\(\left( {{P_1}} \right)\frac{4}{3}\pi {r^3} = \left( {{P_2}} \right)\frac{4}{3}\pi \frac{{125{r^3}}}{{64}}\)
\(\frac{{\rho g\left( {10} \right) + \rho gh}}{{\rho g\left( {10} \right)}} = \frac{{125}}{{64}}\)
\(640 + 64\,h = 1250\)
\(On\,solving\,we\,get\,h = 9.5\,m\)