A solid sphere of radius 4 cm and mass 5 kg is rotating (rotation axis is passing through the centre of the sphere) with an angular velocity of 1200 rpm. It is brought to rest in 10 s by applying a constant torque. The torque applied and the number of rotations it made before it comes to rest are _______ and _______ respectively.

Initial angular velocity, $$\omega_0 = 1200\text{ rpm}$$ 

Final angular velocity, $$\omega = 0\text{ rad/s}$$ (since it is brought to rest) 

Time taken to stop, $$t = 10\text{ s}$$

$$ \omega_0 = 1200 \times \frac{2\pi}{60} $$

$$ \omega_0 = 20 \times 2\pi = 40\pi\text{ rad/s} $$

Using the first equation of rotational kinematics:

$$ \omega = \omega_0 + \alpha t $$

$$ 0 = 40\pi + \alpha (10) $$

$$ 10\alpha = -40\pi $$

$$ \alpha = -4\pi\text{ rad/s}^2 $$

The magnitude of angular deceleration is $$4\pi\text{ rad/s}^2$$.

The formula for the moment of inertia of a solid sphere about an axis passing through its center is:

$$ I = \frac{2}{5}MR^2 $$

$$ I = \frac{2}{5} \times 5 \times (0.04)^2 $$

$$ I = 2 \times 0.0016 $$

$$ I = 0.0032\text{ kg}\cdot\text{m}^2 = 3.2 \times 10^{-3}\text{ kg}\cdot\text{m}^2 $$

Using Newton's second law for rotation (magnitude of torque):

$$ \tau = I |\alpha| $$

$$ \tau = (0.0032) \times (4\pi) $$

$$ \tau = 0.0128\pi\text{ N}\cdot\text{m} $$

$$ \tau = 1.28\pi \times 10^{-2}\text{ N}\cdot\text{m} $$

First, find the total angular displacement ($\theta$):

$$ \theta = \left(\frac{\omega_0 + \omega}{2}\right) t $$

$$ \theta = \left(\frac{40\pi + 0}{2}\right) \times 10 $$

$$ \theta = 20\pi \times 10 = 200\pi\text{ rad} $$

Now, convert radians to the number of rotations:

$$ N = \frac{\theta}{2\pi} $$

$$ N = \frac{200\pi}{2\pi} $$

$$ N = 100 $$