R Radius of orbit of satellite of earth is R. Its K.E. is proportional to (A) 60. –2 mgRe (B) 2 mgRe (C) (A) 1 1 1 (B) (C) R (D) R R R 3 /2 A particle falls towards earth from infinity. It’s velocity reaching the earth would be............. (A) 63. (A) infinity 64. 66. (C) 2 gR 2gR (D) zero The escape velocity of a sphere of mass m from earth having mass M and Radius R is given by (A) 65. (B) 2GM R (B) 2 GM R (C) 2GMm R (D) GM R The escape velocity for a rocket from earth is 11.2 kms–1 value on a planet where acceleration due to gravity is double that on earth and diameter of the planet is twice that of earth will be = ........... kms–1 (A) 11.2 (B) 22.4 (C) 5.6 (D) 53.6 –1 The escape velocity from the earth is about 11 kms . The escape velocity from a planet having twice the radius and the same mean density as the earth is =............ kms–1. (A) 22 (B) 11 (C) 5.5 (C)15.5 159 Downloaded from www.studiestoday.com Downloaded from www.studiestoday.com 67. If g is the acceleration due to gravity at the earth’s surface and r is the radius of the earth, the escape velocity for the body to escape out of earth’s gravitational field is................. (A) gr 68. 69. (B) (D) (B) GM r (C) 2GM r (D) Ve 2 (B) Ve 2 (C) Ve 2 2 (D) 4GM r Ve 4 (B) 12800 (C) 3200 (D) 1600 The escape velocity of a planet having mass 6 times and radius 2 times as that of earth is.......... 3 Ve (B) 3 Ve (C) 2 Ve (D) 2Ve There are two planets, the ratio of radius of two planets is k but the acceleration due to gravity of both planets are g what will be the ratio of their escape velocity. 1/ 2 kg (B) kg 1/ 2 (C) kg 2 (D) kg 2 The escape velocity of a body on the surface of the earth is 11.2 km/sec. If the mass of the earth is increases to twice its present value and the radius of the earth becomes half, the escape velocity becomes = ............... kms–1 (A) 5.6 77. 2GM R2 An artificial satellite is revolving round the earth in a circular orbit. its velocity is half the escape velocity. Its height from the earth surface is = ............... km (A) 76. (C) 2GMR The escape velocity of a body from earth’s surface is Ve. The escape velocity of the same body from a height equal to 7 R from earth’s surface will be (A) 75. 8 8 GR G (B) M 3 3 Two small and heavy sphere, each of mass M, are placed distance r apart on a horizontal surface the gravitational potential at a mid point on the line joining the center of spheres is (A) 6400 74. (D) m-1 The escape velocity of an object from the earth depends upon the mass of earth (M), its mean density (), its radius (R) and gravitational constant (G), thus the formula for escape veloctiy is (A) 73. (C) m0 (B) m (A) zero 72. (D) 1120 kms–1 The escape velocity of a particle of mass m varies as................... (A) R 71. (D) r/g The escape velocity of a projectile from the earth is approximately (A) 11.2 kms–1 (B) 112 kms–1 (C) 11.2 ms–1 (A) m2 70. (C) g/r 2gr (B) 11.2 (C) 22.4 (D) 494.8 1 1 Given mass of the moon is of the mass of the earth and corresponding radius is of 81 4 –1 the earth, If escape velocity on the earth surface is 11.2 kms the value of same on the surface of moon is = ....... kms–1. (A) 0.14 (B) 0.5 (C) 12.5 (D) 5 160 Downloaded from www.studiestoday.com Downloaded from www.studiestoday.com 78. 3 particle each of mass m are kept at vertices of an equilateral triangle of side L. The gravitational field at center due to these particies is ................ (A) zero 79. (B) (C) 9Gm L2 12 Gm (D) 3 L2 Escape velocity on the surface of earth is 11.2 kms–1 Escape velocity from a planet whose masses the same as that of earth and radius 1/4 that of earth is = ....... kms–1 (A) 2.8 80. 3GM L2 (B) 15.6 (C) 22.4 (D) 44.8 The velocity with which a projectile must be fired so that it escapes earth’s gravitational does not depend on .................. (A) mass of earth (B) Mass of the projectile (C) Radius of the projectiles’s orbit (D) Gravitational constant 81. The escape velocity for a body projected vertically upwards from the surface of earth is 11 kms-1. If the body is projected at an angle of 450 with the vertical, the escape velocity will be ............kms-1. (A) 82. 11 2 (B) 11 2 84. (B) 2 Ve (D) Ve/2 (A) 6.67 10-9 J (B) 6.67 10-10 J (C) 13.34 10-10 J (D) 3.33 10-10 J A particle of mass M is situated at the center of a spherical shell of same mass and radius a the magnitude of gravitational potential at a point situated at a / 2 distance from the center will be 4GM a (B) GM a (C) 2GM a (D) 3GM a The mass and radius of the sun are 1.99 x 1030 kg and R = 6.96 x 108 m. The escape velocity of rocket from the sun is =.......... km/sec (A) 11.2 86. (C) 4 Ve A particle of mass 10g is kept on the surface of a uniform sphere of mass 100 kg and radiius 10 cm. Find the work to be done aginst the gravitational force between them to take the particle is away from the sphere (G = 6.67 10-11 SI unit) (A) 85. (D) 11 The acceleration due to gravity on a planet is same as that on earth and its radius is four times that of earth. What will be the value of escape velocity on that planet if it is Ve on the earth (A) Ve 83. (C) 22 (B) 12.38 (C) 59.5 (D) 618 The mass of a space ship is 1000 kg. It is to be lauched from earth’s surface out into free space the value of g and R (radius of earth) are 10ms-2 and 6400 km respectively the required energy for this work will be = ................. J (A) 6.4 1011 (B) 6.4 108 (C) 6.4 109 (D) 6.4 1010 161 Downloaded from www.studiestoday.com Downloaded from www.studiestoday.com 87. 