JEE Physics 2026 Question with Solution

Given: A thin uniform rod of mass $M$ and length $L$ is pivoted at a point such that the pivot is at a height $\dfrac{L}{3}$ from one end. The rod falls from vertical to horizontal.

Find: The angular velocity $\omega$ when the rod hits the table top.

Use conservation of mechanical energy. The loss in gravitational potential energy of the centre of mass becomes rotational kinetic energy about the pivot.

The centre of mass of the rod is at its midpoint. Initially, with the rod vertical, the centre of mass is above the pivot by

$$\frac{2L}{3} - \frac{L}{2} = \frac{L}{6}$$

So the vertical drop of the centre of mass is

$$h = \frac{L}{6}$$

Now apply energy conservation:

$$Mgh = \frac{1}{2} I \omega^2$$

The moment of inertia of the rod about its centre is

$$I_{cm} = \frac{1}{12}ML^2$$

The distance of the centre of mass from the pivot is

$$\frac{L}{6}$$

Using the parallel axis theorem,

$$I = I_{cm} + M\left(\frac{L}{6}\right)^2$$ $$I = \frac{1}{12}ML^2 + M\left(\frac{L}{6}\right)^2 = \frac{1}{9}ML^2$$

Substitute into the energy equation:

$$Mg\left(\frac{L}{6}\right) = \frac{1}{2}\left(\frac{1}{9}ML^2\right)\omega^2$$

Solving,

$$\omega^2 = \frac{3g}{L}$$

Therefore,

$$\omega = \sqrt{\frac{3g}{L}}$$

So, the correct option is C.

Given: The rod rotates about a fixed pivot, not about its centre.

Find: Angular speed at the instant it becomes horizontal.

The key idea is to track the centre of mass and compute the moment of inertia about the pivot.

1. The rod length is $L$, so its centre of mass lies at $\dfrac{L}{2}$ from either end.
2. The pivot is located at a height $\dfrac{L}{3}$ from the lower end when the rod is vertical.
3. Hence the centre of mass is above the pivot by

$$\frac{L}{2} - \frac{L}{3} = \frac{L}{6}$$

4. When the rod becomes horizontal, the centre of mass is level with the pivot, so the drop in height is exactly

$$\frac{L}{6}$$

Thus the loss in potential energy is

$$\Delta U = Mg\left(\frac{L}{6}\right)$$

The rotational kinetic energy is

$$K = \frac{1}{2}I\omega^2$$

Now evaluate $I$ about the pivot:

$$I = \frac{1}{12}ML^2 + M\left(\frac{L}{6}\right)^2$$ $$I = \frac{1}{12}ML^2 + \frac{1}{36}ML^2 = \frac{1}{9}ML^2$$

Now equate energies:

$$Mg\left(\frac{L}{6}\right) = \frac{1}{2}\left(\frac{1}{9}ML^2\right)\omega^2$$

Cancel $M$ and simplify:

$$\frac{gL}{6} = \frac{L^2}{18}\omega^2$$ $$3gL = L^2\omega^2$$ $$\omega^2 = \frac{3g}{L}$$

Hence,

$$\omega = \sqrt{\frac{3g}{L}}$$

Therefore, the angular velocity is $\sqrt{\dfrac{3g}{L}}$.