Question

Depth of sea is maximum at Mariana Trench in West Pacific Ocean. Trench has a maximum depth of about $11km$. At bottom of trench water column above it exerts $1000atm$ pressure. FInd the Percentage change in density of sea water at a such depth.
(Given, $B=2\times {10}^{9}N{m}^{-2}$ and $p_{atm}=1\times {10}^{5}N{m}^{-2})$.

Answer 1

Given,

Maximum depth $=11km\left(h\right)$

Pressure at bottom $=1000atm\left(P\right)$

Percentage of change in density of sea water at a such depth

$\frac{\Delta P}{P}\times 100=?$

Where $B=2\times {10}^{5}N{m}^{-2}{P}_{atm}=1\times {10}^{5}N{m}^{-2}$

$P=h\rho gP+P_{atm}=h\rho gP=h\rho g-P_{atm}100\times 101325=11\times {10}^3\times \frac{\Delta P}{P}\times 100\times 10-1\times {10}^{5}11\times {10}^3\times \frac{\Delta P}{P}\times 1000=101325\times {10}^2+{10}^5\frac{\Delta P}{P}=\frac{101325\times {10}^2+{10}^5}{11\times {10}^3}=930.227=9.3\%$