Page 1 10. Had q been 90°. For every x displacement of wedge ( ) w the vertical fall in mass would have been 2x as the string passes through pulley B. i.e., v v 1 2 2 = For the situation given the speed of mass would thus be v v 1 2 2 90 = °  cos ( ) q i.e., v v 1 2 2 = sin q Option (c) is correct. 11. The cylinder will start rising up the inclined plane if ma mg cos sin q q > i.e., a g > tan q \ a g min tan = q [cylinder will be at the point of rising up the inclined plane] 12. At the position of maximum deflection the net acceleration of bob (towards its mean position will be) g a sin cos q q  as explained in figure. 13. Mg T Ma  = T mg ma  = Þ a M m M m g =  + \ T m g a = + ( ) = +  + é ë ê ù û ú m g M m M m g = +  + é ë ê ù û ú mg M m M m 1 = + mg[ ] 1 1 {as m M << , M m M  » and M m M + » } = 2 mg \ Tension in the string suspended from ceiling = = 2 4 T mg Option (a) is correct. 14. For 0° < £ f q frictional force ( ) sin F mg = q For q ³ f frictional force ( ) cos F mg = m q [f = angle of repose] For example Let m = 1 3 \ tan f = 1 3 f = ° 30 q F (force of friction) Condition 10° mgsin10° = 0.174 mg 20° mgsin20° = 0.342 mg Increase but not linearly only 30° mmgcos30° = 0.500 mg 40° mmgcos40° = 0.442 mg Decrease but not linearly only 60° mmgcos60° = 0.287 mg 90° Zero Option (b) is correct. 15. a mg m B = m 2 = mg 2 Laws of Motion 91 ma cos q ma (Pseudo furce) O a cos q a g cos (90° – q) = g sin q 90°– q q g T T T T mg Mg 2T 2T m v 2 w m q v 1 90° – q v 1 B O 90° q F f Page 2 10. Had q been 90°. For every x displacement of wedge ( ) w the vertical fall in mass would have been 2x as the string passes through pulley B. i.e., v v 1 2 2 = For the situation given the speed of mass would thus be v v 1 2 2 90 = °  cos ( ) q i.e., v v 1 2 2 = sin q Option (c) is correct. 11. The cylinder will start rising up the inclined plane if ma mg cos sin q q > i.e., a g > tan q \ a g min tan = q [cylinder will be at the point of rising up the inclined plane] 12. At the position of maximum deflection the net acceleration of bob (towards its mean position will be) g a sin cos q q  as explained in figure. 13. Mg T Ma  = T mg ma  = Þ a M m M m g =  + \ T m g a = + ( ) = +  + é ë ê ù û ú m g M m M m g = +  + é ë ê ù û ú mg M m M m 1 = + mg[ ] 1 1 {as m M << , M m M  » and M m M + » } = 2 mg \ Tension in the string suspended from ceiling = = 2 4 T mg Option (a) is correct. 14. For 0° < £ f q frictional force ( ) sin F mg = q For q ³ f frictional force ( ) cos F mg = m q [f = angle of repose] For example Let m = 1 3 \ tan f = 1 3 f = ° 30 q F (force of friction) Condition 10° mgsin10° = 0.174 mg 20° mgsin20° = 0.342 mg Increase but not linearly only 30° mmgcos30° = 0.500 mg 40° mmgcos40° = 0.442 mg Decrease but not linearly only 60° mmgcos60° = 0.287 mg 90° Zero Option (b) is correct. 