JEE Physics 2025 Question with Solution

Given: A uniform solid cylinder of mass $m$ and radius $r$ rolls without slipping down a rough incline of angle $\theta = 45^\circ$.

Find: The linear acceleration $a$ of the cylinder axis.

For rolling without slipping, use translation and rotation together. The rolling condition is

$$a = r\alpha$$

The moment of inertia of a solid cylinder about its axis is

$$I = \frac{1}{2}mr^2$$

Along the incline, Newton's second law gives

$$mg\sin\theta - f = ma$$

For rotational motion,

$$fr = I\alpha = \frac{1}{2}mr^2\alpha$$

Using $a = r\alpha$,

$$f = \frac{1}{2}ma$$

Substitute this into the translational equation:

$$mg\sin\theta - \frac{1}{2}ma = ma$$

So,

$$mg\sin\theta = \frac{3}{2}ma$$

Hence,

$$a = \frac{2}{3}g\sin\theta$$

Now substitute $\theta = 45^\circ$ and $\sin 45^\circ = \frac{1}{\sqrt{2}}$:

$$a = \frac{2}{3}g\cdot \frac{1}{\sqrt{2}} = \frac{\sqrt{2}g}{3}$$

Therefore, the linear acceleration of the cylinder axis is $\frac{\sqrt{2}g}{3}$. The correct option is C.

Given: A solid cylinder rolls without slipping down an incline of angle $45^\circ$.

Find: The linear acceleration $a$.

For a rigid body rolling without slipping,

$$a = \frac{g\sin\theta}{1 + \frac{I}{mr^2}}$$

For a solid cylinder,

$$I = \frac{1}{2}mr^2$$

So,

$$a = \frac{g\sin\theta}{1 + \frac{1}{2}} = \frac{2}{3}g\sin\theta$$

Now use $\sin 45^\circ = \frac{1}{\sqrt{2}}$:

$$a = \frac{2}{3\sqrt{2}}g = \frac{\sqrt{2}}{3}g$$

This shortcut works because the standard rolling formula already combines translational dynamics, rotational dynamics, and the no-slipping condition into one expression.

Therefore, the correct option is C.