88. The diagram showing th variation of gravitational potential of earth with distance from the center of earth is (A) (B) (C) (D) A shpere of mass M and Radius R2 has a concentric cavity of Radius R1as shown in figure. The force F exerted by the shpere on a particle of mass m located at a distance r from the center of shhere varies as (O r ) (A) (B) (C) (D) 162 Downloaded from www.studiestoday.com Downloaded from www.studiestoday.com 89. 90. 91. Which one of the following graphs represents correctly the variation of the gravitational field with the distance (r) from the center of spherical shell of mass M and radius a (A) (B) (C) (D) Which of the following graphs represents the motion of a planet moving about the sun. (A) (B) (C) (D) The curves for P.E. (U) K.E. (Ek) of two particle system ae shown in figure . At what points systemwill be bound. (A) Only at point D Ek (B) Only at point A (C) At point D and A (D) At points A, B and C 163 Downloaded from www.studiestoday.com Downloaded from www.studiestoday.com Energy The correct graph representing the variation of total energy (E) kinetic energy (K) and potential energy (U) of a satellite with its distance from the centre of earth is........... Energy 92. Energy (B) Energy (A) (C) 93. 94. (D) A shell of mass M and radius R has a point mass m placed at a distance r from its center. The gravitational potential energy U(r) –v will be (A) (B) (C) (D) Motion of satellite If Ve and Vo are represent the escape velocity and orbital velocity of satelllite correspoinding to a circular orbit of radius r, then (A) Ve = Vo (B) 2 Vo Ve (C) Ve Vo / 2 (D)ve and vo are not related 164 Downloaded from www.studiestoday.com Downloaded from www.studiestoday.com 95. If r represents the radius of the orbit of a saltellite of mass m moving around a planet of mas M, the velocity of the satellite is given by (A) 2 96. gM r 99. 100. 101 102 103. (C) GM r (D) 2 GM r (B) V1 < V2 V1 V2 (D) r r 1 2 (C) V1 > V2 A satellite which is geostationary in a particular orbit is taken to another orbit. Its distance from the centere of earth in new orbit is two times of the earlier orbit. The time period in second orbit is .........hours. (a) 4.8 98. GMm r Two satellites of mass m1 and m2 (m1 > m2 ) are revolving round the earth in cirular orbits of r1 and r2 ( r1 > r2) respectively. Which of the following statement is true regarding their speeds V1 and V2 (A) V1 = V2 97. (B) 2 (C) 24 (B) 48 2 (D) 24 2 As astronaut orbiting the earth in a circular orbit 120 km above the surface of earth,gently drops a spoon out of space-ship. The spoon will (A) Fall vertically down to the earth (B) move towards the moon (C) Will move along with sapace - ship (D) Will move in an irregualr way then fall down to earth The period of a satellite in cirular orbit around a planet is independent of (A) the mass of the planet (B) the radius of the planet (C) mass of the satellite (D) all the three parameters (A), (B) and (C) Two satellites A and B go round a planet p in circular orbits having radii 4 R and R respectively if the speed of the satellite A is 3V, the speed if satellite B will be (A) 12V (B) 6V (C) 4/3 V (D) 3/2 V A small satellite is revolveing near earth’s surface. Its orbital velocity will be nearly = .......... kms–1. (A) 8 (B) 4 (C) 6 (D) 11.2 A satellite revolves around the earth in an elliptical orbit. Its speed (A) is the same at all points in the orbit (B) is greatest when it is closest to the earth (C) is greatest when it is farthest to the earth (D) goes on increasing or decresing continuously depending upon the mass of the satellite If the height of a satellite from the earth is negligible in comparison of the radius of the earth R, the orbital velocity of the stellite is ............ (A) gR (B) gR/2 (C) g/R 165 Downloaded from www.studiestoday.com (D) gR Downloaded from www.studiestoday.com 104 105. 106. 107. 108. 109. 110. 111. A satellite is moving around the earth with speed v in a circular orbit of radius r. If the orbit radius is decreasd by 1% its speed will (A) increase by 1% (B) increase by 0.5% (C) decrrease by 1% (C) Decrrease by 0.5% orbital velocity of an artifiicial satellite does not depend upon (A) mass of earth (B) mass of satellite (C) radius of earth (D) acceleration due to gravity orbital velocity of eath’s satellite near the surface is 7 kms–1. when the radius of orbit is 4 times that of earth’s radius, then orbital velocity in that orbit is =........kms-1 (A) 3.5 (B) 17 (C) 14 (D) 35 Two identical satellites are at R and 7R away from each surface, the wrong statement is (R - Radius of earth) (A) ratio of total energy will be 4 (B) ratio of kinetic energes will be 4 (C) ratio of potential energies will be 4 (D) ratio of total energy will be 4 but ratio of potential and kinetic energies will be 2 Which one of follwoing sttements regarding artificial satellite of earth is incorrect (A) The orbital velocity depends on the mass of the satellite (B) A minimum velocity of 8kms-1 is required by a satellite to orbit quite close to the earth. (C) The period of revolution is large if the radius of its orbit is large (D) The height of geostationary satellite is about 36000 km from earth The weight of an astronuat, in an artificial satellite revolving around the earth is (A) zero (B) Equal to that on the earth (C) more than that on earth (D) less than that on the earth The distance of a geo-stationary satellite from the center of the earth (Radius R= 6400km) is nearest to (A) 5R (B) 7R (C) 10R (D) 18R A