15. a mg m B = m 2 = mg 2 Laws of Motion 91 ma cos q ma (Pseudo furce) O a cos q a g cos (90° – q) = g sin q 90°– q q g T T T T mg Mg 2T 2T m v 2 w m q v 1 90° – q v 1 B O 90° q F f For no slipping a a A B = F mg m g  = m m 2 i.e., F mg = 3 2 m m = 2 3 F mg Slipping will obviously be m 2 there if m is greater than above mentioned value \ m min = 2 3 F mg For no slipping. Option (c) is correct. 16. F net (downward) = + mg ma sin cos q q = + m g a ( sin cos ) q q \ g g a eff = + sin cos q q Time ( ) T required to cover 2L distance along inclined would be T L g = 2 eff = + 2L g a ( sin cos ) q q Option (c) is correct. 17. F net on block along incline in the upward direction =  ma mg cos sin q q =  m a g ( cos sin ) q q \ g a g eff =  cos sin q q \ Time ( ) t to move s distance would be given by s g t = 1 2 2 eff i.e., t s g s a g = =  2 2 eff ( cos sin ) q q Substituting s = 1 m, q = ° 30 , a =10 3 m/s 2 and g = 10 m/s 2 t = ´ ´ æ è ç ö ø ÷ ´ æ è ç ö ø ÷ 2 1 10 3 3 2 10 1 2 _ = 1 5 s Option (b) is correct. 18. N w w w + ° = + 2 30 2 sin f = frictional force N w = 5 4 f N w max = = × m m 5 4 The block will remain stationary if w w 2 30 5 4 cos ° £ m or w w 2 3 2 5 4 £ m or 3 5 £ m or 3 5 £ m \ Block will move if m < 3 5 Option (d) is correct. 92  Mechanics1 2m m A B mmg F F F mmg F F q a ma ma cos q mg sin q q ma cos q mg sin q a w 2 sin 30° w/2 30° w 2 cos 30° w f w/2 N Page 3 10. Had q been 90°. For every x displacement of wedge ( ) w the vertical fall in mass would have been 2x as the string passes through pulley B. i.e., v v 1 2 2 = For the situation given the speed of mass would thus be v v 1 2 2 90 = °  cos ( ) q i.e., v v 1 2 2 = sin q Option (c) is correct. 11. The cylinder will start rising up the inclined plane if ma mg cos sin q q > i.e., a g > tan q \ a g min tan = q [cylinder will be at the point of rising up the inclined plane] 12. At the position of maximum deflection the net acceleration of bob (towards its mean position will be) g a sin cos q q  as explained in figure. 13. Mg T Ma  = T mg ma  = Þ a M m M m g =  + \ T m g a = + ( ) = +  + é ë ê ù û ú m g M m M m g = +  + é ë ê ù û ú mg M m M m 1 = + mg[ ] 1 1 {as m M << , M m M  » and M m M + » } = 2 mg \ Tension in the string suspended from ceiling = = 2 4 T mg Option (a) is correct. 14. For 0° < £ f q frictional force ( ) sin F mg = q For q ³ f frictional force ( ) cos F mg = m q [f = angle of repose] For example Let m = 1 3 \ tan f = 1 3 f = ° 30 q F (force of friction) Condition 10° mgsin10° = 0.174 mg 20° mgsin20° = 0.342 mg Increase but not linearly only 30° mmgcos30° = 0.500 mg 40° mmgcos40° = 0.442 mg Decrease but not linearly only 60° mmgcos60° = 0.287 mg 90° Zero Option (b) is correct. 15. a mg m B = m 2 = mg 2 Laws of Motion 91 ma cos q ma (Pseudo furce) O a cos q a g cos (90° – q) = g sin q 90°– q q g T T T T mg Mg 2T 2T m v 2 w m q v 1 90° – q v 1 B O 90° q F f For no slipping a a A B = F mg m g  = m m 2 i.e., F mg = 3 2 m m = 2 3 F mg Slipping will obviously be m 2 there if m is greater than above mentioned value \ m min = 2 3 F mg For no slipping. Option (c) is correct. 