Power output, $P=50 \%$ of potential energy of water

Received by generator per second

$=\frac{50}{100} \times(\frac{m g h}{t})$

$=05 \times\left(\frac{V}{t}\right) \rho \times g \times h \quad \dot(i)$

Here, $P=1 \times 10^9 \,W$,

$g=10 \,ms ^{-2}, \rho=1000 \,kgm ^{-3}$

and $h=500 \,m$.

Substituting these values in Eq. $(i)$, we get

Volume flow rate of water $=\frac{V}{t}$

$=\frac{1 \times 10^9}{0.5 \times 1000 \times 10 \times 500}=400 \,m ^3 s ^{-1}$