Question

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

Answer 1

Hint Given the depth and bulk modulus of the sea water. Assume the volume at the surface as ${V_1}$and volume at the depth as ${V_2}$ and the density as ${\rho _1}$and ${\rho _2}$. Using Bulk modulus formula, calculate change in volume and use that to calculate density change and hence percentage.

Complete Step By Step Solution
First we calculate the change in volume from the top surface to the bottom surface of the mariana trench using the given bulk modulus formula. Bulk modulus of any substance is defined as its measure of it’s resistance to compression.
Mathematically, Bulk modulus is given as
$B = \dfrac{{\Delta P}}{{\dfrac{{\Delta V}}{V}}}$, where $\Delta P$is change in pressure from top surface to the bottom surface
And $\Delta V$is the change in volume of water from top surface to the bottom.
$\Rightarrow B = \dfrac{{(1000 \times {{10}^5}N{m^{ - 2}})}}{{\dfrac{{\Delta V}}{V}}}$
We know B value , taking $\dfrac{{\Delta V}}{V}$to the other side we get,
$\Rightarrow \dfrac{{\Delta V}}{V} = \dfrac{{(1000 \times {{10}^5}N{m^{ - 2}})}}{B}$
$\Rightarrow \dfrac{{\Delta V}}{V} = \dfrac{{(1000 \times {{10}^5}N{m^{ - 2}})}}{{(2 \times {{10}^9}N{m^{ - 2}})}}$
Now, $V$is initial volume ${V_1}$
$\Rightarrow \dfrac{{{V_1} - {V_2}}}{{{V_1}}} = \dfrac{1}{{20}}$
$\Rightarrow {V_1}(1 - \dfrac{1}{{20}}) = {V_2}$
Now, density at 11km depth is given as
${\rho _2} = \dfrac{{mass}}{{Volum{e_2}}}$
Mass of water is constant throughout ,Hence
$\Rightarrow {\rho _2} = \dfrac{{{\rho _1} \times {V_1}}}{{{V_2}}}$
Substituting for ${V_2}$, we get
$\Rightarrow {\rho _2} = \dfrac{{{\rho _1} \times {V_1}}}{{{V_1}(1 - \dfrac{1}{{20}})}}$
Cancelling out ${V_1}$we get,
$\Rightarrow {\rho _2} = \dfrac{{{\rho _1}}}{{0.95}}$
Now ,we need to calculate change in density percentage given by the formula
$\Rightarrow \dfrac{{{\rho _2} - {\rho _1}}}{{{\rho _1}}} \times 100$
$\Rightarrow \dfrac{{\dfrac{{{\rho _1}}}{{0.95}} - {\rho _1}}}{{{\rho _1}}} \times 100$
$\Rightarrow \dfrac{{{\rho _1}(0.0526)}}{{{\rho _1}}} \times 100$
$\therefore 5\%$

Hence, the percentage change in density sums unto 5%.

Note We can also calculate the percentage change in density by considering the hydrostatic pressure exerted by the water on the bottom surface and by the atmosphere on top. Hydrostatic pressure is the pressure exerted by a fluid which is at a specified equilibrium at any given point on the fluid due to gravitational force. It is mathematically represented as the product of depth, density of the surface and gravitational force g.