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A block 'A' of mass 'm' is attached at one end of a light spring and the other end of the spring is connected to another block 'B' of mass 2 m through a light string as shown in the figure.'A' is held and B is in static equilibrium. Now A is released. The acceleration of A just after that instant is 'a'. In the next case, B is held and A is in static equilibrium. Now when,B is released, its acceleration immediately after the release is 'b'. The value of a/b is: (Pulley, string and the spring are massless)

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a

0

b

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c

2

d

12

answer is C.

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Detailed Solution

For first case tension in spring will beTs = 2mg just after 'A' is released.2mg-mg = ma ⇒ a = gIn second case Ts = mg 2mg - mg = 2mbb = g2ab= 2

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A block 'A' of mass 'm' is attached at one end of a light spring and the other end of the spring is connected to another block 'B' of mass 2 m through a light string as shown in the figure.'A' is held and B is in static equilibrium. Now A is released. The acceleration of A just after that instant is 'a'. In the next case, B is held and A is in static equilibrium. Now when,B is released, its acceleration immediately after the release is 'b'. The value of a/b is: (Pulley, string and the spring are massless)