Question

What is the pressure at the bottom of this ocean at a place where it is 3 km deep? [atmospheric pressure $ = 1.01 \times {10^5}pa$ . Density of seawater $ = 1030\,kg.{m^{ - 3}}$ ]
A. $1.01 \times {10^5}pa$
B. $3 \times {10^3}pa$
C. $1.01 \times {10^7}pa$
D. $3 \times {10^7}pa$

Answer 1

Hint: In order to answer this question, first we will rewrite the given facts, and then we will apply the formula to find the total pressure at the bottom of this ocean at a place where it is 3 km deep.

Formula-used:
\[P = {P_0} + \rho gh\]

Complete step by step answer:
Given that: Atmospheric pressure, ${P_{atm}} = 1.01 \times {10^5}pa$.
Density of the sea water, $\rho = 1030\,kg.{m^{ - 3}}$
Depth of the sea, $h = 3\,km = 3000\,m$
Now, we will apply the formula of the Pressure-
Therefore, the total pressure will be:
\[P = {P_0} + \rho gh\]
where, ${P_0}$ is the atmospheric pressure that is given.
$P = 1.01 \times {10^5} + 1030 \times 10 \times 3000 \\
\therefore P = 310.01 \times {10^5}pa = 3.10 \times {10^7}pa $
Since, the pressure at the bottom of this ocean at a place where it is 3 km deep is $3 \times {10^7}pa$ .

Hence, the correct option is D.

Note: The water column above imposes a pressure of \[1,086{\text{ }}bars\left( {15,750{\text{ }}psi} \right)\] at the bottom of the trench, more than 1,000 times the usual atmospheric pressure at sea level. The density of water increases by \[4.96\% \] at this pressure. The temperature at the bottom ranges from 1 to 4 degrees Celsius (34 to 39 degrees Fahrenheit).