Q60 / 75 · JEE Main 2026 (23 January Shift 1) · Physics
Two small balls with masses $m$ and 2 m are attached to both ends of a rigid rod of length $d$ and negligible mass. If angular momentum of this system is $L$ about an axis ($A$) passing through its centre of mass and perpendicular to the rod then angular velocity of the system about $A$ is :
- A$\frac{2 L}{5 m d^{2}}$
- B$\frac{4}{3} \frac{L}{m d^{2}}$
- Correct answerC$\frac{3}{2} \frac{L}{m d^{2}}$
- D$\frac{2 L}{m d^{2}}$
Solution
CM from mass $m$: $x_{cm} = \frac{2md}{3m} = \frac{2d}{3}$.
Distances: $m$ is $2d/3$ from CM, $2m$ is $d/3$ from CM.
$I_{cm} = m(2d/3)^2 + 2m(d/3)^2 = \frac{4md^2}{9} + \frac{2md^2}{9} = \frac{2md^2}{3}$.
$L = I\omega \Rightarrow \omega = \frac{L}{I} = \frac{3L}{2md^2}$.
Distances: $m$ is $2d/3$ from CM, $2m$ is $d/3$ from CM.
$I_{cm} = m(2d/3)^2 + 2m(d/3)^2 = \frac{4md^2}{9} + \frac{2md^2}{9} = \frac{2md^2}{3}$.
$L = I\omega \Rightarrow \omega = \frac{L}{I} = \frac{3L}{2md^2}$.
Hence, the correct answer is option: 'C'.