16. F net (downward) = + mg ma sin cos q q = + m g a ( sin cos ) q q \ g g a eff = + sin cos q q Time ( ) T required to cover 2L distance along inclined would be T L g = 2 eff = + 2L g a ( sin cos ) q q Option (c) is correct. 17. F net on block along incline in the upward direction =  ma mg cos sin q q =  m a g ( cos sin ) q q \ g a g eff =  cos sin q q \ Time ( ) t to move s distance would be given by s g t = 1 2 2 eff i.e., t s g s a g = =  2 2 eff ( cos sin ) q q Substituting s = 1 m, q = ° 30 , a =10 3 m/s 2 and g = 10 m/s 2 t = ´ ´ æ è ç ö ø ÷ ´ æ è ç ö ø ÷ 2 1 10 3 3 2 10 1 2 _ = 1 5 s Option (b) is correct. 18. N w w w + ° = + 2 30 2 sin f = frictional force N w = 5 4 f N w max = = × m m 5 4 The block will remain stationary if w w 2 30 5 4 cos ° £ m or w w 2 3 2 5 4 £ m or 3 5 £ m or 3 5 £ m \ Block will move if m < 3 5 Option (d) is correct. 92  Mechanics1 2m m A B mmg F F F mmg F F q a ma ma cos q mg sin q q ma cos q mg sin q a w 2 sin 30° w/2 30° w 2 cos 30° w f w/2 N 19. At t = 2 s F 1 4 = N f 1 and f 2 are the frictional forces ( ) max f 1 1 10 = ´ ´ 0.6 = 6 N ( ) max f 2 2 10 = ´ ´ 0.5 = 10 N At t = 2 s Net external force ( ) F net on system = 15 N  4 N = 11 N As F f f net > + ( ) ( ) max max 1 2 , the system will remain at rest and the values frictional forces on the blocks will be given T f = + 4 1 and T f =  15 2 4 15 1 2 + =  f f …(i) f f 1 2 11 + = N Let direction being + ive for Eq. (i) Option (a) f 1 4 =  N, f 2 5 = = N Þ f f 1 2 1 + = N wrong Option (b) f 1 2 =  N, f 2 5 = + N Þ f f 1 2 3 + = N wrong Option (c) f 1 0 = N,f 2 10 = + N Þ f f 1 2 10 + = N wrong Option (d) f 1 1 = + N, f 2 10 = + N Þ f f 1 2 11 + = N correct. OR As the likely movement would be towards right f 2 will be at its maximum. \ f 2 10 = N Þ f 1 1 = N Option (d) is correct. 20. 2 2 mg T ma sin a  = and T ma = \ 2 3 mg ma sin a = Þ a g = ° 2 30 3 sin = g 3 \ T mg = 3 Force ( ) R applied by clamp on pulley would be     T T 1 2 ® ® = = T T 1 3 30 3 30 ® = ° + ° mg mg (cos ) (sin ) ^ ^ i j = + mg mg 3 6 6 i j ^ ^   ^ T 2 3 ® = mg j Force by clamp on pulley P = + ® ® T T 1 2 = + + mg mg mg 3 6 6 3 i j j ^ ^ ^ = + mg mg 3 6 3 6 $ $ i j = + mg 6 3 3 ( ) ^ ^ i j Option (b) is correct. 21. f 1 4 10 (max) = ´ ´ 0.3 = 12 N f 1 and f 2 are frictional forces. f 2 2 10 12 (max) = ´ ´ = 0.6 N As, f f 1 2 16 (max) (max) + < N ( ) F ext Laws of Motion  93 f 1 m = 0.6 1 m = 0.5 2 f 2 F = 15N F = 2t 1 T T 15N 4N T T B A a = 30° m T T T T P T T 2m 60° x i y j ® ® T 1 x y T 2 60° f 1 m = 0.6 1 m = 0.3 2 f 2 16N 2 kg 4 kg Page 4 10. Had q been 90°. For every x displacement of wedge ( ) w the vertical fall in mass would have been 2x as the string passes through pulley B. i.e., v v 1 2 2 = For the situation given the speed of mass would thus be v v 1 2 2 90 = °  cos ( ) q i.e., v v 1 2 2 = sin q Option (c) is correct. 11. The cylinder will start rising up the inclined plane if ma mg cos sin q q > i.e., a g > tan q \ a g min tan = q [cylinder will be at the point of rising up the inclined plane] 12. At the position of maximum deflection the net acceleration of bob (towards its mean position will be) g a sin cos q q  as explained in figure. 