Amidst a heavy rain, hailstones having an average diameter of 1.0...

The volume of a single hailstone is given by:

$V = \dfrac{4}{3} \pi r^3 = \dfrac{4}{3} \pi (0.5 \times 10^{-2})^3 = \dfrac{\pi}{6} \times 10^{-6} \text{ m}^3$

The mass of a single hailstone is:

$m = \rho V = 900 \times \dfrac{\pi}{6} \times 10^{-6} = 150 \pi \times 10^{-6} \text{ kg}$

The total number of hailstones hitting the roof per second is:

$N = n \times A = 2000 \times (10 \times 10) = 2 \times 10^5 \text{ s}^{-1}$

The total mass of hailstones hitting the roof per second is:

$\dfrac{M}{t} = N \times m = 2 \times 10^5 \times 150 \pi \times 10^{-6} = 30 \pi \text{ kg/s}$

Since the hailstones do not rebound, the average force exerted on the roof is equal to the rate of change of momentum:

$F = \dfrac{M}{t} \times v = 30 \pi \times 20 = 600 \pi \text{ N}$

Substituting $\pi \approx 3.1416$:

$F \approx 600 \times 3.1416 = 1884.96 \text{ N} \approx 1885 \text{ N}$