A) \[3.0\times {{10}^{5}}J\]
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 :
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 = 9 m/s (given) Therefore, energy transferred by rain, \[E=\frac{1}{2}m{{v}^{2}}\] \[=\frac{1}{2}\times {{10}^{3}}\times {{(9)}^{2}}\] \[=\frac{1}{2}\times {{10}^{3}}\times 81=4.05\times {{10}^{4}}J\]
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