Question

At a depth of $500\text{m}$ under an ocean, what is the absolute pressure? Given density of sea water is $1.03\times {10}^3{\text{kg/m}}^3$. (Take $g=10{\text{m/s}}^2$ & $1\text{atm}={10}^5\text{Pa}$)

• A. $51.5\text{atm}$
• B. $53.5\text{atm}$
• C. $50.5\text{atm}$
• D. $52.5\text{atm}$

Answer 1

The correct option is D $52.5\text{atm}$

Given height $\left(h\right)=500\text{m}$
Density of ocean, ${\rho }_o=1.03\times {10}^3{\text{kg/m}}^3$
$g=10{\text{m/s}}^2$
The absolute pressure is given by:
Absolute pressure = pressure due to water column + atmospheric pressure
$P={\rho }_{o}gh+P_o$
$=1.03\times {10}^3\times 10\times 500+1\times {10}^5$
[$\because$ atmospheric pressure $P_o={10}^5\text{Pa}$]
$=(51.5+1)\times {10}^5\text{Pa}$
$=52.5\times {10}^5\text{Pa}$
$\Rightarrow P=52.5\text{atm}(\because {10}^5\text{Pa}=1\text{atm})$