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What is the pressure 200 meter below the surface of ocean? If they specific gravity of Seawater is 1.03. (Atmospheric pressure $1.013 \times {10^5}N/{m^2}$ )
A) $21.2 \times {10^5}N/{m^2}$
B) $20.4 \times {10^5}N/{m^2}$
C) $40 \times {10^4}N/{m^2}$
D) $21.2 \times {10^5}N/{m^2}$

seo-qna
Last updated date: 20th Jun 2024
Total views: 405k
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Answer
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Hint: We have a formula for pressure inside the fluid is ${P_{fluid}} = \rho gh$ \[\]
If you want to calculate the total pressure inside the fluid then we have to add atmospheric pressure to the absolute pressure.
The total pressure inside the fluid is given by
 $P = {P_{fluid}} + {P_{atm}}$

Complete step by step answer:
In the question the specific gravity of sea water is given 1.03
Then the question arises here what is the specific gravity?
Definition → the specific gravity or the relative gravity is a dimension less quantity that is defined as the ratio of the density of substance to the density of water at given temperature and pressure
specific gravity=density of substantdensity of water=substantwater
Step 1
By using this formula we can calculate the density of sea water
Specific gravity of sea water is given 1.03
$ \Rightarrow 1.03 = \dfrac{{{\rho _{sea}}}}{{{\rho _{water}}}}$
We know the density of water is $ \Rightarrow {10^3}kg/{m^3}$
Put this value in above equation
$ \Rightarrow 1.03 = \dfrac{{{\rho _{sea}}}}{{{{10}^3}}}$
$ \Rightarrow {\rho _{sea}} = 1.03 \times {10^3}kg/{m^3}$
In this step we get the density of seawater is $ \Rightarrow {\rho _{sea}} = 1.03 \times {10^3}kg/{m^3}$
Step 2
Now we use the formula of pressure inside the sea water.
$ \Rightarrow {P_{sea}} = {\rho _{sea}} \times gh$
Where ${\rho _{sea}} \Rightarrow $ Density of sea water
$g \Rightarrow $ Gravitational acceleration
$h \Rightarrow $ Height of point from free surface of sea water
Put all given values in this formula
$ \Rightarrow {P_{sea}} = 1.03 \times {10^3} \times 9.8 \times 200$
$ \Rightarrow {P_{sea}} = 2018.8 \times {10^3}$
Further solving
$ \Rightarrow {P_{sea}} = 20.188 \times {10^5}N/{m^3}$
Finally we get to the pressure at a point 200 meter inside the seawater
To get the final answer we have to add atmospheric pressure to the pressure due to seawater because atmospheric pressure applied everywhere
$ \Rightarrow P = {P_{sea}} + {P_{atm}}$
$ \Rightarrow P = 20.188 \times {10^5} + 1.013 \times {10^5}$
$\therefore P = 21.201 \times {10^5}N/{m^2}$
So total pressure at a point inside the fluid is $ \Rightarrow 21.20 \times {10^5}N/{m^2}$
Hence in this question option D is correct

Note: Sometimes student make a common mistake, they do not add the atmospheric pressure to the pressure due to fluid after applying the formula of pressure inside the liquid, so always be careful if the total pressure asked in the question or the atmospheric pressure is given in the question then you have to add the atmospheric pressure to the absolute pressure at a point inside a liquid
And sometimes students get confused from a given height. The height given in the formula of pressure is the height from the free surface of fluid. Not from the bottom of the fluid or we can say that it is the depth of the point inside the fluid.