Tiêu đề 08. Gravitation Easy Ans(1).pdf ✅

GRAVITATION [EXPLANATIONS] 1. (a) 2. (b) As it depends on the weight of the body. 3. (b) Due to inertia of direction. 4. (b) 5. (a) 6. (d) . 1 2 r F  If r becomes double then F reduces to 4 F 7. (b) 8. (c) F = G m1m2 r 2 = 6.675 × 1×1 1 2 × 10 −11 = 6.675 × 10 −11N 9. (c) Centripetal force provided by the gravitational force of attraction between two particles i.e. 2 2 (2R) Gm m R mv  = R Gm v 2 1  = 10. (d) m = 6 × 10 24kg, ω = 2 × 10 −7 rad ⥂/⥂ s,R = 1.5 × 10 11m The force exerted by the sun on the earth F = mω2R By substituting the value we can get, F = 36 × 10 21N 11. (d) 12. (a) k represents gravitational constant which depends only on the system of units. 13. (d) Force will be zero at the point of zero intensity d m m m x 1 2 1 + = . 10 9 81 81 D D M M M = + = 14. (a) 15. (d) g = GM R2 = 6.67×10 −11×7.34×10 22 (1.74×10 6) 2 = 1.62N ⥂/⥂ kg 16. (b) Actually gravitational force provides the centripetal force. 17. (c) 18. (a) F ∝ xm × (1 − x)m = xm2 (1 − x) For maximum force dF dx = 0  dF dx = m2 − 2xm2 = 0 ⇒ x = 1 ⥂/⥂ 2 19. (c) 20. (a) 21. (d) 22. (b) The value of g at the height h from the surface of earth        = − R h g g 2 1 The value of g at depth x below the surface of earth        = − R x g g 1 These two are given equal, hence        = −      − R x R h 1 2 1 On solving, we get x = 2h 23. (d) Time period of simple pendulum ' 2 g l T =  In artificial satellite g' = 0  T = infinite. 24. (a) g GR 3 4 = . If  = constant then 2 1 2 1 R R g g = 25. (b) Time of decent g h t 2 = . In vacuum no other force works except gravity so time period will be exactly equal. 26. (a) 27. (b) Because acceleration due to gravity increases 28. (d) Because acceleration due to gravity decreases 29. (b) We know that 2 R GM g = m R O R m m1 m2 x d P
On the planet g g R GM gp 7 4 7 4 / 4 / 7 2 = = = Hence weight on the planet 400 gmwt 7 4 = 700  = 30. (b) In pendulum clock the time period depends on the value of g, while in spring watch, the time period is independent of the value of g. 31. (c) 2 0 0 2 0 0 2 4 ( /2) D GM D GM R GM g = = = 32. (a) 33. (b) g′ g = M′ M ( R R′ ) 2 = ( 2M M ) ( R 2R ) 2 = 1 2 g ′ = g 2 = 9.8 2 = 4.9 m ⥂/⥂ s 2 34. (c) 35. (a) 36. (c) For the condition of weightlessness at equator R g  =  rad/s 800 1 640 10 1 3 =   = 37. (c) 2 r GM g = . Since M and r are constant, so 2 g = 9.8 m/s 38. (c) 2 R GM g = and =    3 3 4 M R RG g R R G g     4 3 . 3 4 2 3  =   = 39. (a) Because value of g decreases when we move either in coal mine or at the top of mountain. 40. (d) g GR 3 4 =  2 2 1 1 2 1   R R g g = 41. (a) 2 R GM g = (Given 81 , 3.5 ) Me = Mm Re = Rm Substituting the above values, = 0.15 e m g g 42. (c) Value of g decreases when we go from poles to equator. 43. (d) 44. (b) Because value of g decreases with increasing height. 45. (a) g′ g = ( R R+h ) 2 = ( 6400 6400+64) 2 ⇒ g′ = 960.40 cm ⥂/⥂ s 2 46. (d) 47. (b)   2 2 g' = g − R cos 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 90o . 48. (b) . 2 R GM g = If radius shrinks to half of its present value then g will becomes four times. 49. (b) Using 2 R GM g = we get gm = g / 5 50. (a)   2 2 g g R cos = p− = − cos 60 2 2 gp  R = 2 4 1 gp − R 51. (b)        = −       = − R d g n g R d g g 1 1 R n n d       −  = 1 52. (a) 2 r GM g   2 1 r g  or g r 1  If g decrease by one percent then r should be increase by % 2 1 i.e. R = 6400 32 km 2 100 1  =  53. (c) g GR 3 4 = 2 2 1 1 2 1 R R g g    = 1 2 1 4 2 1 =  = 54. (b) 2 '       + = R h R g g  2 4       + = R h R g g  R h R + = 2 1  R + h = 2R  h = R 55. (c) Acceleration due to gravity at poles is independent of the angular speed of earth. 56. (a) Mass of the ball always remain constant. It does not depend upon the acceleration due to gravity 57. (d) 2 m m m R GM g = and 2 / 6 9.8 6 m s g g e m = = 2 = 1.63 m/s Substituting 1.768 10 , 6 Rm =  m 2 g 1.63 m/s m = and 11 2 2 G 6.67 10 N-m /kg − =  We get Mm kg 22 = 7.65 10
