
The density of water at the surface of the ocean is $ \rho $ . If the bulk modulus of water is $ B $ , what is the density of ocean water at a depth where the pressure is $ n{{P}_{o}} $ , where $ {{P}_{o}} $ is the atmospheric pressure:
$ \text{A}\text{. }\dfrac{\rho B}{B-(n-1){{P}_{o}}} $
$ \text{B}\text{. }\dfrac{\rho B}{B+(n-1){{P}_{o}}} $
$ \text{C}\text{. }\dfrac{\rho B}{B-n{{P}_{o}}} $
$ \text{D}\text{. }\dfrac{\rho B}{B+n{{P}_{o}}} $
Answer
595.5k+ views
Hint: As we go deep in water, the pressure increases resulting in less volume of water at the depth. For a given amount of water, mass will be conserved. We will apply the mass conservation equation when both density and volume of water will be changed at the depth.
Formula used:
\[\begin{align}
& B=\dfrac{\Delta P}{\dfrac{\Delta V}{V}} \\
& \Delta V=\dfrac{(n-1){{P}_{o}}V}{B}\text{ ,for }P'=n{{P}_{o}} \\
\end{align}\]
Complete step-by-step answer:
Due to the increase in hydrostatic pressure, the force per unit area exerted by liquid on a body, pressure at the depth of ocean is greater than the pressure at the surface. The deeper we go under the ocean, the greater the pressure pushing us down.
Let the volume of mass $ M $ at the surface has volume $ V $
Density of water $ \rho =\dfrac{M}{V} $
Pressure at the surface of water is $ {{P}_{o}} $
Pressure at some depth would be $ P'=n{{P}_{o}} $
Change in pressure $ \Delta P=n{{P}_{o}}-{{P}_{o}}=(n-1){{P}_{o}} $
As given the bulk modulus of water is $ B $
\[\begin{align}
& B=\dfrac{\Delta P}{\dfrac{\Delta V}{V}} \\
& \Delta V=\dfrac{(n-1){{P}_{o}}V}{B} \\
\end{align}\]
As we go deep, pressure increases and thus the volume of water will decrease at that depth, therefore,
Volume of water of mass $ M $ at that depth, $ V'=V-\Delta V $
$ V'=V-\dfrac{(n-1){{P}_{o}}V}{B}=\left( B-(n-1){{P}_{o}} \right)\dfrac{V}{B} $
$ V=\left( B-(n-1){{P}_{o}} \right)\dfrac{V}{B} $
Density of water at that depth $ \rho '=\dfrac{M}{V'}=\rho \dfrac{V}{V'} $
Therefore, $ \rho '=\dfrac{\rho B}{B-(n-1){{P}_{o}}} $
Density of ocean water at a depth where the pressure is $ n{{P}_{o}} $ , $ \rho '=\dfrac{\rho B}{B-(n-1){{P}_{o}}} $
Hence, the correct option is A.
Additional information:
Bulk modulus is a numerical figure, a constant, which describes the elastic properties of a solid or a fluid when it is under high pressure. The pressure applied on the material reduces its volume, however volume returns to original when pressure is removed. Bulk modulus is actually a measure of the ability of a substance or a material to withstand changes in volume when it is under the compression from all sides. The value of bulk modulus is equal to the quotient of the applied force divided by the relative deformation.
Note: Students should keep in mind that as we go deep in water both pressure and density of water will be changed. For a given mass of water, the value of density and volume at surface and at some depth will be different.
Formula used:
\[\begin{align}
& B=\dfrac{\Delta P}{\dfrac{\Delta V}{V}} \\
& \Delta V=\dfrac{(n-1){{P}_{o}}V}{B}\text{ ,for }P'=n{{P}_{o}} \\
\end{align}\]
Complete step-by-step answer:
Due to the increase in hydrostatic pressure, the force per unit area exerted by liquid on a body, pressure at the depth of ocean is greater than the pressure at the surface. The deeper we go under the ocean, the greater the pressure pushing us down.
Let the volume of mass $ M $ at the surface has volume $ V $
Density of water $ \rho =\dfrac{M}{V} $
Pressure at the surface of water is $ {{P}_{o}} $
Pressure at some depth would be $ P'=n{{P}_{o}} $
Change in pressure $ \Delta P=n{{P}_{o}}-{{P}_{o}}=(n-1){{P}_{o}} $
As given the bulk modulus of water is $ B $
\[\begin{align}
& B=\dfrac{\Delta P}{\dfrac{\Delta V}{V}} \\
& \Delta V=\dfrac{(n-1){{P}_{o}}V}{B} \\
\end{align}\]
As we go deep, pressure increases and thus the volume of water will decrease at that depth, therefore,
Volume of water of mass $ M $ at that depth, $ V'=V-\Delta V $
$ V'=V-\dfrac{(n-1){{P}_{o}}V}{B}=\left( B-(n-1){{P}_{o}} \right)\dfrac{V}{B} $
$ V=\left( B-(n-1){{P}_{o}} \right)\dfrac{V}{B} $
Density of water at that depth $ \rho '=\dfrac{M}{V'}=\rho \dfrac{V}{V'} $
Therefore, $ \rho '=\dfrac{\rho B}{B-(n-1){{P}_{o}}} $
Density of ocean water at a depth where the pressure is $ n{{P}_{o}} $ , $ \rho '=\dfrac{\rho B}{B-(n-1){{P}_{o}}} $
Hence, the correct option is A.
Additional information:
Bulk modulus is a numerical figure, a constant, which describes the elastic properties of a solid or a fluid when it is under high pressure. The pressure applied on the material reduces its volume, however volume returns to original when pressure is removed. Bulk modulus is actually a measure of the ability of a substance or a material to withstand changes in volume when it is under the compression from all sides. The value of bulk modulus is equal to the quotient of the applied force divided by the relative deformation.
Note: Students should keep in mind that as we go deep in water both pressure and density of water will be changed. For a given mass of water, the value of density and volume at surface and at some depth will be different.
Recently Updated Pages
Which cell organelles are present in white blood C class 11 biology CBSE

What is the molecular geometry of BrF4 A square planar class 11 chemistry CBSE

How can you explain that CCl4 has no dipole moment class 11 chemistry CBSE

Which will undergo SN2 reaction fastest among the following class 11 chemistry CBSE

The values of mass m for which the 100 kg block does class 11 physics CBSE

Why are voluntary muscles called striated muscles class 11 biology CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Explain zero factorial class 11 maths CBSE

State the laws of reflection of light

Difference Between Prokaryotic Cells and Eukaryotic Cells

Show that total energy of a freely falling body remains class 11 physics CBSE