geo-stationary satellite is orbiting the earth of a height of 6R above the surface of earth, R being the radius of earth. The time period of another satellite at a height of 2.5 R from the surface fo earth is = ...............hr (A) 6 112. (B) 6 2 (C) 10 (D) 6 / 2 1 1 (and not as 2 )where R R R is separation between them, then a particle in circular orbit under such a force would have its orbital speed v proportional to If the gravitational force between two objects were proportional to 1 1 (B) R0 (C) R1 (D) 2 R R A satellite moves around the earth in a circular orbit of radius r with speed v, If mass of the satellite is M , its total energy is (A) 113. (A) 1 MV 2 2 (B) 1 MV 2 2 (C) 3 MV 2 2 (D) MV2 166 Downloaded from www.studiestoday.com Downloaded from www.studiestoday.com 114. A satellite with K.E. Ek is revolving round the earth in a circular orbit. How much more K.E. should be given to it so that it may just escape into outerspace ? (A) Ek 115. 116. 117. 118. 119. 120. 121. (B) 2 Ek (D) 3 Ek Potential energy of a satellite having mass m and rotating at a height of 6.4 x 106 m from the surface of earth (A) -0.5 mg Re (B) -mg Re (C) -2mgRe (D) 4 mgRe When a satellite going round the earth in a circular obrit of radius r and speed v loses some of its energy, then r and v changes as (A) r and v bothe will increase (B) r and v both will decease (C) r will decrease and v will increase (D) r will increase and v will decrease The time period of a satellite of earth is 5 hours If the sepration between the earth and the satellite is increased to four times the previous value, the new time period will become ...... hours (A) 10 (B) 120 (C) 40 (D) 80 A person sitting in a chair in a satellite feels weightless because (A) the earth does not attract the objects in a satellite (B) the normal force by the chair on the person balances the earth’s attraction (C) the normal force is zero (D) the person in satellite is not accelerated Two satellites A and B go round a planet in cirular orbits having radii 4R and R respectively If the speed of satellite A is 3v, then speed of satellite B is (A) 3v/2 (B) 4v/2 (C) 6v (D) 12v A satellite moves in a circle around the earth, the radius of this circlr is equal to one half of the radius of the moon’s orbit the satellite completes one revolution in .........lunar month (A) 1/2 (B) 2/3 (C) 2–3/2 (D) 123/2 The additional K.E. to be provided to a satellite of mass m revolving around a planet of mass M, to transfer it from a circular orbit of radius R1 to another radius R2 (R2 > R1) is 1 1 2 2 R2 R1 (A) G M m 1 1 R2 R1 (C) 2 G M m 122. 1 Ek 2 (C) 1 1 R2 R1 (B) G M m (D) 1 1 1 GMm 2 R2 R1 Rockets are launched in eastward direction to take advantage of (A) the clear sky on eastem side (B) the thiner atmosphere on this side (C) earth’s rotation (D) earth’s tilt 167 Downloaded from www.studiestoday.com Downloaded from www.studiestoday.com 123. 124. A satellite of mass m is orbiting close to the surface of the earth (Radius R = 6400 km) has a K.E. K. The corresponding K.E. of satellite to escape from the earth’s gravitational field is (A) k (B) 2 k (C) mg R (D) m k A planet moving along an elliptical orbit is closest to the sun at a distance r 1 and farthest away at a distance of r2. If v1 and v2 are the liner velocities at these points respectively, then the v1 ratio v is ............ 2 r1 (A) r 2 125. r1 (B) r2 2 r2 (C) r 1 r1 (D) r2 2 The time period T of the moon of planet Mars(Mm) is related to its orbital radius R as (G = Gravitational constant) 4π 2GR 2 (B) T Mm 4 R 2 (A) T GMm 2 2 2 R 2G (d) T2 = 4 Mm GR2 Mm A geostationary satellite is orbiting the earth at a height of 5 R above that of surface of the earth. R being the radius of the earth. The time period of another satellite in hours at a height of 2R from the surface of earth is ............ hr (C) T 2 126. 127. 128. 129. 130. (A) 5 (B) 10 (C) 6 2 (D) 6 / 2 The figure shows ellipticall orbit of a planet m about the sun s. the shaded area SCD is twice the saded area SAB. If t1 is the time for the planet to move from C and D and t2 is the time to move from A to B then (A) t1 > t2 (B) t1 > 4t2 (C) t1 = 2t2 (D) t1 = t2 The period of a satellite in a circular orbit of radius R is T. the period of another satellite in a circular orbit of radius 4R is (A) 4T (B) T/4 (C) 8T (D) T/8 If the earth is at one- fourth of its present distance from the sun the duration of year will be (A) half the pesent Year (B) one-eight the present year (C) one-fourth the present year (D) one-sixth the present year The orbital speed of jupiter is (A) greater than the orbital speed of earth (B) less than the orbital speed of earth (C) equal to the orbital speed of earth (D) zero 168 Downloaded from www.studiestoday.com Downloaded from www.studiestoday.com 131. 132. 133. 134. Kepler’s second law regarding constancy of aerial velocity of a palnet is consequence of the law of conservation of (A) energy (B) angular Momentum (C) linear momentum (D) None of these The largest and shortest distance of the earth from the sun are r1 and r2 its distance from the sun when it is at the perpendicular to the major axis of the orbit drawn from the sun r1 r2 2r1 r2 r1 r2 r1 r2 (B) r r (A) (C) r r (D) 4 3 1 2 1 2 According to keplar, the period of revolution of a planet (T) and its mean distance from the sun (r) are related by the equation (A) T 3 r 3 cons tan t (B) T 2 r 3 cons tan t (C) Tr 3 cons tan t (D) T 2 r cons tan t A satellite of mass m is circulating around the earth with constant angular velocity. If radius of the orbit is Ro and mass of earth M , the angular momentum about the center of earth is (A) m GMRo (C) m 135. 