13. Mg T Ma  = T mg ma  = Þ a M m M m g =  + \ T m g a = + ( ) = +  + é ë ê ù û ú m g M m M m g = +  + é ë ê ù û ú mg M m M m 1 = + mg[ ] 1 1 {as m M << , M m M  » and M m M + » } = 2 mg \ Tension in the string suspended from ceiling = = 2 4 T mg Option (a) is correct. 14. For 0° < £ f q frictional force ( ) sin F mg = q For q ³ f frictional force ( ) cos F mg = m q [f = angle of repose] For example Let m = 1 3 \ tan f = 1 3 f = ° 30 q F (force of friction) Condition 10° mgsin10° = 0.174 mg 20° mgsin20° = 0.342 mg Increase but not linearly only 30° mmgcos30° = 0.500 mg 40° mmgcos40° = 0.442 mg Decrease but not linearly only 60° mmgcos60° = 0.287 mg 90° Zero Option (b) is correct. 15. a mg m B = m 2 = mg 2 Laws of Motion 91 ma cos q ma (Pseudo furce) O a cos q a g cos (90° – q) = g sin q 90°– q q g T T T T mg Mg 2T 2T m v 2 w m q v 1 90° – q v 1 B O 90° q F f For no slipping a a A B = F mg m g  = m m 2 i.e., F mg = 3 2 m m = 2 3 F mg Slipping will obviously be m 2 there if m is greater than above mentioned value \ m min = 2 3 F mg For no slipping. Option (c) is correct. 16. F net (downward) = + mg ma sin cos q q = + m g a ( sin cos ) q q \ g g a eff = + sin cos q q Time ( ) T required to cover 2L distance along inclined would be T L g = 2 eff = + 2L g a ( sin cos ) q q Option (c) is correct. 17. F net on block along incline in the upward direction =  ma mg cos sin q q =  m a g ( cos sin ) q q \ g a g eff =  cos sin q q \ Time ( ) t to move s distance would be given by s g t = 1 2 2 eff i.e., t s g s a g = =  2 2 eff ( cos sin ) q q Substituting s = 1 m, q = ° 30 , a =10 3 m/s 2 and g = 10 m/s 2 t = ´ ´ æ è ç ö ø ÷ ´ æ è ç ö ø ÷ 2 1 10 3 3 2 10 1 2 _ = 1 5 s Option (b) is correct. 18. N w w w + ° = + 2 30 2 sin f = frictional force N w = 5 4 f N w max = = × m m 5 4 The block will remain stationary if w w 2 30 5 4 cos ° £ m or w w 2 3 2 5 4 £ m or 3 5 £ m or 3 5 £ m \ Block will move if m < 3 5 Option (d) is correct. 92  Mechanics1 2m m A B mmg F F F mmg F F q a ma ma cos q mg sin q q ma cos q mg sin q a w 2 sin 30° w/2 30° w 2 cos 30° w f w/2 N 19. At t = 2 s F 1 4 = N f 1 and f 2 are the frictional forces ( ) max f 1 1 10 = ´ ´ 0.6 = 6 N ( ) max f 2 2 10 = ´ ´ 0.5 = 10 N At t = 2 s Net external force ( ) F net on system = 15 N  4 N = 11 N As F f f net > + ( ) ( ) max max 1 2 , the system will remain at rest and the values frictional forces on the blocks will be given T f = + 4 1 and T f =  15 2 4 15 1 2 + =  f f …(i) f f 1 2 11 + = N Let direction being + ive for Eq. (i) Option (a) f 1 4 =  N, f 2 5 = = N Þ f f 1 2 1 + = N wrong Option (b) f 1 2 =  N, f 2 5 = + N Þ f f 1 2 3 + = N wrong Option (c) f 1 0 = N,f 2 10 = + N Þ f f 1 2 10 + = N wrong Option (d) f 1 1 = + N, f 2 10 = + N Þ f f 1 2 11 + = N correct. OR As the likely movement would be towards right f 2 will be at its maximum. \ f 2 10 = N Þ f 1 1 = N Option (d) is correct. 20. 