A) \[3.5\times {{10}^{5}}J\]
B) \[4.05\times {{10}^{4}}J\]
C) \[3.0\times {{10}^{5}}J\]
D) \[9.0\times {{10}^{4}}J\]
Correct Answer: B
Solution :
[b] Total volume of rain drops, received 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 \[v=9m/s\] (given) Therefore, energy transferred by rain, \[\begin{align} & E=\frac{1}{2}m{{v}^{2}}=\frac{1}{2}\times {{10}^{3}}\times {{(9)}^{2}}=\frac{1}{2}\times {{10}^{3}}\times 81 \\ & =4.05\times {{10}^{4}}J \\ \end{align}\]
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