KVPY 2016 SA Question 10

Efficiency of conversion $( \eta ) = \dfrac{\text{Power produced}}{\text{Power given}} \times 100\%$

In this case

$50 = \dfrac{10^9}{P} \times 100 \implies P = 2 \times 10^9 \ \text{W}$

So $2 \times 10^9 \ \text{W}$ of power must be given by the falling water per second which also further implies that falling water should do an amount of work equal to $2 \times 10^9 \ \text{J}$ on the turbine per second. Thus

$W = F \cdot s \\ \implies 2 \times 10^9 = m g s \\ \implies 2 \times 10^9 = m \times 10 \times 500 \\ \implies m = \dfrac{2 \times 10^9}{5 \times 10^3} \\ \implies m = 4 \times 10^5 \\ \implies V \rho_{\text{water}} = 4 \times 10^5 \\ \implies V \times 10^3 = 4 \times 10^5 \\ \implies V = \boxed{400 \text{ m}^3}$