2 2 mg T ma sin a  = and T ma = \ 2 3 mg ma sin a = Þ a g = ° 2 30 3 sin = g 3 \ T mg = 3 Force ( ) R applied by clamp on pulley would be     T T 1 2 ® ® = = T T 1 3 30 3 30 ® = ° + ° mg mg (cos ) (sin ) ^ ^ i j = + mg mg 3 6 6 i j ^ ^   ^ T 2 3 ® = mg j Force by clamp on pulley P = + ® ® T T 1 2 = + + mg mg mg 3 6 6 3 i j j ^ ^ ^ = + mg mg 3 6 3 6 $ $ i j = + mg 6 3 3 ( ) ^ ^ i j Option (b) is correct. 21. f 1 4 10 (max) = ´ ´ 0.3 = 12 N f 1 and f 2 are frictional forces. f 2 2 10 12 (max) = ´ ´ = 0.6 N As, f f 1 2 16 (max) (max) + < N ( ) F ext Laws of Motion  93 f 1 m = 0.6 1 m = 0.5 2 f 2 F = 15N F = 2t 1 T T 15N 4N T T B A a = 30° m T T T T P T T 2m 60° x i y j ® ® T 1 x y T 2 60° f 1 m = 0.6 1 m = 0.3 2 f 2 16N 2 kg 4 kg The system will remain at rest. For the equilibrium of 4 kg mass : \ 16 1 = + T f …(i) As f 1 will be at its maximum value f 1 12 = N \ T =  16 12 = 4 N [from Eq. (i)] Further, for the equilibrium of 2 kg mass. T f = 1 \ f 1 4 = N Option (c) is correct. 22. For the rotational equilibrium of rod Taking moment about O. R l s l ´ = 2 2 cos sin q q or mg ma cos sin q q = Þ a g = cotq Option (d) is correct. 23. v t = 2 2 \ a dv dt d dt t t = = = ( ) 2 4 2 At t = 1 s, a =  4 2 ms As a g s = m m s a g = = = 4 10 0.4 Option (c) is correct. 24. Just at the position of tipping off, R 1 will be zero. \ Taking moment about point Q ( )( ) ( )( ) 10 4 80 g g x = \ x = 1 2 m Option (a) is correct. 25. N ma mg = + sin cos q q …(i) Now, as the block does not slide ma mg cos sin q q = i.e., a g = tan q Substituting the found value of a in Eq. (i) N m g mg = + ( tan ) sin cos q q q = + é ë ê ù û ú = mg mg sin cos cos 2 q q q q sec When, the block stops a = 0, the value of normal force will be N mg ¢ = cos q \ N N mg mg ¢ = cos q q sec Option (c) is correct. 26. For the ro ta tional equi lib rium of the block Taking moment about O. Nx f a = 2 or ( cos ) ( sin ) mg x mg a q q = 2 or x a = 2 tan q or x a / tan 2 = q or tan tan f = q or f = q 94  Mechanics1 q O mg ma s (= ma) R (= mg) a A 8 m P R 1 R 2 1 m B A 1 m x 10 g mg cos q a ma ma sin q N mg sin q q q = 45° mg N mg sin q B O f mg cos q x A f f 1 f 2 16N T T 4 kg 2 kg Page 5 10. Had q been 90°. For every x displacement of wedge ( ) w the vertical fall in mass would have been 2x as the string passes through pulley B. i.e., v v 1 2 2 = For the situation given the speed of mass would thus be v v 1 2 2 90 = °  cos ( ) q i.e., v v 1 2 2 = sin q Option (c) is correct. 11. The cylinder will start rising up the inclined plane if ma mg cos sin q q > i.e., a g > tan q \ a g min tan = q [cylinder will be at the point of rising up the inclined plane] 12. At the position of maximum deflection the net acceleration of bob (towards its mean position will be) g a sin cos q q  as explained in figure. 13. Mg T Ma  = T mg ma  = Þ a M m M m g =  + \ T m g a = + ( ) = +  + é ë ê ù û ú m g M m M m g = +  + é ë ê ù û ú mg M m M m 1 = + mg[ ] 1 1 {as m M << , M m M  » and M m M + » } = 2 mg \ Tension in the string suspended from ceiling = = 2 4 T mg Option (a) is correct. 