1. Given data

• Mass of disk: $M = 20\,\text{kg}$
• Radius of disk: $R = 0.2\,\text{m}$
• Force applied through string: $F = 20\,\text{N}$
• Initial angular speed: $\omega_0 = 0$
• Final angular speed: $\omega = 50\,\text{rad s}^{-1}$

We need the number of revolutions $n$ made by the disk to reach this angular speed.

2. Torque on the disk

Since the force is applied tangentially at the rim,

$\tau = F R = 20 \times 0.2 = 4\,\text{N m}$

3. Moment of inertia of the disk

For a uniform circular disk about its center,

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

So,

$I = \frac{1}{2}(20)(0.2)^2 = 10 \times 0.04 = 0.4\,\text{kg m}^2$

4. Angular acceleration

Using rotational equation,

$\tau = I\alpha$

$\alpha = \frac{\tau}{I} = \frac{4}{0.4} = 10\,\text{rad s}^{-2}$

5. Angular displacement needed to reach $\omega = 50\,\text{rad s}^{-1}$

Use rotational kinematics:

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

Since $\omega_0 = 0$,

$50^2 = 2(10)\theta$

$2500 = 20\theta$

$\theta = 125\,\text{rad}$

6. Convert angular displacement to number of revolutions

Given: one revolution $= 6.28\,\text{rad}$

$n = \frac{\theta}{6.28} = \frac{125}{6.28} \approx 19.9$

To the nearest integer,

$n = 20$

7. Final answer

The required number of revolutions is

$\boxed{20}$