A raindrop of mass 1 g falling from a height of 1 km hits the ground with a speed of 50 ${\text{ms}}^{-1}$. If the resistive force is proportional to the speed of the drop, then the work done by the resistive force is (Take g : 10 ${\mathrm{ms}}^{-2}$)

Here, m = 1 g = ${10}^{-3} \mathrm{kg}$

    h = 1 km = 1000 m = ${10}^3 \mathrm{m}$

The change in kinetic energy of the drop is

   $∆\mathrm{K} = \frac{1}{2}{\mathrm{mv}}^2-0 (\because \mathrm{u} = 0)$

   = $\frac{1}{2}\times {10}^{-3}\times 50\times 50 = 1.25 \mathrm{J}$

The work done by the gravitational force is

   ${\mathrm{W}}_{\mathrm{g}} = \mathrm{mgh} = {10}^{-3}\times 10\times {10}^3 = 10 \mathrm{J}$

According to work-energy theorem

$∆\mathrm{K} = {\mathrm{W}}_{\mathrm{g}} + {\mathrm{W}}_{\mathrm{r}}$

W_r is the work done by the resistive force on the raindrop

 ${\mathrm{W}}_{\mathrm{r}} = ∆\mathrm{K} - {\mathrm{W}}_{\mathrm{g}} = 1.25 \mathrm{J}-10 \mathrm{J} = -8.75 \mathrm{J}$