To solve this problem, we need to calculate the work done by the gravitational force and the resistive force of air.$\\$Given:$\\$• Mass of the raindrop $m = 1$ g$= 0.001$ kg$\\$• Height $h = 1$ km$= 1000$ m$\\$• Final speed $v = 50$ m/s$\\$• Acceleration due to gravity $g = 10$ m/s$^2 \\$1. Work done by gravitational force:$\\$The work done by gravity is given by $W_g = mgh. \\ W_g = 0.001 \times 10 \times 1000 \\ W_g = 10$ J$\\$2. Work done by resistive force of air:$\\$The work-energy principle states that the total work done on an object is equal to the change in its kinetic energy.$\\$The initial kinetic energy $KE_i = 0$ (since it starts from rest).$\\$The final kinetic energy $KE_f = \frac{1}{2}mv^2. \\ KE_f = \frac{1}{2} \times 0.001 \times (50)^2 \\ KE_f = \frac{1}{2} \times 0.001 \times 2500 \\ KE_f = 1.25$ J$\\$According to the work-energy principle:$\\ W_{total} = KE_f - KE_i \\ W_{total} = 1.25 - 0 \\ W_{total} = 1.25$ J$\\$The total work done is the sum of the work done by gravity and the resistive force:$\\ W_{total} = W_g + W_{resistive} \\ 1.25 = 10 + W_{resistive} \\$Solving for $W_{resistive}: \\ W_{resistive} = 1.25 - 10 \\ W_{resistive} = -8.75$ J$\\$Therefore, the work done by the gravitational force is $10$ J$, \\$and the work done by the resistive force of air is $-8.75$ J$. \\$This corresponds to Option 3.