Answer
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Hint Density is given by mass per unit volume, find the relation between the Bulk modulus and the change in density. For that you have to first find a relation between change in pressure to change is density.
Complete step-by-step solution
We know that the formula for bulk modulus of the material is given as
\[B = - \dfrac{{\Delta P}}{{\Delta V/V}}\]
\[ - \dfrac{{\Delta V}}{V} = \dfrac{{\Delta P}}{B}\]
We also know that the when the volume of a body decreases, it density increases, so
\[ - \dfrac{{\Delta V}}{V} = \dfrac{{\Delta \rho }}{\rho }\]
Substituting this in the above equation,
\[\dfrac{{\Delta \rho }}{\rho } = \dfrac{{\Delta P}}{B}\]
\[\Delta \rho = \dfrac{{\rho \Delta P}}{B}\]
As we move from the surface of the water to a height of depth h, pressure is increased by \[h\rho g\], Therefore,
\[\Delta \rho = \dfrac{{{\rho ^2}hg}}{B}\]
So the correct answer is option D.
Note You can check the authenticity of your answer with the help of dimensional analysis. The dimensions on the right side of equation should be equal to the right side which is the case only with option D
Complete step-by-step solution
We know that the formula for bulk modulus of the material is given as
\[B = - \dfrac{{\Delta P}}{{\Delta V/V}}\]
\[ - \dfrac{{\Delta V}}{V} = \dfrac{{\Delta P}}{B}\]
We also know that the when the volume of a body decreases, it density increases, so
\[ - \dfrac{{\Delta V}}{V} = \dfrac{{\Delta \rho }}{\rho }\]
Substituting this in the above equation,
\[\dfrac{{\Delta \rho }}{\rho } = \dfrac{{\Delta P}}{B}\]
\[\Delta \rho = \dfrac{{\rho \Delta P}}{B}\]
As we move from the surface of the water to a height of depth h, pressure is increased by \[h\rho g\], Therefore,
\[\Delta \rho = \dfrac{{{\rho ^2}hg}}{B}\]
So the correct answer is option D.
Note You can check the authenticity of your answer with the help of dimensional analysis. The dimensions on the right side of equation should be equal to the right side which is the case only with option D
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