Tiêu đề 08. Gravitation Hard Ans.pdf ✅

1. (c) If two particles of mass m are placed x distance apart then force of attraction F x Gmm = 2 (Let) Now according to problem particle of mass m is placed at the centre (P) of square. Then it will experience four forces FPA = force at point P due to particle F x Gmm A = = 2 Similarly F x G mm FPB 2 2 2 = = , F x G mm FPC 3 3 2 = = and F x G mm FPD 4 4 2 = = Hence the net force on P Fnet = FPA + FPB + FPC + FPD = 2 2 F  2 2 2 x Gmm Fnet = 2 2 ( / 2) 2 2 a Gm = [ = = 2 a x half of the diagonal of the square] 2 2 4 2 a Gm = . 2. (a) Acceleration due to gravity g GR 3 4 =  g   R or e m e m e m R R g g .   = [As 6 1 = e m g g and 3 5 = m e   (given)]  3 5 6 1 =                  = m e e m e m g g R R    Rm Re 18 5 = 3. (c) We know 2 2 2 4 ( / 2) D GM D GM R GM g = = = If mass of the planet = M 0 and diameter of the planet = D0 . Then 2 0 4 0 D GM g = . 4. (b) Acceleration due to gravity 2 R GM g =  2 2 2 1 4 80 1 .             = = moon earth earth moon earth moon R R M M g g 80 5 16 g g g moon = earth  = . 5. (c) We know 2 1 R g  [As R decreases, g increases] So % change in g = 2 (% change in R ) = 2  1% = 2%  acceleration due to gravity increases by 2%. 6. (b) We know 2 R GM g =  1.4 6.67 10 7.34 10 1 1 2 2    = = − g GM R m 6 = 1.87 10 . 7. (b) At equator the value of g is minimum so it is profitable to purchase sugar at this position. 8. (a) 9 1 2 2 2  =      +  =      + =  R R R R h R g g  9 g g  = . 9. (b) Acceleration due to gravity at height h is given by 2       +  = R h R g g 2 100       +  = R h R g g 10 1 = +  R h R  h = 9R . 10. (c) Weight of the body at height R, 2       +  = R h R W W 2 2             + = R R R W 72 32 . 9 4 9 4 3 2 2 W  = W =  = N      = 11. (b) Percentage change in g when the body is raised to height h , 1% 2 100 100 % =   =  R h g g Percentage change in g when the body is taken into depth d, 100% = 100 % = 100%  R h R d g g [As d = h ] D C A B m 2m 4m 3m P m FPD FPA FPC FPB
 Percentage decrease in weight (1%) 0.5% 2 1 100 2 2 1  = =      =  R h . 12. (b)        = − R d g g 1        = − R d g g 1 4 4 3R d = 13. (a) Acceleration due to gravity at depth d,        = − R d g g 1       = − 6400 100 g 1       = − 64 1 9.8 1 2 9.66 / 64 63 = 9.8  = m s . 14. (b)        = − R d g g 1        = − R d g n g 1 R n n d R n d       −  = −  = 1 1 1 15. (a) Effective acceleration due to gravity due to rotation of earth   2 2 g  = g − R cos  o 0 g R cos 60 2 2 = −  g R = 4 2   sec rad R g R g 800 2 2 4  = = = [As g  = 0 and o  = 60 ]  sec 2.5 10 400 1 −3 rad  = =  . 16. (a) When earth stops suddenly, centrifugal force on the man becomes zero so its effective weight increases. 17. (a) Time period for the given condition hr hr g R T = 2 = 1.40  1.5 nearly. 18. (b)           +   +   = − k z V j y V i x V I ˆ ˆ ˆ ( i j k) ˆ 12 ˆ 4 ˆ = − 3 + + [As V = (3x + 4y + 12z) (given)] It is uniform field Hence its value is same every where 2 2 2 1 | | 3 4 12 13 − I = + + = Nkg . 19. (d) Intensity at the origin ....... I = I 1 + I 2 + I 3 + I 4 + .......... 2 4 2 3 2 2 2 1 = + + + + r GM r GM r GM r GM       = + + + + ....... 8 1 4 1 2 1 1 1 2 2 2 2 GM       = + + + + ......... 64 1 16 1 4 1 GM 1             − = 4 1 1 1 GM [As sum of G.P. r a − = 1 ] 3 4 = GM  G 4G 3 4 =  3  = [As M = 3kg given] 20. (b) Gravitational field on a mass m due to outer shell (radius 2 r ) will be zero because the mass is placed inside this shell. But the inner shell (radius 1 r ) behaves like point mass placed at the centre so 2 1 r GM I = for 1 2 r  r  r 21. (d) Intensity due to uniform circular ring at a point on its axis 2 2 3 / 2 (a r ) Gmr I + =  Force on sphere 2 2 3 / 2 (a r ) GMmr F + = 2 2 3 / 2 ( ( 3 ) ) 3 a a GMm a + = 2 3 / 2 2 8 3 (4 ) 3 a GMm a GMm a = = [As r = 3a ] 22. (b) As dx dV I = − , if I = 0 then V = constant. 23. (d) V E dx  = − dx x K  = − 3 2 2x K = . 24. (c) 3kg 3kg 3kg 3kg 1m 2m 4m 8m O
