Question

The Mariana trench is located in the Pacific Ocean, and at one place it is nearly $11Km$ beneath the surface of water. The water pressure at the bottom of the trench is about $1.1 \times {10^8}Pa$ . A steel ball of initial volume $0.32{m^3}$ is dropped into the ocean and falls to the bottom of the trench. What is the change in the volume of the ball when it reaches to the bottom?

Answer 1

Hint : In order to solve this question, we are going to first take the pressure, and volume as given in the question and the general value of the bulk modulus of the steel, then, taking the relation for bulk modulus which is inversely proportional to the change in the volume, we get the required answer.
Formula used: The bulk modulus is given by
$B = \dfrac{p}{{\dfrac{{\Delta V}}{V}}}$

Complete Step By Step Answer:
As it is given in the question that the water pressure at the bottom of the trench is $p = 1.1 \times {10^8}Pa$
The volume of the steel ball is $V = 0.32{m^3}$
The bulk modulus of steel is, $B = 1.6 \times {10^{11}}N{m^{ - 2}}$
The ball falls at the bottom of the Pacific Ocean, about $11Km$ below the surface
Let the volume change of the ball on reaching the bottom of the trench be $\Delta V$ .
The bulk modulus is given by
$B = \dfrac{p}{{\dfrac{{\Delta V}}{V}}}$
This implies that the change in the volume is given by
$\Delta V = \dfrac{{pV}}{B}$
Now, putting the values in the above equation
$\Delta V = \dfrac{{1.1 \times {{10}^8} \times 0.32}}{{1.6 \times {{10}^{11}}}} = 2.2 \times {10^{ - 4}}{m^3}$
Therefore, the change in the volume of the ball on reaching the bottom of the trench is $2.2 \times {10^{ - 4}}{m^3}$ .

Note :
It is important to note that the bulk modulus depends on the pressure and is inversely proportional to the ratio of the change in the volume and the volume of the ball. Knowing the constant value for the bulk modulus gives us an easy way to calculate the change in the volume.