Question

1${{m}^{3}}$ water is brought inside the lake upto 200m depth from the surface of the lake. What will be changed in the volume when the bulk modulus of elasticity of water is 22000atm?
(density of water is $1\times {{10}^{3}}kg{{m}^{-3}}$, atmosphere pressure = ${{10}^{5}}N{{m}^{-2}}$ and g = 10$m{{s}^{-2}}$)
$\text{A}\text{. }8.9\times {{10}^{-3}}{{m}^{3}}$
$\text{B}\text{. 7}.8\times {{10}^{-3}}{{m}^{3}}$
$\text{C}\text{. }9.1\times {{10}^{-4}}{{m}^{3}}$
$\text{D}\text{. }8.7\times {{10}^{-4}}{{m}^{3}}$

Answer 1

Hint: Use the formula $P=\rho gh$ and find the pressure at the depth of 200m. Then use the formula for bulk modulus of a body i.e. $\beta =\dfrac{P}{\dfrac{\Delta V}{V}}$. Substitute the calculated value of P and the given values of $\beta $ and V to find the change in volume.
Formula used:
$\beta =\dfrac{P}{\dfrac{\Delta V}{V}}$
$P=\rho gh$

Complete answer:
When we apply a pressure on a body from all the direction, the body compresses and thus the volume of the body changes.
To understand the change in volume of the body we defined the bulk modulus of the body and it is given as $\beta =\dfrac{P}{\dfrac{\Delta V}{V}}$ …. (i),
where P is the pressure applied on the body and $\dfrac{\Delta V}{V}$ is the relative change in the volume of the body.
In the given case, when 1${{m}^{3}}$ of water is taken to a depth of 200m, the surrounding water will exert a pressure on this volume of water. Therefore, the volume of water will reduce.
Let us first calculate the pressure P. The pressure inside a liquid at a depth of h from the top surface is given as $P=\rho gh$,
where $\rho $ is the density of the liquid and g is acceleration due to gravity.
It is given that h=200m, $\rho ={{10}^{3}}kg{{m}^{-3}}$ and g = 10$m{{s}^{-2}}$.
Therefore, $P=\rho gh={{10}^{3}}\times 10\times 200=2\times {{10}^{6}}N{{m}^{-2}}$.
The initial volume of water is given to be V=1${{m}^{3}}$.
The bulk modulus of water is given as 22000atm. It is given that 1atm = ${{10}^{5}}N{{m}^{-2}}$.
$\Rightarrow \beta =22000atm=22000\times {{10}^{5}}N{{m}^{-2}}.$
Substitute the values of P, V and $\beta $ in equation (i).
$\Rightarrow 22000\times {{10}^{5}}=\dfrac{2\times {{10}^{6}}}{\dfrac{\Delta V}{1}}$
$\Rightarrow \Delta V=\dfrac{2\times {{10}^{6}}}{22000\times {{10}^{5}}}=9.1\times {{10}^{-4}}{{m}^{3}}$.
Therefore, the change in the volume of the water is $9.1\times {{10}^{-4}}{{m}^{3}}$.

So, the correct answer is “Option C”.

Note:
The bulk modulus is similar to Young’s modulus and modulus of rigidity.
Young’s modulus is the ratio of linear stress to the linear strain in the body.
Modulus of rigidity is the ratio of shear stress to the shear strain in the body.
Similarly bulb modulus is also a ratio of stress to the strain. The pressure on the body is equal to the stress in the body and relative change in the volume of the body is the strain in the body.