A uniform rod of mass m and length l suspended by means of two identical inextensible light strings as shown in figure. Tension in one string immediately after the other string is cut, is \_\_\_\_ . (g acceleration due

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Correct answer: A. m g / 4

When string at B is cut, rod rotates about A.
Torque about A:

\tau = mg \times \frac{l}{2}

Moment of inertia about A:

I_A = \frac{ml^2}{3}

Angular acceleration:

\alpha = \frac{mg \cdot l/2}{ml^2/3} = \frac{3g}{2l}

Acceleration of COM:

a_{cm} = \alpha \times \frac{l}{2} = \frac{3g}{4}

Newton's second law:

mg - T = ma_{cm}

T = mg - \frac{3mg}{4} = \frac{mg}{4}

A uniform rod of mass m and length l suspended by means of two...

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