1. Identify the wheel’s moment of inertia

Since the wheel is supported by spokes of negligible mass, its mass is effectively concentrated at the rim. So the wheel behaves like a ring.

Hence, $I = MR^2$

Given: $M = 10\,\text{kg}, \quad R = 10\,\text{cm} = 0.1\,\text{m}$

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

2. Torque due to the pull

The applied pull is tangential, so torque is $\tau = FR$

Given: $F = 20\,\text{N}, \quad R = 0.1\,\text{m}$

Thus, $\tau = 20 \times 0.1 = 2\,\text{N m}$

3. Angular acceleration

Using rotational equation, $\tau = I\alpha$

So, $\alpha = \frac{\tau}{I} = \frac{2}{0.1} = 20\,\text{rad/s}^2$

4. Angular displacement corresponding to 1 m unwinding

The cord unwinds by length $s = 1\,\text{m}$

Since $s = R\theta$

we get $\theta = \frac{s}{R} = \frac{1}{0.1} = 10\,\text{rad}$

5. Use rotational kinematics

Initially, the wheel is at rest, so $\omega_0 = 0$

Using $\omega^2 = \omega_0^2 + 2\alpha\theta$

$\omega^2 = 0 + 2(20)(10) = 400$

Thus, $\omega = 20\,\text{rad/s}$

6. Check options

• A: $20\,\text{rad/s}$ ✅
• B: $30\,\text{rad/s}$ ❌
• C: $10\,\text{rad/s}$ ❌
• D: $0\,\text{rad/s}$ ❌

So the correct answer is: $\boxed{20\,\text{rad/s}}$