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# The pulley (figure) is fixed, light and smooth. The string is light and mass m < mass M. At t = 0, M and m are at rest, when the system is released. At time t = t1 the mass 'M' is displaced by 'd'. The speed of the big block M at time t1 with respect to the pulley is:√2MgdM+m √2(M+m)gdM √2(M−m)dgM+m √2(M−m)dgM

A
2MgdM+m
B
2(Mm)dgM
C
2(M+m)gdM
D
2(Mm)dgM+m
Solution
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#### Let a be the acceleration of the system as well as of the block of mass M.for mass M, Ma=Mg−T....(1)for mass m, ma=T−mg.....(2)(1)+(2),a=(M−m)g(M+m)now using the formula, v2−u2=2ashere, u=0,s=d so , v2=2ad=2(M−m)gd(M+m)∴v=√2(M−m)gd(M+m)Ans:(C)

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