(a) For seawater of density 1.03 g/cm³, find the weight of water on top of a submarine at a depth of $\mathbf{255}\mathbf{m}$if the horizontal cross-sectional hull area is $\mathbf{2200}\mathbf{.}\mathbf{0}{\mathbf{m}}^{\mathbf{2}}$. (b) In atmospheres, what water pressure would a diver experience at this depth?

• The density of seawater is, $\rho =1.03\mathrm{g}/{\mathrm{cm}}^3=1030\mathrm{kg}/{\mathrm{m}}^3$

• The submarine has depth $h=255\mathrm{m}$below the surface of water

• The cross-sectional hull area is, $A-2200{\mathrm{m}}^2$.

Archimedes Principle states that when a body is fully or partially submerged in a fluid, a buoyant force ${\stackrel{\mathbf{F}}{\mathbf{F}}}_{\mathbf{b}}$from the surrounding fluid acts on the body. The force is directed upward and has a magnitude given by,

${\mathit{F}}_{\mathbf{b}}\mathbf{=}{\mathit{m}}_{\mathbf{f}}\mathit{g}$

Where ${\mathit{m}}_{\mathbf{f}}$is the mass of the fluid that has been displaced by the body.

Using Archimedes Principle, we can write the force of buoyancy equation. We can find the weight of the water from this equation. Writing the equation for the pressure in terms of density, gravitational acceleration, and depth, we can find the pressure

Formulae:

Archimedes’ principle gives that when an object floats in a liquid, then

$\begin{array}{r}{F}_o=F_g\\ m=\rho V\end{array}$

The pressure at a point in a fluid in static equilibrium is $p=p_0+\rho gh$

According to Archimedes principle,

$\begin{array}{r}{F}_g=F_b\\ F_g=\rho gV\\ =\rho gAh\end{array}$

Substituting the given values, we get,

$\begin{array}{r}{F}_g=\left(1030\mathrm{kg}/{\mathrm{m}}^3\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(2200{\mathrm{m}}^2\right)\left(255\mathrm{m}\right)\\ =5.66\times {10}^9\mathrm{N}\end{array}$

Therefore, the weight of the water on the top of the submarine is $5.66\times {10}^9\mathrm{N}$.

The water pressure experienced by a diver in the atmosphere is the gauge pressure, therefore,

$p_g=\rho gh$

Substituting the given values in the above equation, we get

$\begin{array}{r}{p}_g=\left(1030\mathrm{kg}/{\mathrm{m}}^3\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(255\mathrm{m}\right)\left(\frac{1\mathrm{atm}}{1.01\times {10}^5\mathrm{Pa}}\right)\\ =25.48\mathrm{atm}\\ \approx 25.5\mathrm{atm}\end{array}$

Therefore, the water pressure experienced by the diver in the atmosphere is $25.5\text{atm}$.