136. 137. 138. 139 (B) M GMRo GM Ro (D) M GM Ro The earth E moves in an elliptical orbit with the sun s at one of the foci as shown in figure. Its speed of motion will be maximum at a point .......... (A) C (B) A (C) B (D) D The period of revolution of planet A around the sun is 8 times that of B. The distance of A from the sun is how many times greater than that of B from the sun. (A) 2 (B) 3 (C) 4 (D) 5 The earth revolves round the sun in one year. If distace between then becomes double the new period will be .......... years. (A) 0.5 (B) 2 2 (C) 4 (D) 8 The maximum and minimum distance of a comet from the sun are 8 x 1012m and 1.6 x 1012 m. If its velocity when nearest to the sun is 60 ms-1, What will be its velocity in ms-1 when it is farthest ? (A) 6 (B) 12 (C) 60 (D) 112 The period of moon’s rotation around th earth is nearly 29 days. If moon’s mass were 2 fold its present value and all other things remained unchanged the period of moons’s rotation would be nearly .....days (A) 29 2 (B) 29 / 2 (C) 29 2 (D) 29 169 Downloaded from www.studiestoday.com Downloaded from www.studiestoday.com 140. 141. 142. 143. 144. 145. If the velocity of planet is given by G a M b R c then (A) a =1/3 b = 1/3 c = -1/3 (B) a =1/2 b = 1/2 c = -1/2 (C) a =1/2 b = -1/2 c = 1/2 (D) a =1/2 b = -1/2 c = -1/2 The radius of orbit of a planet is two times that of earth. The time period of planet is.........years. (A) 4.2 (B) 2.8 (C) 5.6 (D) 8.4 If r denotes the distance between the sun and the earth, then the angular momentum of the earth around the sun is proportional to (A) r3/2 (B) r (C) r1/1 (D) r2 What does not change in the field of central force (A) potential energy (B) Kinetic energy (C) linear momentum (D) Angular momentum A thin uniform annular disc (See figure) of mass M has outer radius 4R and inner radius 3R. The work required to take a unit mass from point P on its axix to infinity is.............. (A) 2GM (4 2 5) 7R (B) 2GM (4 2 5) 7R (C) GM 4R (D) 2GM ( 2 1) 5R Suppose the gravitational force varies inversely as the nth power of distance the time period of planet in circular orbit of radius R around the sun will be proportional to n 1 2 146. 147. Rn n 2 2 (D) R (A) R (B) R (C) If the radius of the earth were to shrink by 1% its mass remaing the same, the accelration due to gravity on the earth’s surface would (A) decrease by 2% (B) remain Unchanged (C) increase by 2% (D) increases by 1% A body of mass m is taken from earth surface to the height h equal to radius of earth, the increase in potential energy will be (A) mg R 148. n 1 2 1 (B) 2 mgR (C) 2 mgR 1 (D) 4 mgR An artificial satellite moving in a circular orbit around earth has a total (kinetic + potential lenergy) Eo, its potential energy is (A) -Eo (B) 1.5 Eo (C) 2 Eo (D) Eo 170 Downloaded from www.studiestoday.com Downloaded from www.studiestoday.com 149. Two bodies of masses m1 and m2 are initially at rest at infinite distance apart. They are then allowed to move towards each other under mutual agravitational attraction Their relative velocity of apporach at sepration distance r between them is 2 G ( m1 m 2 ) (A) r 1 1 2 1 2 r (C) 2 G m1 m 2 150. 1 2 G ( m1 m 2 ) 2 (B) r (D) 2 G m1 m 2 2 r A geostationary satellite orbits around the earth in a circular orbit of radus 3600 km the time period of a satellite orbiting a few hundred kilometers above the earth’s surface (Rearth = 6400 km) will approximately be = ...... h (A) 1/2 (B) 1 (C) 2 (D) 4 KEY NOTE 1.C 2.A 3. B 4. A 5. B 6. C 7. A 8. D 9. C 10. D 11. B 12. D 13. A 14. A 15. A 16. C 17. C 18. C 19. D 20. C 21. B 22. A 23. B 24. B 1.C 25. B 2 . A 25. B 26. A 3. B 27. C 26. A 4. A 28. A 5. B 27. C 29. C 6. C 28. A 30. C 7. A 29. C 31. A 8. D 32. C 9. C 30. C 33. B 10. D 31. A 34. D 11. B 32. C 35. A 12. D 26. B 13. A 33. B 27. C 14. A 34. D 38. C 15. A 35. A 39. A 16. C 40. D 17. C 26. B 41. A 18. C 27. C 42. B 19. D 38. C 43. B 20. C 44. C 21. B 39. A 45. D 40. D 22. A 46. B 23. B 41. A 47. A 24. B 48. A 42. B 43. B 44. C 45. D 46. B 47. A 48. A 49. C 50.C A 49. 51. C 50. A 52. A 51. 53.C A 54.A A 52. 55. 53. AC 56. C 54. 57.A A 58.CB 55. 59.CD 56. 60. A 57. 61.A D 58. 62.B A 63. 59. DB 64. A 60. A 65.B 61. 66. D A 67. A B 62. 68. C 63. 69. BC 64. 70. A A 71. 65.BD 72. C 66. A 67. B 68. C 69. C 70. A 71. D 72. C 73. A 74.73. A A 75. A 74. A 76. C 77.75. C A 78.76. A C 79.77. C C 80. B 81.78. D A 82.79. B C 83.80. B B 84. D 85.81. D D 86.82. D B 87.83. C B 88. B 89.84. D D 90.85. C D 91.86. D D 92. C 93.87. C C 94.88. B B 95.89. D D 96. B 90. C 91. D 92. C 93. C 94. B 95. D 96. B 97. B 121 D 98. C97. B122. C 99. C 98. C123 B 100. B 124. C 99. C 101. A 125. A 102. B100.126. B C 103. D 127. 101. A C 104. B 128. C B B 105. B102.129. 106. A103.130. D B 107. D 104.131. B B 108. A 132. C B B 109. A105.133. 110. B106.134. A A 111. D107.135. D B 112. B 136. C A B 113. A108.137. A A 114. A109.138. 115. A110.139. B D 116. C 140. B D C 117. D111.141. B C 118. C112.142. 119. C113.143. A D 120. C 144.A 114. A 115. A 116. C 117. D 118. C 119. C 120. C 145. A 146.DC 121 147. B 122. C 148. C 123 149.BB 150.CC 124. 