Answer

Verified

405k+ views

**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.

Recently Updated Pages

Cryolite and fluorspar are mixed with Al2O3 during class 11 chemistry CBSE

Select the smallest atom A F B Cl C Br D I class 11 chemistry CBSE

The best reagent to convert pent 3 en 2 ol and pent class 11 chemistry CBSE

Reverse process of sublimation is aFusion bCondensation class 11 chemistry CBSE

The best and latest technique for isolation purification class 11 chemistry CBSE

Hydrochloric acid is a Strong acid b Weak acid c Strong class 11 chemistry CBSE

Trending doubts

Difference Between Plant Cell and Animal Cell

The Buddhist universities of Nalanda and Vikramshila class 7 social science CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Name 10 Living and Non living things class 9 biology CBSE

Which are the Top 10 Largest Countries of the World?

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE

Who founded the Nalanda University 1 Mauryan 2 Guptas class 6 social science CBSE