(D) Given:
Mass of a raindrop,$m_d = 3.0 \times 10^{-5} \ kg$
Terminal velocity,$v = 9 \ m/s$
Rainfall depth,$h = 100 \ cm = 1 \ m$
Area,$A = 1 \ m^2$
Density of water,$\rho = 10^3 \ kg/m^3$
Step $1$: Calculate the total volume of water falling on $1 \ m^2$ area.
$V = A \times h = 1 \ m^2 \times 1 \ m = 1 \ m^3$
Step $2$: Calculate the total mass of water $(M)$ falling on $1 \ m^2$ area.
$M = V \times \rho = 1 \ m^3 \times 10^3 \ kg/m^3 = 10^3 \ kg$
Step $3$: Calculate the kinetic energy $(E)$ transferred to the surface.
Since the raindrops fall at terminal velocity,the energy transferred is the kinetic energy of the total mass of water.
$E = \frac{1}{2} M v^2$
$E = \frac{1}{2} \times 10^3 \times (9)^2$
$E = 0.5 \times 1000 \times 81$
$E = 40500 \ J = 4.05 \times 10^4 \ J$