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The average mass of rain drops is $3.0\times10^{-5}\, kg$ and their avarage terminal velocity is $9\, m/s$. Calculate the energy transferred by rain to each square metre of the surface at a place which receives $100\, cm$ of rain in a year
$3.5\times10^5\, J$
$4.05\times10^4\, J$
$3.0\times10^5\, J$
$9.0\times10^4\, J$
Solution
Total volume of rain drops, recrived $100\,cm$ in a year by area $1\,m^2$
$ = 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$
