The average mass of rain drops is 3.0 × 10^ -5 ~kg and their...

Total volume of rain drops, received $100 \mathrm{~cm}$ in a year by area $1 \mathrm{~m}^2$ =1 \mathrm{~m}^2 $\times \frac{100}{100} \mathrm{~m}=$1 \mathrm{~m}^3 As we know, density of water, d=10^3 $\mathrm{~kg}$/ \mathrm{m}^3 Therefore, mass of this volume of water $M=d \times v=10^3 \times 1=10^3 \mathrm{~kg}$ Average terminal velocity of rain drop $v=9 \mathrm{~m} / \mathrm{s}$ (given) Therefore, energy transferred by rain, $$\begin{aligned}&E=\frac{1}{2}m v² \\ &=\frac{1}{2} \times10^3 \times(9)^2 \\ &=\frac{1}{2} \times10^3 \times81=4.05 \times10^4 \mathrm{~J} \end{aligned}$$