91Ó°ÊÓ

The water pressure at a depth'$d$" in the Ocean

$p=p_o+p_{n}gd$

$P_o$=Atmospheric pressure =$101325pas$

${\mathrm{ÒÏ}}_w$= Density of Ocean water =$1.025kg/L$

$=1025\mathrm{kg}/{\mathrm{m}}^3$

$\left(âˆ\mu 1000\mathrm{L}=1{\mathrm{m}}^3\right)$

$g$= Acceleration due to gravity

$d$= Ocean depth

$p=101325\mathrm{pa}+1025\mathrm{kg}/{\mathrm{m}}^3×9.8\mathrm{m}/\sec ×5000\mathrm{m}$

$=50326325pasclas$

$≈497\mathrm{atm}$

B) Fractional volume change =$\frac{∆V}{V}$=

$\frac{\mathrm{Δ}V}{V}=-\frac{p}{B}$

$B$=Bulk modulas of water =$0.2×{10}^{10}\mathrm{N}/{\mathrm{m}}^3$

$\frac{\mathrm{Δ}V}{V}=-\frac{50326325pas}{0.2×{10}^{10}\mathrm{N}/{\mathrm{m}}^2}$

$=-0.025163\left(Or\right)-2.5\%$of initial volume

$\frac{\mathrm{Δ}V}{V}=-2.5\%$.........$\left(1\right)$

C) Density of sea water at this pressure is more

Let ${\mathrm{ÒÏ}}_e$be the density of sea water at surface

And ${\mathrm{ÒÏ}}_d$be the density at the depth'$d$'

${\mathrm{ÒÏ}}_o=\frac{m}{V_n}$and ${\mathrm{ÒÏ}}_d=\frac{m}{V_d}$

Let us consider the mass of water be $m=1kg$

Taking the ratios of ${\mathrm{ÒÏ}}_d$and ${\mathrm{ÒÏ}}_o$

$\frac{{\mathrm{ÒÏ}}_d}{{\mathrm{ÒÏ}}_o}=\frac{\frac{m}{V_d}}{\frac{m}{V_o}}$

$=\frac{V_o}{V_d}$

$⇒\frac{{\mathrm{ÒÏ}}_d}{{\mathrm{ÒÏ}}_o}=\frac{V_o}{V_d}$..........$\left(1\right)$

We know that from equation (1)

$\frac{\mathrm{Δ}V}{V_u}=-2.5\%$of initial volume

$⇒\frac{V_d-V_e}{V_e}=-0.025$

$⇒V_d=V_o(1-0.025)$

$V_d=0.975{\mathrm{V}}_e$From equation$\left(2\right)$

$\frac{{\mathrm{ÒÏ}}_d}{{\mathrm{ÒÏ}}_e}=\frac{V_v}{0.975V_e}$

${\mathrm{ÒÏ}}_s={\mathrm{ÒÏ}}_e(1.02564)$

As a conclusion, average volume at a depth '$d$' is $1.025$times the same at the Ocean's surface.

$V_d=0.975V_e$

$\left(1\right)$