Gravitational intensity at point P, 2 r GM I = and gravitational potential r GM V = −  V = I  r = 6 N / kg  8000 km kg 7 Joule = 4.8  10 . 25. (b) Potential increases by +10 J / kg every where so it will be +10 − 5 = +5J / kg at P 26. (b) Net potential at origin       = − + + + ......... 1 2 3 r Gm r Gm r Gm V       = − + + + 8 1 4 1 2 1 1 1 Gm Gm 2Gm 2 1 1 1 = −             − = − 27. (a) 28. (a) 2 3 2 8 GR R Gm ve = =  ve  R if  = constant. Since the planet having double radius in comparison to earth therefore the escape velocity becomes twice i.e. 22 km/ s . 29. (c) Escape velocity ve  R  and if density remains constant ve  R So if the radius reduces by 4% then escape velocity also reduces by 4%. 30. (c) Kinetic energy given to rocket at the surface of earth = Change in potential energy of the rocket in reaching from ground to highest point  h R mgh mv 2 1 / 1 2 + =  h R v g 2 1 1 2 + =  2 1 1 2 v g h R + =  v R g h 1 2 1 2 = −  v R gR v h 2 2 1 2 − =  2 2 2gR v v R h − =        − = 1 2 2 v gR R h 31. (c) Potential energy of the body at a distance 4Re from the surface of earth U = − mgRe 1+h/Re = − mgRe 1+4 = − mgRe 5 [As h = 4Re (given)] So minimum energy required to escape the body will be mgRe 5 . 32. (a) According to Kepler’s law T ∝ R 3/2 ∴ Tplanet = (5) 3/2Tearth = 5 (3/2) × 1year = 5 3/2years. 33. (b) Orbital velocity r Gm v =  r v 1  [If r decreases then v increases] Percentage change in 2 1 v = (Percentage change in r) 2 1 = (1%) = 0.5%  orbital velocity increases by 0.5%. 34. (b) If n R F 1  then 1 1 −  n R v ; here n = 1 0 1 1 1 R R v   − . 35. (a) Orbital velocity 3 11 24 38400 10 6.67 10 6 10     = = − r GM v v = 1.02 km / sec = 1km / sec (Approx.) 36. (d) Orbital radius of second satellite is 2% more than first satellite So from 3 / 2 T  (r) , Percentage increase in time period 2 3 = (Percentage increase in orbital radius) 2 3 = (2%) = 3%. 37. (b) m 1m 2m 4m 8m O m m m
GM R h T 3 ( ) 2 + =  2 3 ( ) 2 gR R + R =  g 8R = 2 g R = 4 2 [As h = R (given)] 38. (b) Orbital radius of satellite s c r = 4r (given) From Kepler’s law 3 / 2 T  r  3 / 2 3 / 2 = (4)         = c s c s s r T T  Ts Tc = 8 = 8 1 day = 8 days. 39. (b) 2 3 2 gR r T =  2 3 (9 ) 2 gR R =  g 3 / 2 R = 2(9) g R = 27  2 [As r = 9R (given)]. 40. (c) Both the particles moves diametrically opposite position along the circular path of radius R and the gravitational force provides required centripetal force 2 2 (2R) Gmm R mv =  R Gm v 2 1 = 41. (a) Because in beam type balance effect of less gravitation force works on both the Pans. So it is neutralizes but in spring balance weight of the body decreases so apparent weight varies with actual weight. 42. (c) If body falls from height h then time of descent g h t 2 =  = = 6 moon earth earth moon g g t t  t t moon = 6 . 43. (d) Applying conservation of mechanical energy between A and B point       − = + − R GMm mv r GMm 2 2 1 ; r GMm R GMm mv = − 2 2 1 r Gm R Gm v 2 2 2 = − 2 0 2 v 2v = e − 2 0 2 v v 2v  = e − [As escape velocity R Gm ve 2 = , orbital velocity r Gm v0 = ] 44. (b) Gravitational potential at the surface of the earth R GM Vs = − Gravitational potential at the centre of earth R GM Vc 2 3 = − By the conservation of energy ( ) mv = m Vs − Vc 2 2 1       = − 1 2 3 2 2 R GM v gR R GM = = 2 2 e v = [As ve = 2gR ]  2 e v v = 45. (c) T 2  r 3 8 1 r r/4 T T 3/2 2 1  =      = 46. (d) Weight W = 2 R GMm = ( ) 20.8N. 2.5 10 6.67 10 10 2 2 7 11 26 =     − 47. (d) v0 = r GM or r = 2 v0 GM and T = GM 2 r 3/2  = 3 v0 2GM T = ( ) 175min. 6280 6280 60 2 6.67 10 6 10 2 11 24 =        − 48. (c) F = 2 1 2 4R Gm m shell exerts no force as its gravitational field inside the shell is zero. Oil does not play any role. 49. (a) B A v u = 0 r m m 2R