[tex]\begin{gathered} \rho=1000\text{ kg/m}^3 \\ g=9.81\text{ m/s}^2 \\ h=? \\ P1=P_{atm}=101293Pa \\ P2=P_{atm}+\rho gh \\ P2=6.41P_{atm} \\ P_{atm}+\rho gh=6.41P_{atm} \\ \rho gh=6.41P_{atm}-P_{atm} \\ \rho gh=5.41P_{atm} \\ h=\frac{5.41P_{atm}}{\rho g} \\ h=\frac{5.41(101293Pa)}{(1000\text{ kg/m}^3)(9.81\text{ m/s}^2)} \\ h=55.861m \\ At\text{ 55.861m deep the pressure will be 6.41 times the pressure } \\ \text{at the surface} \end{gathered}[/tex]
