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

           h = 1 km = 1000 m

          g = $10 {\mathrm{ms}}^{-2}$, v = 50 ${\mathrm{ms}}^{-1}$

Loss of potential energy of the drop

    = mgh =  (${10}^{-3} \mathrm{kg}$)($10 {\mathrm{ms}}^{-2}$)(1000 m) = 10 J -----(i)

Hence, option (a) is correct.

Gain in kinetic energy of the drop

  = $\frac{1}{2}{\mathrm{mv}}^2 = \frac{1}{2}\times \left({10}^{-3}\mathrm{kg}\right){(50 {\mathrm{ms}}^{-1})}^2 = 1.25 \mathrm{J} -----\left(\mathrm{ii}\right)$

Hence, option (b) is correct.

From (i) and (ii), we get

Loss of potential energy of the drop $\ne$ Gain in kinetic energy of the drop.

Hence, option c) is also correct