Q70 / 75 · JEE Main 2026 (21 January Shift 1) · Physics
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 to gravity)
- Correct answerA$m g / 4$
- B$m g / 2$
- C$m g / 3$
- Dmg
Solution
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: $mg - T = ma_{cm}$
$T = mg - \frac{3mg}{4} = \frac{mg}{4}$
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: $mg - T = ma_{cm}$
$T = mg - \frac{3mg}{4} = \frac{mg}{4}$
Hence, the correct answer is option: 'A'.