If the atmospheric pressure is 76cm of mercury, at what depth of water the pressure will become 4 atmospheres

${\text{P}}_0=\text{h}\rho \text{g}=76\times 13.6\times 980\text{dynes}$

$\text{P}={\text{P}}_{\text{0}}+\text{h}\rho \text{g}$

${\text{4P}}_0={\text{P}}_{\text{0}}+\text{h}\rho \text{g}$

${\text{3P}}_0=\text{h}\rho \text{g}$

$\text{3}\times 76\times 13.6\times 980=\text{h}\times 1\times 980$

$\text{h}=3\times 76\times 13.6$

$\text{h}=3100.8\text{cm}$

$\text{h}=\text{31m}$