14. For 0° < £ f q frictional force ( ) sin F mg = q For q ³ f frictional force ( ) cos F mg = m q [f = angle of repose] For example Let m = 1 3 \ tan f = 1 3 f = ° 30 q F (force of friction) Condition 10° mgsin10° = 0.174 mg 20° mgsin20° = 0.342 mg Increase but not linearly only 30° mmgcos30° = 0.500 mg 40° mmgcos40° = 0.442 mg Decrease but not linearly only 60° mmgcos60° = 0.287 mg 90° Zero Option (b) is correct. 15. a mg m B = m 2 = mg 2 Laws of Motion 91 ma cos q ma (Pseudo furce) O a cos q a g cos (90° – q) = g sin q 90°– q q g T T T T mg Mg 2T 2T m v 2 w m q v 1 90° – q v 1 B O 90° q F f For no slipping a a A B = F mg m g  = m m 2 i.e., F mg = 3 2 m m = 2 3 F mg Slipping will obviously be m 2 there if m is greater than above mentioned value \ m min = 2 3 F mg For no slipping. Option (c) is correct. 16. F net (downward) = + mg ma sin cos q q = + m g a ( sin cos ) q q \ g g a eff = + sin cos q q Time ( ) T required to cover 2L distance along inclined would be T L g = 2 eff = + 2L g a ( sin cos ) q q Option (c) is correct. 17. F net on block along incline in the upward direction =  ma mg cos sin q q =  m a g ( cos sin ) q q \ g a g eff =  cos sin q q \ Time ( ) t to move s distance would be given by s g t = 1 2 2 eff i.e., t s g s a g = =  2 2 eff ( cos sin ) q q Substituting s = 1 m, q = ° 30 , a =10 3 m/s 2 and g = 10 m/s 2 t = ´ ´ æ è ç ö ø ÷ ´ æ è ç ö ø ÷ 2 1 10 3 3 2 10 1 2 _ = 1 5 s Option (b) is correct. 18. N w w w + ° = + 2 30 2 sin f = frictional force N w = 5 4 f N w max = = × m m 5 4 The block will remain stationary if w w 2 30 5 4 cos ° £ m or w w 2 3 2 5 4 £ m or 3 5 £ m or 3 5 £ m \ Block will move if m < 3 5 Option (d) is correct. 92  Mechanics1 2m m A B mmg F F F mmg F F q a ma ma cos q mg sin q q ma cos q mg sin q a w 2 sin 30° w/2 30° w 2 cos 30° w f w/2 N 19. At t = 2 s F 1 4 = N f 1 and f 2 are the frictional forces ( ) max f 1 1 10 = ´ ´ 0.6 = 6 N ( ) max f 2 2 10 = ´ ´ 0.5 = 10 N At t = 2 s Net external force ( ) F net on system = 15 N  4 N = 11 N As F f f net > + ( ) ( ) max max 1 2 , the system will remain at rest and the values frictional forces on the blocks will be given T f = + 4 1 and T f =  15 2 4 15 1 2 + =  f f …(i) f f 1 2 11 + = N Let direction being + ive for Eq. (i) Option (a) f 1 4 =  N, f 2 5 = = N Þ f f 1 2 1 + = N wrong Option (b) f 1 2 =  N, f 2 5 = + N Þ f f 1 2 3 + = N wrong Option (c) f 1 0 = N,f 2 10 = + N Þ f f 1 2 10 + = N wrong Option (d) f 1 1 = + N, f 2 10 = + N Þ f f 1 2 11 + = N correct. OR As the likely movement would be towards right f 2 will be at its maximum. \ f 2 10 = N Þ f 1 1 = N Option (d) is correct. 20. 2 2 mg T ma sin a  = and T ma = \ 2 3 mg ma sin a = Þ a g = ° 2 30 3 sin = g 3 \ T mg = 3 Force ( ) R applied by clamp on pulley would be     T T 1 2 ® ® = = T T 1 3 30 3 30 ® = ° + ° mg mg (cos ) (sin ) ^ ^ i j = + mg mg 3 6 6 i j ^ ^   ^ T 2 3 ® = mg j Force by clamp on pulley P = + ® ® T T 1 2 = + + mg mg mg 3 6 6 3 i j j ^ ^ ^ = + mg mg 3 6 3 6 $ $ i j = + mg 6 3 3 ( ) ^ ^ i j Option (b) is correct. 