Question

What is the approximate change in density of water in a lake at a depth of 400 m below the surface? The density of water at the surface is 1030 kg/$m^3$ and bulk modulus of water is $2\times {10}^{9}N/m^3$

• A. 4 kg/$m^3$
• B. 2 kg/$m^3$
• C. 6 kg/$m^3$
• D. 8 kg/$m^3$

Answer 1

The correct option is B 2 kg/$m^3$

$\text{At a depth}x\text{m below the surface the pressure}P=\rho gx$

$\begin{array}{cc}\text{Bulk strain}& =\frac{P}{B}=\frac{\rho gx}{B}\\ & \Rightarrow \frac{\Delta V}{V}=\frac{\rho gx}{B}\\ & \Rightarrow \Delta V=\frac{\rho gxV}{B}\end{array}$

$\begin{array}{cc}& \text{Density at}x\text{m below the surface}{\rho }^{\prime }=\frac{\rho V}{v^{\prime }}\\ & (v^{\prime }=v-\Delta v)\\ & \Rightarrow {\dot{\rho }}^{\prime }=\frac{\beta V}{V-\frac{\rho gxV}{\beta }}\\ & \Rightarrow s^{\prime }=\frac{\rho B}{B\rho gx}\end{array}$

$\begin{array}{c}\Rightarrow {\rho }^{\prime }=\frac{1030\times 2\times {10}^9}{2\times {10}^9-1030\times 9.8\times 400}\\ \Rightarrow {\rho }^{\prime }=1032kg/m^3\\ \Delta \rho =1032-1030=2kg/m^3\\ \text{Ans:}B\end{array}$