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Q.

A uniform rod of length 2m and mass $M = 20 k g$ 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 $\frac{M g}{2}$

b

Tension in the string is $\frac{5 M g}{12}$

c

Vertical component of hinge reaction on the rod is $\frac{M g}{2}$

d

Horizontal component of hinge reaction on the rod is zero

answer is A, C, D.

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Detailed Solution

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Force balance $N + T = M g$
Torque balance about $C O M \Rightarrow N = T$

So, $N = T = \frac{M g}{2}$

No component of hinge reaction in horizontal direction.

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