A grindstone of moment of inertia 5 kg m^2 is rotating at 600 rpm ....

Given, initial angular velocity $\omega_i = 600 \text{ rpm} = 600 \times \dfrac{2\pi}{60} \text{ rad/s} = 20\pi \text{ rad/s}$
Final angular velocity $\omega_f = 0$
Moment of inertia $I = 5 \text{ kg m}^2$
Retarding torque $\tau = 10\pi \text{ N m}$

The angular deceleration $\alpha$ is:
$\alpha = \dfrac{\tau}{I} = \dfrac{10\pi}{5} = 2\pi \text{ rad/s}^2$

Using the third equation of rotational kinematics:
$\omega_f^2 = \omega_i^2 - 2\alpha\theta$
$0 = (20\pi)^2 - 2(2\pi)\theta$
$4\pi\theta = 400\pi^2$
$\theta = 100\pi \text{ rad}$

The number of complete revolutions $n$ is:
$n = \dfrac{\theta}{2\pi} = \dfrac{100\pi}{2\pi} = 50$

Answer: $50$