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Q.

A uniform rod of length  2m and mass  M=20kg is hinged at point  O such that it is free to rotate in a vertical plane about a horizontal axis passing through point  O and perpendicular to the plane of the paper. The rod is kept in equilibrium with the help of an ideal string. Then

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a

Tension in the string is Mg2

b

Tension in the string is 5Mg12

c

Vertical component of hinge reaction on the rod is Mg2

d

Horizontal component of hinge reaction on the rod is zero

answer is A.

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Detailed Solution

Force balance  N+T=MgTorque balance about  COM⇒N=TSo, N=T=Mg2No component of hinge reaction in horizontal direction.
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