125. A 126. C 127. C 128. C 129. B 130. B 131. B 132. C 133. B 134. A 135. B 136. C 137. B 138. A 139. D 140. B 141. C 142. C 143. D 144.A 171 Downloaded from www.studiestoday.com 145. A 146. C 147. B 148. C 149. B 150. C Downloaded from www.studiestoday.com HINT 4 G ( R 3. )2 4 3 2 2 R4 2 4R 3 (1) ) (c) F G (m)(m 2 (2 R) (4) (5) (6) (A) Gravitational force does not depend upon the medium (B) Due to inertia to direction (D) the force exerted by sun on the earth F = m2R (6 1024 ) (2 107 ) 2 (1.5 1011 ) 36 1021 N (7) F xm ( x 1) m m 2 x ( x 1) for maximum force dF 0 dx dF m2 2 m2 x 0 x 1/ 2 dx (C) cenripetal force provided by the gravitational force of attraction between two particites (9) m 2 G ( m )( m ) r (2 r ) 2 Gm r 1 2 (D - X) (10) (D) For will be zero at the point of zero intensity x 11. 12. m1 m1 m2 81m 9 D D 10 81m m B mg GM e m where Me and Re is the mass and radius of the earth respectivel Re 2 Me g Re 2 G (D) Here FOA = FOB = FOC = F FOA FOB FOC G (m) (2m) r2 172 Downloaded from www.studiestoday.com Downloaded from www.studiestoday.com = 14. = 0 (A) let a point mass C is placed at a distance of x m from the point mass A as show in figure mA here x m C (1–x) mB ma 4 3 mb G (m)(mA) .........(i ) x2 G (m)(mB) FBC .........(ii) (1 x )2 FAC According to given problem FAC 1 FBC 4 with the help of equ (i) and (ii) G (m)(mA) 1 G (m)(mB ) x2 3 (1 x) 2 mA x2 4 x2 x2 4 mB 3(1 x) 2 3 3(1 x) 2 (1 x) 2 2 x 2 2x x 1 x x 23 m 3x 2 GM 4 and M R 3 2 R 3 17. (c) g 18. G 4 3g . R3 2 R 3 4 RG (c) for condition of weight lessness at equator g g 19. R 10 1 rad sec 3 6400 10 800 (D) Time peripd of simple pendylym T 2 l g' In artificial satellite g’ = 0 T 173 Downloaded from www.studiestoday.com Downloaded from www.studiestoday.com 20. GM GMo 4GMo (C) g R 2 ( Do 2) 2 Do 2 21. (B) we know that g on the planet gp GM R2 GM 7 4 7g R2 4 Hence the weight on the planet = 700 4 = 400 gm t 7 2 2 22. g ' R 6400 2 g ' 960.400 ms g ( R h) 6400 64 23. g 1 g 2 R Cos 2 Rotation of the earth results in the decreased weight apparently.this decrease in weight is not felt at the poles as the angle of latitude is 90’ 24. GM (B) using g R 2 25. 1 (B) g g (1 26. 4 (A) g πGR 3 we get gm = g/5 d (n 1) )d R R n and g' 4/3πGR' g' R' 0.2 g' 0.2g g R 27. (C) mass does not vary from place to place 28. R 4 R (a) g ' g g g Rh R R2 9 2 2 4 W' 4/9 w (72) 32 N 9 2 29. R 9 / 4 by solving h R (c) g ' g Rh 30. Mp Re 1 2 9.8 0.49ms 2 (c) g p g e 9.8 (2) Me Rp 80 20 31. d 100 2 (A) g' g 1 9.8 1 9.66 ms R 6400 2 174 Downloaded from www.studiestoday.com Downloaded from www.studiestoday.com 32. (C) g GM 2 and k = L 2I 2 R 1 1 k 2 2 and R R i.e. if radius of earth decreases by 2% then g and k both increase by 4% (B) Weight on surface of earth, mg = 500N and weight below the surface of earth at If mass of the earth and its angular momentum ramains constant then g 33. d = R mg d 250N , mg1 = mg 1 mg 1 1/2 2 2 R 34. 4 (D) g π GR 3 35. (A) g α g1 g2 R1 1 R 2 2 GM g α 1 or r α 1 r2 g r2 If g decreases by one percent then r should be increase by 1/2 % i.e. R 1 6400 32 km 2 100 2 37. 1 R R R h 2 R h (C) g ' g 2 Rh Rh 2 1 R 0.414 R Hence distance form center = R + 0.414R = 1.414R 4 π GR g α d R ( d given in the problem) 3 38. (C) g 39. (A) insided the earth g' 40. (D) g 4 π Gr g ' r 3 Rp g p 4 π GR Re g e 3 Rp 41. (A) g 42. (B) For height Re R 2 2 Δg 24 100% 100 % g 4 For depth 43. e 1 (1) 2 p Δg d h 100% 1/2 0.5 g R R (B) g R 175 Downloaded from www.studiestoday.com Downloaded from www.studiestoday.com 44. (C) H V2 1 H g Hα B A 2g g HA gB Now g B gA as g R 12 HB g A 12 HB 12 HA 12 1.5 18 m HA gB 46. 2 R R 2 (B) g' g R h g 3 R 4/9.g(g 10 ms ) 2 47. 4 g e e Re (A) g 3 πGR g α R g . Rm m m 6 5 Re . 1 3 Rm Rm 5 Re 18 48. R R (A) g ' g g g /9 R 2R R 2R 49. (C) g ' g (1 h / R) 2 g g 16 (1 h / R ) 2 2 h 1 16 R 1 h 4 R h 3 R h 3R 50. (A) Gravitational attraction fore on particle B Fg GMpm (Dp / 2) 2 Acceleration of paritcle due to particle due to gravity a Fg 4GMp m Dp 2 176 Downloaded from www.studiestoday.com Downloaded from www.studiestoday.com 51. (C) g α r it(r r) and g α 52. (A) I 53. (A) U 1 (it r R) r2 dv dx GMm r 6.67 10 11 7 10 22 6 10 24 r 7.79 10 28 r 3.8 108 m 54. (A) V GM 1 GM 2 d /2 d /2 Now P.E mv 2GM (m1 m2 ) (m mass of particie) d K.E = P.E 1 2 2GM m (m1 m2 ) 2 d 2 55. (C) If the body is projected with velocity υ (υ Ve) then height up to where it rises h 57. G(M1 M 2 ) d R 2 Ve 1 V2 11.2 10 2 (A) Ve 2GM R 100 58. R 4 R(approx) P.E. U GMm 5000 J R GM R GM 5000 R (B) P.E. U Uinitial = – GMm GMm r Rh GMm GMm and Ufinal = – 3R 2R loss of P.E. = gain in K.E = GMm GMm GMm – = 2R 3R 6R 177 Downloaded from www.studiestoday.com Downloaded from www.studiestoday.com 59. (D) ΔU U 2 U1 mgh mgRe mgRe 1 h/Re 1 Re 2 Re U 2 ( mg Re) 60. mg Re 2 (A) If body is projected with velocity υ (υ Ve) then height up to which it will rise R but u ve/2 h 2 Ve 1 Ve/2 61. 1 U 2 mgRe 2 R R/3 4 1 (D) change in potential energy in displacing a body from r 1and r2 is given by 1 1 1 GMm 1 U GMm GMm 6R 2 R 3 R r1 r2 GMm 1 K.E. α 2R R 62. (A) K.E = 63. (B) this should be equal to escape velocity i.e. = 2gR 64. (A) A Escape velocity does not depend on the mass of the projectile 65 (B) Ve Vp g p Rp . 22 2 g e Re Ve 2Ve 2 11.2 22.4Kms 1 2GM R R G 66. (A) Ve 69. since the planet is having double radius incomparision to earth therefore the escape velocity becomes twice i.e. 22 kms-1 (C) Because it does not depend on the mass of projetile 71. (D) 8 3 Ve α R if constant Gravitational potential of A at 0 GM 2GM r /2 r of B at 0 GM 2GM r /2 r Total potential at 0 4GM r 178 Downloaded from www.studiestoday.com Downloaded from www.studiestoday.com 72. V1 V2 73. 1 where r is the position of body from the surface r (C) Ve α r2 r1 R 7R V 2 2 V2 1 R 2 2 GM 1 2GM R h 2 R (A) V 4R 2(R h) h R 6400 km Vp Ve 1 Mp Re 6 3 Ve 3 Ve Me Rp 2 74. (A) 75. (A) 76. (C) Ve 77. (C) 78. 