$\begin{array}{l}\mathrm{v}=\sqrt{2\mathrm{gh}}=\sqrt{2\times 10\times 1000}=100\sqrt{2}\mathrm{m}/\mathrm{s}\\ =100\sqrt{2}\times \frac{60\times 60}{1000}\mathrm{km}/\mathrm{h}\\ =360\sqrt{2}\mathrm{km}/\mathrm{h}\approx 510\mathrm{km}/\mathrm{h}\end{array}$

$\begin{array}{l}\text{ (b) Diameter of the drop }\left(\mathrm{d}\right)=2\mathrm{r}=4\mathrm{mm}\\ \therefore \text{ Radius of the }\mathrm{drop}\left(\mathrm{r}\right)=2\mathrm{mm}=2\times {10}^{-3}\mathrm{m}\end{array}$

Step 2: Find the mass of the raindrop

Mass of a raindrop (m) = volume of drop x density of drop

$\begin{array}{l} =\frac{4}{3}{\mathrm{\pi }}^3\mathrm{\rho }\\ =\frac{4}{3}\times \frac{22}{7}\times (2\times {10}^{-3}{)}^3\times {10}^3(\because \text{ Density of water }={10}^3\mathrm{kg}/{\mathrm{m}}^3)\\ =3.4\times {10}^{-5}\mathrm{kg}\end{array}$

Step 3: Find the momentum of the raindrop when it hits the ground

The momentum of the raindrop (p) = mv

$\begin{array}{l}= 3.4\times {10}^{-5}\times 100\sqrt{2}\\ = 4.7\times {10}^{-3} \mathrm{kg}-\mathrm{m}/\mathrm{s}\\ = 5\times {10}^{-3} \mathrm{kg}-\mathrm{m}/\mathrm{s}\end{array}$

Step 4: Find the time required to flatten the drop 

(c)Time required to flatten the drop = time taken by the drop to travel the distance equal to the diameter of the drop near the ground

$\begin{array}{l}\mathrm{t}=\frac{\mathrm{d}}{\mathrm{v}}=\frac{4\times {10}^{-3}}{100\sqrt{2}}=0.028\times {10}^{-3}\mathrm{s}\\ =2.8\times {10}^{-5}\mathrm{s}\approx 30\mathrm{ms}\end{array}$

Step 5: Force exerted by a raindrop

(d) 

$\begin{array}{l}\mathrm{F}=\frac{\text{ Change in momentum }}{\text{ Time }}=\frac{\mathrm{p}-0}{\mathrm{t}}\\ =\frac{4.7\times {10}^{-3}}{2.8\times {10}^{-5}}\approx 168\mathrm{N}\end{array}$

Step 6: Find the area of the umbrella

(e) Radius of the umbrella $\left(\mathrm{R}\right)=\frac{1}{2}\mathrm{m}$

$\therefore \text{ Area of the umbrella }\left(\mathrm{A}\right)={\mathrm{\pi R}}^2=\frac{22}{7}\times (\frac{1}{2}{)}^2=\frac{22}{28}=\frac{11}{14}\approx 0.8{\mathrm{m}}^2$

Step 7: Find the force exerted on the umbrela.

Number of drops striking the umbrella simultaneously with an average separation of 5 cm $=5\times {10}^{-2}\mathrm{m}$

$=\frac{0.8}{(5\times {10}^{-2}{)}^2}=320$

$\therefore$ Net force exerted on umbrella = 320 x 168 = 53760N 54000 N.