The pressure at the bottom of a lake, due to water is$4.9\times {10}^6{\text{N/m}}^{\text{2}}$ . Then the depth of the lake is

$\rho =1000{\text{kg/m}}^{\text{3}}=10\text{kg}$

$\text{h}=\text{?g}=\text{9}{\text{.8m/s}}^2$

We know $\text{p}=\text{h}\rho \text{g}$

$\therefore \text{h=}\frac{p}{\rho g}=\frac{4.9\times {10}^6}{{10}^3\times 9.8}=\frac{{10}^3}{2}$

$=0.5\times {10}^3=500\text{m}$