Fluid Mechanics

JEE Main level, single correct, medium

74 of 192

Question 74 of 192

The average mass of rain drops is 3.0×10−5 kg3.0 \times 10^{-5} \mathrm{~kg} and their avarage terminal velocity is 9 m/s9 \mathrm{~m} / \mathrm{s}. Calculate the energy transferred by rain to each square metre of the surface at a place which receives 100 cm100 \mathrm{~cm} of rain in a year.

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Answer: (B) 4.05×104 J4.05 \times 10^4 \mathrm{~J}

Total volume of rain drops, received 100 cm100 \mathrm{~cm} in a year by area 1 m21 \mathrm{~m}^2
=1 m2×100100 m=1 m3=1 \mathrm{~m}^2 \times \frac{100}{100} \mathrm{~m}=1 \mathrm{~m}^3
As we know, density of water,
d=103 kg/m3d=10^3 \mathrm{~kg} / \mathrm{m}^3
Therefore, mass of this volume of water M=d×v=103×1=103 kgM=d \times v=10^3 \times 1=10^3 \mathrm{~kg}
Average terminal velocity of rain drop v=9 m/sv=9 \mathrm{~m} / \mathrm{s} (given)
Therefore, energy transferred by rain,
E=12mv2=12×103×(9)2=12×103×81=4.05×104 J\begin{aligned} &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 \mathrm{~J} \end{aligned}

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