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Question

# In the figure shown, the pulleys and strings are massless. The acceleration of the block of mass 4m just after the system is released from rest is (θ=sin−135)

A
2g5 downwards
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B
2g5 upwards
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C
5g11 upwards
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D
5g11 downwards.
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Solution

## The correct option is D 5g11 downwards.The FBD of blocks is as shown From Newton's second law 4mg−2Tcosθ=4mA....(1) and T - mg = ma.....(2) cosθ=45 and from constraint we get a=Acosθ......(3) Solving equation (1), (2) and (3) we get acceleration of block of mass 4m , A=5g11 downwards.

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