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Question

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:
132875_f06a6561d65141ec9324d33f835c1782.png
  1. 2MgdM+m
  2. 2(M+m)gdM
  3. 2(Mm)dgM+m
  4. 2(Mm)dgM

A
2MgdM+m
B
2(Mm)dgM
C
2(M+m)gdM
D
2(Mm)dgM+m
Solution
Verified by Toppr

Let a be the acceleration of the system as well as of the block of mass M.
for mass M,
Ma=MgT....(1)
for mass m,
ma=Tmg.....(2)
(1)+(2),a=(Mm)g(M+m)
now using the formula, v2u2=2as
here, u=0,s=d so , v2=2ad=2(Mm)gd(M+m)
v=2(Mm)gd(M+m)
Ans:(C)

148255_132875_ans_896c19c8b83d41c2af32ca556622c92b.png

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