JEE Main 2026 — Jan 23, Shift 1 — Physics solution

JEE Main2026Jan 23, Shift 1Physics

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
$\frac{3}{2} \frac{L}{m d ^{2}}$
B
$\frac{2 L}{m d ^{2}}$
C
$\frac{4}{3} \frac{L}{m d ^{2}}$
D
$\frac{2 L}{5 m d ^{2}}$

Solution

The centre of mass divides the rod such that $m$ is at distance $\frac{2 d}{3}$ and $2 m$ is at distance $\frac{d}{3}$ from the CoM.

Moment of inertia about the perpendicular axis through CoM:

$I = m (\frac{2 d}{3})^{2} + 2 m (\frac{d}{3})^{2} = \frac{4 m d ^{2}}{9} + \frac{2 m d ^{2}}{9} = \frac{6 m d ^{2}}{9} = \frac{2 m d ^{2}}{3}$

$ω = \frac{L}{I} = \frac{L}{\frac{2 m d ^{2}}{3}} = \frac{3 L}{2 m d ^{2}}$

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