JEE Physics 2025 Question with Solution

Given: Mass of the wheel = $10 \, \text{kg}$, radius = $0.1 \, \text{m}$, applied force = $20 \, \text{N}$, unwound length = $1 \, \text{m}$. The wheel starts from rest and the spokes have negligible mass, so the wheel is treated as a ring.

Find: The angular velocity of the wheel after the cord is unwound by $1 \, \text{m}$.

Using the work-energy theorem for rotational motion, the work done by the steady pull becomes rotational kinetic energy.

The work done by the force is

$$W_f = F \cdot d = 20 \times 1 = 20 \, \text{J}$$

This equals the change in kinetic energy of the wheel:

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

For a ring,

$$I = MR^2$$

Substituting the given values:

$$I = 10 \times (0.1)^2 = 0.1 \, \text{kg m}^2$$

Now equating work done to rotational kinetic energy:

$$20 = \frac{1}{2} \times 0.1 \times \omega^2$$ $$\omega^2 = 400$$ $$\omega = 20 \, \text{rad/s}$$

Therefore, the angular velocity of the wheel is $20 \, \text{rad/s}$. The correct option is A.

Given:

• Mass of the wheel, $M = 10 \, \text{kg}$
• Radius of the wheel, $R = 10 \, \text{cm} = 0.1 \, \text{m}$
• Applied force, $F = 20 \, \text{N}$
• Length of unwound cord, $L = 1 \, \text{m}$
• Initial angular velocity, $\omega_i = 0 \, \text{rad/s}$

Find: Final angular velocity $\omega_f$.

This solution uses the work-energy theorem for rotational motion.

1. Since the spokes have negligible mass, the wheel is treated as a ring. Hence its moment of inertia is

$$I = MR^2$$

Substituting values,

$$I = (10 \, \text{kg}) \times (0.1 \, \text{m})^2 = 10 \times 0.01 = 0.1 \, \text{kg} \cdot \text{m}^2$$

2. The work done by the applied force over the unwound length is

$$W = F \times L$$

So,

$$W = (20 \, \text{N}) \times (1 \, \text{m}) = 20 \, \text{J}$$

3. Apply the work-energy theorem:

$$W = KE_{\text{rot, final}} - KE_{\text{rot, initial}}$$

Since the wheel starts from rest, initial rotational kinetic energy is zero. Therefore,

$$W = \frac{1}{2} I \omega_f^2$$

Substituting the calculated values,

$$20 = \frac{1}{2} (0.1) \omega_f^2$$ $$\omega_f^2 = \frac{2 \times 20}{0.1} = \frac{40}{0.1} = 400$$ $$\omega_f = \sqrt{400} = 20 \, \text{rad/s}$$

Therefore, the angular velocity of the wheel after the cord is unwound by $1 \, \text{m}$ is $20 \, \text{rad/s}$.