2GM 4 2 2GM 2 11.2 . 2.5 kms1 R R 9 R 9 (A) Due to three particles net intensity at the center υ 2gR V1 V2 g1 R1 . gk (kg)1/2 g2 R2 2GM M Ve α R R If M becomes double and R becomes half then escape velocity; becomes two times on earth Ve = 2GM 11.2 kms 1 R on moon Vm I I A IB I C 0 because out of these three intensities ARE equal in magnitude and between each other is 120’ 80. 1 1 if R becomes then Ve will be 2 times 4 R (D) Escape velocity does not depends upon the angle of projection 82. (B) 83. Ve = 2Ve (B) potential energy of system of two mass 79. (C) Ve α 2gR g Rp Vp p. 1 4 2 Ve g e Re GMm 6.67 10 11 100 10 10 3 U 6.67 10 10 J 2 R 10 10 so, the amount of work done to take the particle up to infinte will be 6.67 10-10 J 179 Downloaded from www.studiestoday.com Downloaded from www.studiestoday.com 84. Vp = Vsphere + Vpartical (D) 85. (D) Ve = 86. (D) 2GM = R GM GM 3GM a a/2 a 2 6.67 1011 1.99 1030 618 kms1 6.98 108 m GMm GMm W 0- gR 2 mgR R R R 1000 10 6400 103 64 109 6.4 1010 J GM R 87. (C) Vin 88. (B) F 0 When 0 r R1 Vsurface GM R Vout GM r beccause intensity is zero inside Caviy Fincrease When R1 r R 2 Fα 89. (D) 1 when r R 2 R Intensity will zero inside the spherical shell I = upto r = a and I 1 when r > a r2 90. 91. (C) kepler’s law T2 α r3 (D) System will be bound at points where total energy is negative. In the given curve at point A, B, and C the P.E. is more than K.E. 92. (C) 93. U- GMm GMm GMm K and E r 2r 2r For satellite U, K, and E varies with r and also U and E remains negativw where K reamin always positive (C) Gravitational P.E. = m x gravitational potential U = mv so the graph of U will be same as that of V for a spherical shell 94. (B) Ve 2gR 96. (B) υ and V0 gR Ve 2 V0 GM it r1 r2 then V1 V2 r 180 Downloaded from www.studiestoday.com Downloaded from www.studiestoday.com 97. (B) T α r 3 2 if r becomes double then time period will become (2)3/2 times so new time period will be 24 x 2 2 hr i.e. = 48 2 98. (C) The velocity of the spoon will be equal to the orbital velocity when dropped out of the space ship 100. (B) VA 3V VB VA 104. (B) GM V A R VB υ υα RB RA R 1 4R 2 VB 6V 1 r % increase in speed = = 1 % decrease in radius 2 1 (1%) = 0.5 % 2 105. (B) υ GM r 106. (A) υα 1 it orbital radius becomes 4 times then orbital velocity will becomes halt r 107. (D) orbital rudius of satellites U1 K1 110. 111. (B) (D) r1 = R+R = 2R r2 = R+7R = 8R GMm GMm and U 2 r1 r2 GMm 2r1 and K2 GMm 2r2 U1 K 1 E 1 4 U2 K 2 E2 6R from the surface of earth and 7R from the center Distance of satellite from the center are 7R and 3.5R respectively 3/2 3/2 T2 r2 3.5R T2 24 T1 r1 7R 112. (B 6 24 hr Gravitational force provides the required centripetal force for orbiting the satellite mυ 2 K R R becuse F α 1 R υ α Ro 181 Downloaded from www.studiestoday.com Downloaded from www.studiestoday.com Total energy = -K.E. = 1 2 2 m 113. (A) 114. (A) B.E. = -K.E. And it this amount of energy(Ek) given to satellite it will escape into outer space 1 GMm g Re 2 m GMm GMm = – = – = – = – mg Re = 0.5 mg Re P.E. = 2 Re h 2 Re 2R e r 115. (A) 116. (C) B. E. = 117. R (D) T2 T1 2 R1 118. B (C) V A 120. time period of revolution of moon around the earth = 1 lunar month GMm 1 if B.E. decreases the r also decreases and V increases as r r 3/2 V r Te e Tm rm 121 123. 124. 125. 3 /2 1 2 3 /2 Te 2 2 /3 lunar month GMm 1 1 2 R1 R2 Because Earth rotation from west to east direction GMm 2R (C) v1 r1 = v2r2 (angular momentum is constant) (B) k (A) Time period T T 2 2R 2R GMm GM R 3/ 2 4 2 R 3 GMm 2 126. 4R 2 R GMm GMm (D) 2R K.E. 2R 1 2 K .E 122. rA rB T1 (4)3/2 8T1 40 hr 3 T1 R1 (6R ) 3 8 (C) 2 3 (3R ) 3 T2 R2 2 T2 24 24 72 8 T2 6 2 182 Downloaded from www.studiestoday.com Downloaded from www.studiestoday.com 127. (C) ΔA L ΔAαΔt Δt 2m ΔASCD t1 2A t1 2t 2 ΔASAS t 2 A 3 R T1 1 T2 R2 2 R 4R 3 2 128. (C) 129. T1 1 1 1 2 3 (B) since T α r T 4 T 8 T 130. VJ re (B) orbital radius of Jupiter > orbital radius of Earth V r As rj < re there fore Vj < Vee e j 131. dA L (B) dt 2m = constant 132. (C) The earth moves around the sun in elliptical path, so by using the properties of ellipse 2 T2 8T1 3 r1 (1 e) a and r2 = (1-e) r2, a = r1 r2 2 r1r2 (1 e 2 ) a 2 where a= semi major axis b= semi minor axis e= eccentircity Now required distance = sem latysrectum = b2/9 a2 133. 134. 135. (1 e 2 ) r1r2 2r r 12 a (r1 r2 )/2 r1 r2 (B) T2/ r3 = constant T2 r-3 = constant (A) Angular momentum = Mass orbital velocity x Radius m GM R0 R0 m GMR 0 (B) speed at the earth will be maximum when its distance from the sun is minimum because m r = constant 3 136 (C) TA rA TB rB (B) T2 r2 T1 r1 2 3 137. 3 4 8 A 4B 2 (2) 3 2 2 rA 4B 8 3 2 4rB 2 2 T2 2 2 years 183 Downloaded from www.studiestoday.com Downloaded from www.studiestoday.com 138. (A) By conservation of angular momentum (mr) = constant Vmin X rmax = Vmax X rmin 60 1.6 1012 60 12 ms1 12 8 10 5 (D) Time period does not depends upon the mass of satellite Vmin 139. 1 1 1 GM G 2 M 2R 2 R 140. (B) υ 141. R2 (B) T2 T1 R1 3 142. 2 3 1 2 2 2.8 yar (C) Angular momentum of earth around the sun L = MEVo r GMS r ME 2 .r M E .GMS .r ME = mass of earth MS = mass of Sun r = Distance between sun and the earth 143. L α r (D) For central force toraqe is zero 144. dL 0 L Constant dt (A) Wext U U F Gdm 0 (1) x G M π 7R 2 2π r dr 16R 2 r 2 2 GM 7R2 GM 7R2 2 GM 7R2 r2 2 GM 4 2R 5R 7R 2 GM 4 2 5 7R rdr 16 R2 r2 GM π ad2 7R (2) 2 16R 2 4R 3R 184 Downloaded from www.studiestoday.com Downloaded from www.studiestoday.com 145. (A) m 2 Rα T α R (C) g 147. (B) ΔU α 1 T 2 α R n 1 Rn n+1 2 GM R2 146. 