Question

The average depth of Indian Ocean is about 3000m. The value of fractional compression $\dfrac{\Delta V}{V}$ of water at the bottom of the ocean is: [Given that the bulk modulus of the water is $2.2\times 10^{9} N/m^{2},\;g=9.8m/s^{2}$ and $\rho_{water}=1000kg/m^{3}$]

& A.3.4\times {{10}^{-2}} \\\ & B.1.34\times {{10}^{-2}} \\\ & C.4.13\times {{10}^{-2}} \\\ & D.13.4\times {{10}^{-2}} \\\ \end{aligned}$$

Answer 1

The bulk modulus is defined as the ratio of volumetric stress to volumetric strain of the substance, and degree of compression is the bulk modulus of elasticity i.e. $K=-\dfrac{\Delta P}{\dfrac{\Delta V}{V}}$. To find $\dfrac{\Delta V}{V}$

Formula used:
$K=\dfrac{Vol \;stress}{Vol\;Strain}$ or $K=-\dfrac{\Delta P}{\dfrac{\Delta V}{V}}$

Complete step by step answer:
We know that the Indian Ocean has salt water, which is dominantly liquid. Pressure is applied on water and it gets compressed. The degree of compression is the bulk modulus of elasticity. The bulk modulus is defined as the ratio of volumetric stress to volumetric strain of the substance, here water.
Then$K=\dfrac{Vol \;stress}{Vol\;Strain}$, where$K$ is a constant, which explains the elasticity of the fluid, water.
Then, $K=-\dfrac{\Delta P}{\dfrac{\Delta V}{V}}$ where $\Delta P$change in pressure is, $V$ is the initial volume and $\Delta V$ is the change in volume. Where –ve sign indicates that the substance is compressed.
Thus, $\dfrac{\Delta V}{V}=\dfrac{\Delta P}{K}$.
Given $K=2.2\times 10^{9} N/m^{2},\;g=9.8m/s^{2},\rho_{water}=1000kg/m^{3},h=3000m$
We can write $\Delta P=h\rho g$, assuming the initial height of the sea as $0$. Then the initial pressure is $0$ and change in pressure is the pressure at the final height$h$, clearly pressure varies with height and density of the material.
Then, substituting the values we get, $\dfrac{\Delta V}{V}=\dfrac{3\times 10^{3} \times 9.8\times 10^{3} }{2.2\times 10^{9}}=1.34\times 10^{-2}m$

**Thus, the answer is B.$1.34\times 10^{-2}m$
**

Note:
Instead of giving the change in pressure, here, $h,\rho$ is given which is the only trick here. Then$\Delta P=h\rho g$, assuming the initial height of the sea as $0$. Then the initial pressure is $0$ and change in pressure is the pressure at the final height $h$, clearly pressure varies with height and density of the material. Also, in $K=-\dfrac{\Delta P}{\dfrac{\Delta V}{V}}$, –ve sign indicates that the substance is compressed.