JEE Physics 2026 Question with Solution

Given: Two masses $m$ and $2m$ are fixed at the ends of a rigid rod of length $d$. The angular momentum about the centre of mass axis perpendicular to the rod is $L$.

Find: The angular velocity $\omega$ about the same axis.

Using the relation

$$L = I\omega$$

we first calculate the moment of inertia about the centre of mass.

The distance of the centre of mass from mass $m$ is

$$r_1 = \frac{2m \cdot d}{m+2m} = \frac{2d}{3}$$

and the distance from mass $2m$ is

$$r_2 = d - \frac{2d}{3} = \frac{d}{3}$$

Therefore, the moment of inertia is

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

Now,

$$\omega = \frac{L}{I} = \frac{L}{\frac{2}{3}md^2} = \frac{3L}{2md^2}$$

Therefore, the angular velocity is $\frac{3L}{2md^2}$ and the correct option is C.

Given: Two point masses $m_1 = m$ and $m_2 = 2m$ are separated by distance $d$.

Find: The angular velocity $\omega$ when angular momentum is $L$.

For a two-body system about the centre of mass,

$$I = \mu d^2$$

where the reduced mass is

$$\mu = \frac{m_1m_2}{m_1+m_2} = \frac{m \cdot 2m}{m+2m} = \frac{2m}{3}$$

Hence,

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

Using

$$L = I\omega$$

we get

$$\omega = \frac{L}{I} = \frac{L}{\frac{2}{3}md^2} = \frac{3L}{2md^2}$$

This shortcut works because the moment of inertia of two masses about their centre of mass can be written directly in terms of reduced mass. Therefore, the correct option is C.