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Q.

In the figure  M1=4M2The pulleys are smooth and light. At time t = 0, the system is at rest. If

            the system is released, the acceleration of mass M1 is

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a

g/8

b

g/2

c

g

d

g/4

answer is C.

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Detailed Solution

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M1g−2T=M1a      →1 T−M2g=M22a→2 solving1&2        1⇒M1g−2T=M1a  2×2⇒2T−2M2g=2M22a M1g−2M2g=M1a+4M2a ∵M1=4M2 4M2g−2M2g=4M2a+4M2a 2M2g=8M2a a=g4

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