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If the average terminal velocity of rain drop is $2 \mathrm{~ms}^{-1}$, then the energy transferred by rain to each square metre of the surface at a place which receives $100 \mathrm{~cm}$ of rain in a year is
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The correct answer is:
$2 \times 10^3 \mathrm{~J}$
Given that,
Terminal velocity of rain, $v=2 \mathrm{~m} / \mathrm{s}$
Height, $h=100 \mathrm{~cm}=1 \mathrm{~m}$
Surface area, $A=1 \mathrm{~m}^2$
Volume, $V=A h=1 \mathrm{~m}^3$
Density of water, $\rho=10^3 \mathrm{~kg} \mathrm{~m}^{-3}$
Mass of water (rain), $m=V \rho=10^3 \mathrm{~kg}$
Now, energy transferred by rain per unit area
$$
\begin{aligned}
& =\frac{\text { Kinetic energy }}{\text { Area }}=\frac{\frac{1}{2} m v^2}{1} \\
& =\frac{1}{2} m v^2=\frac{1}{2} \times 10^3 \times(2)^2 \\
& =2 \times 10^3 \mathrm{~J} / \mathrm{m}^2
\end{aligned}
$$
Hence, energy transferred by rain to each square metre is $2 \times 10^3 \mathrm{~J}$.
Terminal velocity of rain, $v=2 \mathrm{~m} / \mathrm{s}$
Height, $h=100 \mathrm{~cm}=1 \mathrm{~m}$
Surface area, $A=1 \mathrm{~m}^2$
Volume, $V=A h=1 \mathrm{~m}^3$
Density of water, $\rho=10^3 \mathrm{~kg} \mathrm{~m}^{-3}$
Mass of water (rain), $m=V \rho=10^3 \mathrm{~kg}$
Now, energy transferred by rain per unit area
$$
\begin{aligned}
& =\frac{\text { Kinetic energy }}{\text { Area }}=\frac{\frac{1}{2} m v^2}{1} \\
& =\frac{1}{2} m v^2=\frac{1}{2} \times 10^3 \times(2)^2 \\
& =2 \times 10^3 \mathrm{~J} / \mathrm{m}^2
\end{aligned}
$$
Hence, energy transferred by rain to each square metre is $2 \times 10^3 \mathrm{~J}$.
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