$ = 1{m^2} \times \frac{{100}}{{100}}m = 1\,{m^3}$
As we know, density of water,
$d = {10^3}\,kg/{m^3}$
Therefore, mass of this volume of water,
$M = d \times v = {10^3} \times 1 = {10^3}\,kg$
Average terminal velocity of rain drop
$v = 9\,m/s\,\left( {given} \right)$
Therefore, energy transferred by rain,
$E = \frac{1}{2}m{v^2}$
$ = \frac{1}{2} \times {10^3} \times {\left( 9 \right)^2}$
$ = \frac{1}{2} \times {10^3} \times 81 = 4.05 \times {10^4}J$


[Given: Surface tension of the liquid is $0.075 \mathrm{Nm}^{-1}$, atmospheric pressure is $10^5 \mathrm{~N} \mathrm{~m}^{-2}$, acceleration due to gravity $(g)$ is $10 \mathrm{~m} \mathrm{~s}^{-2}$, density of the liquid is $10^3 \mathrm{~kg} \mathrm{~m}^{-3}$ and contact angle of capillary surface with the liquid is zero]
