Torque and Rotations to Stop Rotating Solid Sphere

To find the torque applied and the number of rotations made by the solid sphere before coming to rest, we can proceed step by step using the principles of rotational dynamics.

1. Given Data:

• Mass of the solid sphere, $M = 5\text{ kg}$
• Radius of the sphere, $R = 4\text{ cm} = 0.04\text{ m} = 4 \times 10^{-2}\text{ m}$
• Initial angular velocity, $\omega_0 = 1200\text{ rpm} = \frac{1200 \times 2\pi}{60} = 40\pi\text{ rad/s}$
• Final angular velocity, $\omega = 0\text{ rad/s}$
• Time taken to come to rest, $t = 10\text{ s}$

---

2. Angular Retardation ($\alpha$):

Using the first equation of rotational motion: $\omega = \omega_0 - \alpha t$ $0 = 40\pi - \alpha (10)$ $\alpha = \frac{40\pi}{10} = 4\pi\text{ rad/s}^2$

---

3. Moment of Inertia ($I$):

The moment of inertia of a solid sphere about an axis passing through its centre is: $I = \frac{2}{5} M R^2$

Substitute the given values: $I = \frac{2}{5} \times 5 \times (4 \times 10^{-2})^2$ $I = 2 \times 16 \times 10^{-4} = 32 \times 10^{-4}\text{ kg}\cdot\text{m}^2 = 0.0032\text{ kg}\cdot\text{m}^2$

---

4. Applied Torque ($\tau$):

The magnitude of the torque applied to stop the sphere is given by: $\tau = I \alpha$ $\tau = (32 \times 10^{-4}) \times (4\pi)$ $\tau = 128\pi \times 10^{-4}\text{ Nm} = 0.0128\pi\text{ Nm}$

---

5. Number of Rotations ($N$):

The total angular displacement $\theta$ before coming to rest can be found using the average angular velocity: $\theta = \left(\frac{\omega_0 + \omega}{2}\right) \times t$ $\theta = \left(\frac{40\pi + 0}{2}\right) \times 10 = 200\pi\text{ rad}$

The number of complete rotations $N$ is: $N = \frac{\theta}{2\pi} = \frac{200\pi}{2\pi} = 100$

---

Conclusion:

• Torque applied = $0.0128\pi\text{ Nm}$
• Number of rotations = $100$

This corresponds to Option D.