A uniform rod AB of mass 2 kg and length 30 cm is at rest on a smooth horizontal surface. An impulse of force 0.2 N s is applied to end B. The time taken by the rod to turn through at right angles will be π/x s, where x = _______. Numerical answer type. Work it out, then open the answer.

Answer: 4. Rod is free, so it spins about its centre: JL/2 = (ML²)/(12)ω ⇒ ω = 6J/ML. ω = (6(0.2))/(2(0.3)) = 2 rad/s Time to turn π/2: t = (π/2)/(2) = π/4 s Answer: x = 4

A uniform rod AB of mass 2 kg and length 30 cm is at rest on a smooth… | JEE Main 2024 (1 Feb, Shift 2)

Q33JEE Main 2024 · 1 FEB · SHIFT 2Difficult

Angular Momentum & its Conservation

A uniform rod $AB$ of mass $2\text{ kg}$ and length $30\text{ cm}$ is at rest on a smooth horizontal surface. An impulse of force $0.2\text{ N s}$ is applied to end B. The time taken by the rod to turn through at right angles will be $\frac{\pi}{x}\text{ s}$ , where $x =$ _______.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 4

Rod is free, so it spins about its centre: $J\frac{L}{2} = \frac{ML^2}{12}\omega \Rightarrow \omega = \frac{6J}{ML}$ .
$\omega = \frac{6(0.2)}{2(0.3)} = 2\ \text{rad/s}$
Time to turn $\frac{\pi}{2}$ : $t = \frac{\pi/2}{2} = \frac{\pi}{4}\ \text{s}$
Answer: $x = 4$

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