Rotational Motion

JEE Main level, single correct, medium

143 of 280

Question 143 of 280

A uniform rod of mass mm and length ll 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(g acceleration due to gravity)

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Answer: (A) mg/4m g / 4

When string at B is cut, rod rotates about A.
Torque about A: τ=mg×l2\tau = mg \times \frac{l}{2}
Moment of inertia about A: IA=ml23I_A = \frac{ml^2}{3}
Angular acceleration: α=mg⋅l/2ml2/3=3g2l\alpha = \frac{mg \cdot l/2}{ml^2/3} = \frac{3g}{2l}
Acceleration of COM: acm=α×l2=3g4a_{cm} = \alpha \times \frac{l}{2} = \frac{3g}{4}
Newton's second law: mg−T=macmmg - T = ma_{cm}
T=mg−3mg4=mg4T = mg - \frac{3mg}{4} = \frac{mg}{4}

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