148. 4π 2 1 m 2 R Rn T is mass ramains constant then 1 R2 mgh 1 mgR 2 1 h R (C) P.E. = 2. total energy = 2E0 GMM GMM and Eo = – r 2r (B) let velocities of these masses at r distance from each other be v1 and v2 respectively By conservation of momentum Because we know U = – 149. m 1 V1 m 2 V2 ____________ 1 m 1 V1 m 2 V2 0 By conservation of energy change in P.E. = change in K.E. Gm1m 2 1 1 2 2 m1V1 m 2 V2 r 2 2 2 2 m1V1 m 2 V2 2 Gm1m 2 _______ ii m1 m2 r on solving (i) and (ii) 1 2Gm 2 2 r m1 m2 and V ap p V1 V 2 3 150. T2 r2 (C) T1 r1 2 υ2 2Gm 1 2 r m1 m 2 2G m 1 m 2 r 3 6400 2 T2 24 2 hours 3600 185 Downloaded from www.studiestoday.com Downloaded from www.studiestoday.com Assertion - Reason Type Questions 1. 2. Direction (Read the following uestions and choose) (A) If both Assertion and Reason are true and the Reason is correct explanation of assertion (B) If both Assertion and Reason are true, but reason is not correcte explanation of the Assertion (C) If Assertion is true, but the Reason is false (D) If Assertion is faluse, but the Reason is true Assertion : The value of acc. due to gravity (g) does not depend upon mass of the body GM Reason : This follows from g 2 , where M is mass of planet (earth) and R is radius of R planet (earth) (a) A (b) B (c) C (d) D -2 Assertion : Unit of gravitational field intensity is N/kg or ms Reason : Gravitational field intensity Force N kg.m/sec 2 ms 2 mass kg kg 3. 4. 5. (a) A (b) B (c) C (d) D Assertion : The time period of a geostationary satellite is 24 hours Reason : Such a satellite must have the same time period as the time taken by the earth to complete one revolution about its axis (a) A (b) B (c) C (d) D Assertion : Even when orbit of a satellite is elliptical, its plane of rotaiion passes through the center of earth Reason : This is in accordance with the principle of conservation of angular momentum (a) A (b) B (c) C (d) D Assertion : The time Period of pendulum, on a satlellite orbiting the earth is infinity Reason : Time period of a pendulum is inversely proportional to (a) A 6. (b) B (c) C g (d) D Assertion : The escape velocity on the surface of a planet of the same mass but 1 times the 4 radius of earth is 5.6 kms-1 7. 8. 9. Reason : (a) A Assertion : Reason : (a) A Assertion : The escape velocity Ve = 2gR (b) B (c) C (d) D The comet does not obey kepler’s law of planetaty motion The comet does not have ellitical orbit (b) B (c) C (d) D The square of the period of revolution of a planet is proportional to the cube of its distance from the sun. Reason : Sun’s gravitational field is inversely proportional to the square of its distance from the planet (a) A (b) B (c) C (d) D Assertion : Space ship while entering the earth’s atmoshere is likely to catch fire 186 Downloaded from www.studiestoday.com Downloaded from www.studiestoday.com 10 Reason : Temperature of upper atomosphere is very high (a) A (b) B (c) C (d) D Assertion : The earth is slowing down and as a result the moon is coming nearer to it Reason : The angular momentum of earth - moon system is not conserved (a) A (b) B (c) C (d) D 187 Downloaded from www.studiestoday.com Downloaded from www.studiestoday.com Assertion - Reason Type question : 1. A 2. A 3. A Both the asseration and reason are true and the latter is correct expalnation of the former. Both the asseration and reason are true and the latter is correct expalnation of the former. As the satellite is to be stationary over a particular place, its time period of revolution = 24 hours= time peroid of revolution of earth about its axis. 4. A T 2π on a satellite, there is a weightless -ness, so g= 0 hence T g 5. A Thus both the asseration and reason are correct As no torque is acting on the planet, its angular momentum must stay constant in a magnitude as well as direction there for planet of rotation must pass throught the center of earth 6. D Ve 2 GM R Ve' = Ve R i.e. R Ve α 1 R 2 Ve 2 11.2 22.4 kms1 4 thus assertion is wrong but Reason is correct 7. A Both the asseration and reason are correct and the Reason is correct explanation of Asseration 8. A A Both the asseration and reason are correct and the Reason is correct explanation of Asseration 9. C Here Asseration is correct but Reason is wrong because the space ship while entering the earth’s atmosphere may catch fire due to atmoshperic air friction 10. D Here Both the asseration and reason are wrong because the angular momentum of earth moon system is conserved in the absence of extermal touque. 188 Downloaded from www.studiestoday.com Downloaded from www.studiestoday.com Comperehensions Type Questions (1) If a smooth tunnel is dug across a diameter of earth and a particle is related from the surface of earth, the particle oscillates simple harmonically along it. (1) Time period of the particle is not equal to 3 2π R 2 GM (C) 84.0 min (D) Nome of these (2) Maximum speed of the these (A) 2π R g (B) 2 GM R (B) (A) 2. 