Question

$1\text{ cc}$ of water is taken from the surface to the bottom of a lake having depth $100m$. If bulk modulus of water is $2.2\times {{10}^{9}}N/{{m}^{2}}$, then decrease in the volume of water will be
A.)$4.5\times {{10}^{-4}}cc$
B.)$8.8\times {{10}^{-4}}cc$
C.)$2.2\times {{10}^{-4}}cc$
D.)$1.1\times {{10}^{-4}}cc$

Answer 1

Hint: The bulk modulus of water is the ratio of the change in pressure to the change in relative volume of a body. When the volume of water is taken from the bottom of the lake to the surface, there is a pressure change and hence there will be a change in the volume of the water. The change in the volume can be found out using the bulk modulus value.

Formula used:
The bulk modulus $B$ of a material is

$B=-\dfrac{\Delta P}{\dfrac{\Delta V}{V}}$

where $\Delta P$ is the change in the net pressure exerted on the body, $\Delta V$ is the change in volume of the body and $V$ is the initial volume of the body. The negative sign implies that if the pressure increases (positive pressure change), the volume decreases (negative volume change) and vice versa.
The net normal pressure $P$ applied on a body by a liquid of density $\rho $ at a depth $h$ below its surface is given by

$P=h\rho g$
where $g$ is the acceleration due to gravity.

Complete step by step answer:
As explained in the hint, we will find out the volume change of the water when it is brought from the bottom to the surface using the bulk modulus of water.
Therefore, let us analyze the question.
The given depth of the lake is ${{h}_{i}}=100m$.
Let the pressure in this initial state be ${{P}_{i}}$.
The depth at the surface is ${{h}_{f}}=0m$
Let the pressure on the body in this final state be ${{P}_{f}}$.
Let the density of water be $\rho ={{10}^{3}}kg/{{m}^{3}}$.
The net normal pressure $P$ applied on a body by a liquid of density $\rho $ at a depth $h$ below its surface is given by
$P=h\rho g$ --(1)
where $g$ is the acceleration due to gravity.
Therefore, using (1), we get,

${{P}_{i}}={{h}_{i}}\rho g$
${{P}_{f}}={{h}_{f}}\rho g$
Therefore the change in pressure on the body will be $\Delta P$.
$\Delta P={{P}_{f}}-{{P}_{i}}$
$\therefore \Delta P={{h}_{f}}\rho g-{{h}_{i}}\rho g=\left( {{h}_{f}}-{{h}_{i}} \right)\rho g$ --(2)

Now, the volume of the water is $V=1cc$.
Let the change in its volume be $\Delta V$.
The given bulk modulus of water is $B=2.2\times {{10}^{9}}N/{{m}^{2}}$.
Now, the bulk modulus $B$ of a material is
$B=-\dfrac{\Delta P}{\dfrac{\Delta V}{V}}$ --(3)

where $\Delta P$ is the change in the net pressure exerted on the body, $\Delta V$ is the change in volume of the body and $V$ is the initial volume of the body. The negative sign implies that if the pressure increases (positive pressure change), the volume decreases (negative volume change) and vice versa.
Therefore, using the above information and (2) in (3), we get,

$2.2\times {{10}^{9}}N/{{m}^{2}}=-\dfrac{\left( {{h}_{f}}-{{h}_{i}} \right)\rho g}{\dfrac{\Delta V}{1cc}}$
$\therefore 2.2\times {{10}^{9}}N/{{m}^{2}}=-\dfrac{\left( 0-100 \right)\times {{10}^{3}}\times 9.8}{\dfrac{\Delta V}{1cc}}=-\dfrac{\left( -100 \right)\times {{10}^{3}}\times 9.8}{\dfrac{\Delta V}{1cc}}=\dfrac{{{10}^{5}}\times 9.8}{\dfrac{\Delta V}{1cc}}$
$\therefore \dfrac{\Delta V}{1cc}=\dfrac{9.8\times {{10}^{5}}}{2.2\times {{10}^{9}}}=4.45\times {{10}^{-4}}$
$\therefore \Delta V=4.45\times {{10}^{-4}}\times 1cc=4.45\times {{10}^{-4}}cc\approx 4.5\times {{10}^{-4}}cc$

Hence, the change in volume is $4.5\times {{10}^{-4}}cc$.
Therefore, the correct option is A) $4.5\times {{10}^{-4}}cc$.

Note: It is not very intuitive to students that this is a problem involving bulk modulus and pressure change of the body since some students cannot catch that the pressure changes at different depths in the water body. However, this is a typical problem of this topic.

It is essential that the negative sign in the formula for bulk modulus is written since it signifies that the volume decreases if pressure on the body increases. For example, in the above problem, if the negative sign was not written properly, the volume change would have come out to be negative which would mean that the volume of the body decreases however, this is absolutely wrong since the volume of the body is actually increasing. Hence, the negative sign is very significant in this regard.