21. f 1 4 10 (max) = ´ ´ 0.3 = 12 N f 1 and f 2 are frictional forces. f 2 2 10 12 (max) = ´ ´ = 0.6 N As, f f 1 2 16 (max) (max) + < N ( ) F ext Laws of Motion  93 f 1 m = 0.6 1 m = 0.5 2 f 2 F = 15N F = 2t 1 T T 15N 4N T T B A a = 30° m T T T T P T T 2m 60° x i y j ® ® T 1 x y T 2 60° f 1 m = 0.6 1 m = 0.3 2 f 2 16N 2 kg 4 kg The system will remain at rest. For the equilibrium of 4 kg mass : \ 16 1 = + T f …(i) As f 1 will be at its maximum value f 1 12 = N \ T =  16 12 = 4 N [from Eq. (i)] Further, for the equilibrium of 2 kg mass. T f = 1 \ f 1 4 = N Option (c) is correct. 22. For the rotational equilibrium of rod Taking moment about O. R l s l ´ = 2 2 cos sin q q or mg ma cos sin q q = Þ a g = cotq Option (d) is correct. 23. v t = 2 2 \ a dv dt d dt t t = = = ( ) 2 4 2 At t = 1 s, a =  4 2 ms As a g s = m m s a g = = = 4 10 0.4 Option (c) is correct. 24. Just at the position of tipping off, R 1 will be zero. \ Taking moment about point Q ( )( ) ( )( ) 10 4 80 g g x = \ x = 1 2 m Option (a) is correct. 25. N ma mg = + sin cos q q …(i) Now, as the block does not slide ma mg cos sin q q = i.e., a g = tan q Substituting the found value of a in Eq. (i) N m g mg = + ( tan ) sin cos q q q = + é ë ê ù û ú = mg mg sin cos cos 2 q q q q sec When, the block stops a = 0, the value of normal force will be N mg ¢ = cos q \ N N mg mg ¢ = cos q q sec Option (c) is correct. 26. For the ro ta tional equi lib rium of the block Taking moment about O. Nx f a = 2 or ( cos ) ( sin ) mg x mg a q q = 2 or x a = 2 tan q or x a / tan 2 = q or tan tan f = q or f = q 94  Mechanics1 q O mg ma s (= ma) R (= mg) a A 8 m P R 1 R 2 1 m B A 1 m x 10 g mg cos q a ma ma sin q N mg sin q q q = 45° mg N mg sin q B O f mg cos q x A f f 1 f 2 16N T T 4 kg 2 kg Thus, the normal force ( ) N will pass through point A. Option (a) is correct. [Note : The cube will be just at the point of tilting (about point A). The cube will tilt if q is made greater than 45°]. 27. For the rotational equilibrium of the cube Moment of couple ( , ) N mg = Moment of couple ( , ) F f \ mgx Fa = or mgx mg a = × 3 i.e., x a = 3 Option (b) is correct. 28. Taking moment about point O. N l l N l l 1 2 2 4 2 6  æ è ç ö ø ÷ =  æ è ç ö ø ÷ N l N l 1 2 4 3 = \ N N 1 2 4 3 : : = Option (c) is correct. 29. ( / ) T ma 2 = ® Box T ma mg + = ® 2 2 Pendulum with respect to box \ 2 2 2 ma ma mg + = \ a g = / 3 30. N m a B B sinq= …(i) a a A B = tanq …(ii) Þ N m a B A sin tan q q = N m a B A = sin tan q q \ Force on rod by wedge N m a B A cos cos sin tan q q q q = = m a B A tan 2 q = ´ æ è ç ö ø ÷ = 10 9 3 4 160 2 N Option (c) is correct. 31. Net downward force on ring =  mg ma m =  m g a ( ) m \ g g a eff = m t L g = 2 eff =  2L g a m = ´  ´ 2 1 10 4 ( ) 0.5 = = 1 2 0.5 s Laws of Motion  95 l/6 l/4 N 1 N 2 mg a T T ma mg mg q = 37° A N sin q N N B –2 9 ms q a B a A ® ® a a A A B B = = ® ®     a a L = 1m –2 a = 4 ms f = mN = mma mg T N = ma a F ( ) = mg 3 N (= mg) mg x f = F 3 4 37° 5Read More
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