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Q.

The given figure shows a block of mass m placed on a smooth wedge of mass M. Calculate the minimum value of M' and tension in the string, so that the block of mass m will move vertically downward with acceleration 10 ms-2.

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a

The value of M′ is Mcot⁡θ1−cot⁡θ.

b

The value of M is Mtan⁡θ1−tan⁡θ.

c

The value of tension in the string is Mgtan⁡θ.

d

The value of tension is Mgcot⁡θ.

answer is A.

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Detailed Solution

M′g−T=M′a     ........(i)T=Ma                ........(ii)M′g=aM+M′⇒ a=M′gM+M′masin⁡θ=mgcos⁡θ→so that normal force is zero.      a=gcot⁡θ     gcot⁡θ=M′gM+M′⇒cot⁡θM+cot⁡θM′=M′or  M′=Mcot⁡θ(1−cot⁡θ),T=Ma=Mgcot⁡θ=Mg/tan⁡θ
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