JEE Physics 2026 Question with Solution

Given: Bob A of mass $m$ is attached to a rigid rod of length $1 \, \text{m}$ and released from $60^\circ$ with the vertical. Bob B of equal mass is initially at rest. The collision is elastic, and bob B just completes a vertical circle of radius $R$.

Find: The value of $R$ and the correct option.

From the swing of bob A, the loss in height is

$$h = L(1 - \cos \theta)$$

With $L = 1 \, \text{m}$ and $\theta = 60^\circ$,

$$h = 1(1 - \cos 60^\circ) = 1\left(1 - \frac{1}{2}\right) = \frac{1}{2} \, \text{m}$$

Using conservation of mechanical energy for bob A,

$$\frac{1}{2}mv_A^2 = mgh$$ $$v_A = \sqrt{2gh} = \sqrt{2g \cdot \frac{1}{2}} = \sqrt{g}$$

In an elastic collision between two equal masses, when one mass is initially at rest, the velocities are exchanged. Therefore bob B acquires speed

$$v_B = \sqrt{g}$$

For a particle to just complete a vertical circle of radius $R$, the minimum speed required at the lowest point is

$$v_B = \sqrt{5gR}$$

Hence,

$$\sqrt{g} = \sqrt{5gR}$$

Squaring both sides,

$$g = 5gR$$ $$R = \frac{1}{5} \, \text{m}$$

Therefore, the radius is $\frac{1}{5} \, \text{m}$ and the correct option is C.

Given: The masses of A and B are equal, and the collision is elastic.

Find: Use the quickest route to determine $R$.

First find the speed of bob A at the bottom from the vertical drop

$$h = 1(1 - \cos 60^\circ) = \frac{1}{2}$$

So,

$$v_A = \sqrt{2gh} = \sqrt{g}$$

Now use the key trick: in a head-on elastic collision of two equal masses with one initially at rest, they exchange velocities. Hence immediately after collision,

$$v_B = \sqrt{g}$$

For just complete looping,

$$v_{\text{bottom,min}} = \sqrt{5gR}$$

So,

$$\sqrt{g} = \sqrt{5gR} \Rightarrow R = \frac{1}{5}$$

This works because the equal-mass elastic collision removes the need to separately solve momentum and restitution equations. Therefore, the correct option is C.