A submarine is designed to withstand an absolute pressure of 100 atm. How deep can it go below the water surface? (Consider the density of water = 1000 kg m³, 1 atm = 1 × 10 Pa and g =10m/s²)

The submarine can endure an absolute pressure of 100 atm. We want the maximum depth $h$ below the water surface.

At the surface the pressure is 1 atm (the atmospheric pressure). At depth the absolute pressure is the sum of atmospheric pressure and the pressure from the water column:

$P\_\{\backslash text\{abs\}\}\; =\; P\_\{\backslash text\{atm\}\}\; +\; \backslash rho\; g\; h$

Given $P\_\{\backslash text\{abs\}\}\; =\; 100\backslash \; \backslash text\{atm\}$ and $P\_\{\backslash text\{atm\}\}\; =\; 1\backslash \; \backslash text\{atm\}$, the pressure contributed by the water is:

$\backslash Delta\; P\; =\; P\_\{\backslash text\{abs\}\}\; -\; P\_\{\backslash text\{atm\}\}\; =\; (100\; -\; 1)\backslash \; \backslash text\{atm\}\; =\; 99\backslash \; \backslash text\{atm\}$

Convert the pressure difference to pascals using $1\backslash \; \backslash text\{atm\}\; =\; 1\; \backslash times\; 10^\{5\}\backslash \; \backslash text\{Pa\}$:

$\backslash Delta\; P\; =\; 99\; \backslash times\; 10^\{5\}\backslash \; \backslash text\{Pa\}\; =\; 9.9\; \backslash times\; 10^\{6\}\backslash \; \backslash text\{Pa\}$

Now solve for the depth $h$ using the hydrostatic relation $\backslash Delta\; P\; =\; \backslash rho\; g\; h$ where $\backslash rho\; =\; 1000\backslash \; \backslash text\{kg m\}^\{-3\}$ and $g\; =\; 10\backslash \; \backslash text\{m s\}^\{-2\}$:

$h\; =\; \backslash frac\{\backslash Delta\; P\}\{\backslash rho\; g\}\; =\; \backslash frac\{9.9\; \backslash times\; 10^\{6\}\backslash \; \backslash text\{Pa\}\}\{1000\backslash \; \backslash text\{kg m\}^\{-3\}\; \backslash times\; 10\backslash \; \backslash text\{m s\}^\{-2\}\}$

Carrying out the division:

$h\; =\; \backslash frac\{9.9\; \backslash times\; 10^\{6\}\}\{1.0\; \backslash times\; 10^\{4\}\}\; =\; 9.9\; \backslash times\; 10^\{2\}\backslash \; \backslash text\{m\}\; =\; 990\backslash \; \backslash text\{m\}$

Therefore the submarine can safely descend to a depth of 990 m.