As given that height $(h)=1 \mathrm{~km}=1000 \mathrm{~m}$, $u=0 \mathrm{~m} / \mathrm{s}, g=10 \mathrm{~m} / \mathrm{s}^2, \mathrm{~d}=4 \mathrm{~mm}$
$$
\Rightarrow r=\frac{4}{2} \mathrm{~mm}=2 \mathrm{~mm}
$$
(a) Velocity attained by the rain drop in freely falling through a height $h$.
$$
\begin{aligned}
v^2 &=u^2+2 g h, u=0 \\
v &=\sqrt{2 g h}=\sqrt{2 \times 10 \times 1000}=100 \sqrt{2} \mathrm{~m} / \mathrm{s} \\
&=100 \sqrt{2} \times \frac{60 \times 60}{1000} \mathrm{~km} / \mathrm{hr} \\
&=360 \sqrt{2} \mathrm{~km} / \mathrm{h} \approx 510 \mathrm{~km} / \mathrm{hr} .
\end{aligned}
$$
(b) Diameter of the drop $(d)=2 r=4 \mathrm{~mm}$
So, radius of the drop $(r)=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m}$
Mass of a rain drop $(m)$
$=$ Volume $(v) \times$ density $(\rho)$
$=\frac{4}{3} \pi r^3 \rho=\frac{4}{3} \times \frac{22}{7} \times\left(2 \times 10^{-3}\right)^3 \times 10^3$
$\left(\because\right.$ Density of water $\left.=10^3 \mathrm{~kg} / \mathrm{m}^3\right)$ $=33.5 \times 10^{-6} \mathrm{~kg}$
Momentum of the rain drop $(\mathrm{P})$
$$
\begin{aligned}
&=m v=33.5 \times 10^{-6} \times 100 \sqrt{2} \\
&=47.37 \times 10^{-4} \mathrm{~kg}-\mathrm{m} / \mathrm{s} \\
&=4.7 \times 10^3 \mathrm{~kg}-\mathrm{m} / \mathrm{s}
\end{aligned}
$$
(c) Time to required drop of $4 \mathrm{~mm}$ diameter spherical drop i.e., time to reach upper part of spherical drop on ground, i.e. time taken by the drop to travel the distance equal to the diameter $(d)=4 \mathrm{~mm}$ of the drop near the ground.
$$
\begin{aligned}
\text { Time } &=\frac{\text { distance }}{\text { speed }}=\frac{d}{v}=\frac{4 \times 10^{-3}}{100 \sqrt{2}} \\
&=0.028 \times 10^{-3} \mathrm{~s}=2.8 \times 10^{-5} \mathrm{sec} . \\
&=2.8 \times 10^{-5} \mathrm{~s} \approx 30 \mathrm{~ms}
\end{aligned}
$$
(d) Force exerted by a rain drop
$$
\begin{aligned}
F &=\frac{\text { Change in momentum }}{\text { Time }} \\
&=\frac{d P}{d t} \\
&=\frac{4.7 \times 10^{-3}}{2.8 \times 10^{-5}}=168 \mathrm{~N}
\end{aligned}
$$
(e) Radius of the umbrella $(R)=\frac{1}{2} m$
Area of the umbrella $(A)$
$$
=\pi R^2=\frac{22}{7} \times\left(\frac{1}{2}\right)^2=\frac{22}{28}=\frac{11}{14}=0.78 \mathrm{~m}^2
$$
Number of drops falling on umbrella simultaneously with average separation of $5 \mathrm{~cm}=5 \times 10^{-2} \mathrm{~m}$.
$$
=\frac{\pi R^2}{\left(5 \times 10^{-2}\right)^2}=\frac{0.78}{\left(5 \times 10^{-2}\right)^2}=314 \text { drops }
$$
$\therefore$ Net force exerted on umbrella by 314 drops $=314 \times 168=52752 \mathrm{~N}$.
It is equivalent to $5275 \mathrm{Kg}$ wt. which again not possible on umbrella.