JEE Physics 2026 Question with Solution

Given: A uniform bar of length $12\,\text{cm}$ and mass $20m$ lies on a smooth table. Two point masses $2m$ and $m$ strike the bar simultaneously and stick to it at distances $2\,\text{cm}$ and $4\,\text{cm}$ from the center of the bar, respectively.

Find: The ratio $\frac{v}{\omega}$ after collision.

Since the table is smooth, there is no external torque on the system about the vertical axis. Therefore, angular momentum is conserved.

Step 1: Initial angular momentum

From the figure, the perpendicular distances from the center are:

$$r_1 = 2\,\text{cm}, \qquad r_2 = 4\,\text{cm}$$

The initial angular momentum is

$$L_i = (2m)v(2) + (m)v(4) = 4mv + 4mv = 8mv$$

Step 2: Moment of inertia after collision

Moment of inertia of the bar about its center:

$$I_{\text{bar}} = \frac{1}{12}(20m)(12)^2 = 240m$$

Moment of inertia of the point masses:

$$I_1 = 2m(2)^2 = 8m, \qquad I_2 = m(4)^2 = 16m$$

So total moment of inertia is

$$I = 240m + 8m + 16m = 264m$$

Step 3: Apply conservation of angular momentum

$$L_i = I\omega$$

So,

$$8mv = 264m\omega$$

Hence,

$$\frac{v}{\omega} = \frac{264}{8} = 33$$

Therefore, the correct option is D and the ratio is $33$.