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

The correct option is A $-8.75J$
Here, $m=1g={10}^{-3}kg$
$h=1km=1000m={10}^{3}m$
The change in kinetic energy of the drop is
$△K=\frac{1}{2}m{v}^2-0=\frac{1}{2}\times {10}^{-3}\times 50\times 50=1.25J(\because u=0)$
The work done by the gravitational force is
$W_g=mgh={10}^{-3}\times 10\times {10}^3=10J$
According to work-energy theorem
$△K=W_g+W_r$
where $W_r$ is the work done by the resistive force on the raindrop. Thus
$W_r=△K-W_g=1.25J-10J=-8.75J$.