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Consider a drop of rain water having mass 1 g falling from a height of 1 km. It hits the ground with a speed of 50 $m s^{- 1}$ . Take 'g' constant with a value 10 $m s^{- 2}$ . The work done by the (i) gravitational force and the (ii) resistive force of air is

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(i) 1.25 J (ii) -8.25 J

(i) 100 J (ii) 8.75 J

(i) 10 J (ii) -8.75 J

(i) -10 J (ii) 8.25 J

answer is C.

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Detailed Solution

Work done by gravitational force = mgh = 10^-3 x 10 x 10³ = 10 J
Gain in kinetic energy = $\frac{1}{2} {mv}^{2} = \frac{1}{2} (10^{- 3}) (50)^{2} = 1.25 J$

Gain in kinetic energy = work done by gravitational force + work done by resistance force.

$\begin{aligned} \Rightarrow 1.25 = W + w_{R} \\ W_{R} = - 8.75 J \end{aligned}$

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