3. GM R (C) 3 GM 2R (D) GM 2R When a paricle is projected from the surface of earth, it mechanincal energy and angular momentum about center of earth at all time is constant (i) A particle of mass m is projected from the surface of earth with velocity vo at angle with horizontal suppose h be the maximum height of particle from surface of earth and v its speed at that point them v is (A) v0coso (B) >v0coso (C) <v0coso (D) zero (ii) Maximum height h of the paritcle is (A) Vo 2 sin 2θ 2g (C) Vo 2 sin 2θ 2g (B) Vo 2 sin 2θ 2g Vo 2 sin 2θ (D) can be greater than or less than 2g A solid sphere of mass M and radius R is surrounding by a spherical shell of same mass M and raius 2R as shown. A small particle of mass m is relased from rest from a height (h <<R) aobove the shell there is a hole in the shell. (i) in what time will it enter the hole at A hR 2 GM (A) 2 hR 2 GM (C) (B) 2hR 2 GM (D) None of these (ii) what time will it take to move from A to B ? (A) (B) R2 (C) GMh R2 R2 GMh (D) None of these GMh 189 Downloaded from www.studiestoday.com Downloaded from www.studiestoday.com (iii) with what apporximate speed will it colide at B ? 4. 5. (A) 2 GM R (C) GM 2R (B) 3 GM 2R (D) GM R A planet is revolving round the sun in elliptical orbit. Velocity at perigee position (nearest) is v1| and at apogee position (farthest) is v2 Both these velocities are perpendicular to the joinging center of sun and planet r is the minimum distance and r2 the maximum distance. (i) when the planet is at perigee position, it wants to revolve in a circular orbit by itself. For this value of G (A) Should increase (B) Should decrease (C) data is in sufficeint (D) will not depend on the value of G (ii) At apogee position suppose speed of planer is slightly decreased from v2, then what will happen to minimum distance r1 in the subsequent motion (A) r1 and r2 bothe will dicreases (B) r1 and r2 bothe will increases (C) r2 will remain as it is while r1 will increase (D) r2 will remain as it is while r1 will decrease Garvitational potential at any point inside a spherical shall is uniform GM where M is the mass of shell and R its and is given by R 3GM radius. At the center solid sphere, potential is 2R (i) There is a concentric hole of radius R in a solid sphere of radius 2R Mass of the remaining portion is M What is the gravitation potential at center? (A) 3GM 7R (B) 5GM 7R (C) 7GM 14 (D) 9GM 14 R 190 Downloaded from www.studiestoday.com Downloaded from www.studiestoday.com Solution (1) (i) T 2π R g putting g GM we have R2 T 3 2π R2 Gm (ii)maximum speed is at centre from conservation mechanicall energy(from cerface to center) increase in K.E. = decrease in P.E. 1 mn 2 U i U f m(Vi - Vf ) 2 υ 2(vi vf ) GM 3 GM 2 12 R (2) (i) GM R From conservation of angular momentum at A and B mVocosθ m R h R V0 cosθ R h Vo cosθ (ii) From consercation of mechanical energy, Decrease in K.E. = increase in P.E. 1 2 m V0 V 2 2 2 V0 V 2 2 gh 1 h R 2 mgh 1 h R but here V V0cos θ 2 2 V0 V 2 V0 sin 2θ so let V0 V 2 x here x V0 x h 2 sin 2θ 2gh 1 h R x 2g x R 2 x O sin 2 i.e. h > i.e. h > 2g 2g 191 Downloaded from www.studiestoday.com Downloaded from www.studiestoday.com (3) (i) Acceleration due to gravity near the surface of shell can be assumed to be uiform g G(2M) GM (2R) 2 2R 2 From h t = (ii) 2h hR 2 2 g GM GMh GM VA 2 gh 2 2 h R2 2R From A to B field due to shell is zero, but field due to sphere is not-zero hence tAB (iii) 1 2 gt 2 R R2 VA GMh KA=0 polential between A and B due to shell is From energy conservation KA + UA = KB + UB KB = (UA = UB) = m(VA – VB) 1 mVB2 = m(VA – VB) 2 VB (4) 2(VA VB ) = GM R At perigee position v1 > v0 where v0 is the orbital velocity for circular motion V0 GM G r so value of G should incease , so that v0 will increase for this position and which will become equal to v1 (5) (ii) path will becomemore elliptical , keeping r2 constant and r1 to decrease Density of given material M 3M 3 3 4 28πR 3 3 π[(2R) R ] Vwhole = Vhole + Vremaing Vremaing = Vwhole – Vhole 3 GM1 GM2 2 2R R 192 Downloaded from www.studiestoday.com Downloaded from www.studiestoday.com 4 8 3 here M1 π (2R) M 3 7 4 M M 2 ( ) R 3 3 7 Vremaining 9GM 14R Match the Column (1) (2) on the surface of earth acceleration due to gravity is g and gravitational potential is V match the following Table - 1 Table -2 (A) At height h = R value of g (P) decreases by a factor 1 4 (B) At height h = R/2 value of g (Q) decreases by a factor 1 2 (C) (D) At height h = R value of v At depth h = R/2 value of V (R) increases by a factor 11/8 (S) increases by a factor 2 (T) None Density of planet is two times the density of earth Radius of this planet is half (As compared to earth) Match the following Table-1 Table-2 (A) Acceleration due to gravity on (P) Half this planet’s surface (B) Gravitational potential (Q) same on the surface (C) (3) Gravitational potential (R) Two times at centre (D) Gravitational field strength (S) four times at centre let V and E denote the gravitational potential and gravitational field at a point. Then the match the follwing Table-1 Table-2 (A) E = 0, V = 0 (P) At center of Spherical shell (Q) At centre of solid sphere (B) E 0, V= o (C) V 0, E -0 (R) At centre of circular ring (D) V 0, E 0 (S) At centre of two point masses of equal magnitude (T) None 193 Downloaded from www.studiestoday.com Downloaded from www.studiestoday.com (4) (5) Match the following Table-1 (A) time period of an earth satellite in circular orbit (B) Orbital velocity of satellite (C) Mechnical energy of stellite Table -2 (P) Independent of mas of satellite (Q) independent of radius of orbit (R) independent of mass of earth (S) none Match the following (for a satellite in circular orbit) Table-1 Table-2 (A) kinetic energy (P) GMm 2r (B) potential energy (Q) GM r (C) Total energy (R) (D) orbital velocity (S) GMm 2r GMm r solution : (1) A-p, B-Q, C-s D-t (2) A-Q, B-p, C-p D-o (3) A-t, B-t, C-p,q,r,s, D-t (4) A-q, B-t, C-r, D-s (5) A-s, s-B, c-p, D-q • • • 194 Downloaded from www.studiestoday.com