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Question

A smooth light pulley fixed to the ceiling of a lift carries a light thread whose ends are attached to the loads to masses m1 and m2. The lift starts going up with acceleration a. The forceexerted by the pulley on the ceiling is.(m1>m2)

A
4m1m2(m1+m2)(g+a)
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B
2m1m2(m1+m2)(ga)
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C
4m1m2(m1+m2)(g)
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D
3m1m2(m1+m2)(g+a)
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Solution

The correct option is A 4m1m2(m1+m2)(g+a)
For the motion as per above figure,
Let's assume (m1>m2) so that acceleration of m2 is A downwards and m1 is A upwards,

Then we have,
m2(g+a) - T=m2A
T-m1(g+a) =m1A

Solving we get T=2m2m1(m1+m2)(g+a)

Thus force exerted by pulley on ceiling =2T=4m2m1(m1+m2)(g+a)

1303563_1094018_ans_4590323cd4eb4509a21731063c53ab1e.jpg

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