JEE Physics 2025 Question with Solution

Given: A wheel is rolling without slipping on a plane surface. The speed of the particle at the highest point of the rim is $8 \, \text{m/s}$.

Find: The speed of the particle on the rim at the same level as the center of the wheel.

For pure rolling, the speed of the highest point is twice the speed of the center.

$$2V = 8 \, \text{m/s}$$ $$V = 4 \, \text{m/s}$$

At the point on the rim which is at the same level as the center, the translational velocity $V$ and the rotational velocity $V$ are perpendicular to each other.

$$v = \sqrt{V^2 + V^2}$$

Substituting $V = 4 \, \text{m/s}$,

$$v = \sqrt{4^2 + 4^2}$$ $$v = \sqrt{16 + 16}$$ $$v = \sqrt{32} = 4\sqrt{2} \, \text{m/s}$$

Therefore, the speed of the particle is $4\sqrt{2} \, \text{m/s}$. The correct option is A.

Given: The speed at the highest point of the rim is $8 \, \text{m/s}$.

Find: The speed at the point on the rim level with the center.

In rolling without slipping, the highest point has speed $2V$, so the center moves with speed $V = 4 \, \text{m/s}$. At the side point of the rim, the translational and rotational velocity vectors have equal magnitude $V$ and are perpendicular. Hence the resultant speed is

$$v = \sqrt{2}V = \sqrt{2} \times 4 = 4\sqrt{2} \, \text{m/s}$$

Therefore, the correct option is A.