58. (b) 2 '       + = R h R g g  when h = R then 4 ' g g = So the weight of the body at this height will become one- fourth. 59. (c) 2 R GM g = and I L K 2 2 = If mass of the earth and its angular momentum remains constant then 2 1 R g  and 2 1 R K  i.e. if radius of earth decreases by 2% then g and K both increases by 4%. 60. (b) Weight is least at the equator. 61. (c) 2 1 R g  Percentage change in g = 2(percentage change in R) = 2 1.5 = −3% 62. (b) 2 1 R g  . If radius of earth decreases by 2% then g will increase by 4% i.e. weight of the body at earth surface will increase by 4% 63. (c) Mass does not vary from place to place. 64. (b) 2 R GM g =  g GM R = Substituting the above values we get 1.87 10 . 6 R =  m 65. (c) Weight of the body at equator = 3 5 of initial weight  g′ = 3 5 g (because mass remains constant) g′ = g − ω2R cos2 λ 3 5 g = g − ω2R cos2 ( 0°)  ω2 = 2g 5R  ω = √ 2g 5R = √ 2×10 5×6400×10 3 = 7.8 × 10 −4 rad sec 66. (b) h = 32 km , R = 6400 km , so h  R        = − R h g g 2 1        = − 6400 2 32 g 1  g g 0.99 g 100 99  = = 67. (a) Same change in the value of g can be observed at a depth x and height 2x given d = x =10 k m  h = 2x = 20 k m 68. (a) 69. (b) h R R h R R h R g g 9 100 1 2 2   =      +   =      + =  70. (a) g ′ = g ( R R+h ) 2 = g ( R R+ R 2 ) 2 = 4 9 g ∴ W′ = 4 9 × W = 4 9 × 72 = 32N 71. (a) g ′ = g − ω2R cos2 λ0 = g − ω2R cos2 6 0 o 0 = g − ω2R 4 ⇒ ω = 2√ g R = 1 400 rad sec = 2.5 × 10 −3 rad sec 72. (a) 2 9.66 / 6400 100 ' 1 9.8 1 m s R d g g  =       = −      = − 73. (c) 4 ' 2 g R h R g g  =      + = . By solving h = R 74. (a) g = 4 3 πρGR g ∝ rρ  ge gm = R r × ρe ρm 75. (c) 2 2 (2) 80 1 9.8       =                 = p e e p p e R R M M g g 2 = 9.8 / 20 = 0.49 m /s 76. (d) Range of projectile g u R sin 2 2 = if u and  are constant then g R 1  0.2 0.2 1 e m e m m e e m R R R R g g R R =  =  =  Rm = 5Re 77. (a) For condition of weightlessness of equator s rad R g 3 1.25 10 800 1 −  = = =  78. (d) R h R R h R g g +   =      + = 2 1 ' 2  R + h = 2 R h = ( 2 − 1)R = 0.414 R 79. (b) g   R 80. (c) g H g u H 1 2 2 =    B A A B g g H H = Now g R g g A B = as   12
 B A A B g g H H = =12 HB = 12  H A = 12 1.5 = 18m 81. (b) 82. (a) 2 2 1       +  =      +  = R h g R h R g g 83. (c) g = GR g  dR 3 4 ( = d given in the problem) 84. (a) Inside the earth g Gr 3 4 ' =  g' r 85. (c) g W W R h R g g 9 4 ' 9 4 ' 2  =  =      + = 86. (a) g   87. (d) 88. (d) 89. (a) g R R g R h R g g 9 4 3 / 2 ' 2 2  =       =      + =  W mg 880 N 9 4 200 9.8 9 4 ' =   =  = 90. (a) g = 4 3 πGρR ⇒ g ∝ ρR ⇒ ge gm = ρe ρm × Re Rm ⇒ 6 1 = 5 3 × Re Rm ⇒ Rm = 5 18 Re 91. (a) 2 9 ' 2 2 g R R R g R h R g g  =      +  =      + = 92. (b) d R R d g g R d g g 4 3 1 4 ' 1  =        = −      = − 93. (b) For height 1%; 2 100 % = =  R h g g For depth % 0.5% 2 1 100 % = = = =  R h R d g g 94. (b) As 2 R GM g = therefore 1% decrease in mass will decreases the value of g by 1%. But 1% decrease in radius will increase the value of g by 2%. As a whole value of g increase by 1%. 95. (d) g GR 3 4 =  ( )       =                  = 2 1 1 p e e p e p g g R R    2 2 R R R e p = = 96. (a) g1 g2 = ρ1 ρ2 × R1 R2 = 3 2 × 2 3 = 1 97. (d) Because the body weighs zero in satellite 98. (a) Radius of earth R = 6400 km  4 R h = Acceleration due to gravity at a height h 2       + = R h R gh g 2 4             + = R R R g g 25 16 = At depth 'd' value of acceleration due to gravity gd gh 2 1 = (According to problem)  gd g      = 25 16 2 1        − R d g 1  g      = 25 16 2 1 By solving we get d m 6 = 4.3 10 99. (a)   2 2 g  = g − R cos For weightlessness at equator  = 0 and g' = 0  g R 2 0 = −  800 sec 1 rad R g  = = 100.(b) Weight on surface of earth, mg = 500 N and weight below the surface of earth at 2 R d = N mg mg R d mg mg 250 2 2 1 1 1  = =       = −       = − 101.(a) g GR 3 4 = and g  = GR 3 4  = 0.2  =  R R g g  g  = 0.2 g 102. (a)