Question

A ball falling in a lake of depth 200m shows 0.1% decrease in its volume at the bottom. What is the bulk modulus of the material of the ball?
A. $19.6\times {{10}^{8}}N{{m}^{-2}}$
B.$19.6\times {{10}^{-10}}N{{m}^{-2}}$
C.$19.6\times {{10}^{10}}N{{m}^{-2}}$
D.$19.6\times {{10}^{-8}}N{{m}^{-2}}$

Answer 1

Hint: Firstly, we could recall the expression for the pressure on a body at depth h and then substitute accordingly to find that pressure. Then we have the percentage decrease in volume given. Now we could substitute them in the expression for bulk modulus and thus find the answer.

Formula used:
Bulk modulus,
$B=\dfrac{\Delta P}{\left( \dfrac{\Delta V}{V} \right)}$
Pressure at depth h,
$\Delta P=P-{{P}_{0}}=\rho gh$

Complete Step by step solution:
In the question, we are given a ball that is falling in a lake of depth 200m. This lake is said to show 0.1% decrease in volume at the bottom. We are asked to find the bulk modulus of the material of the ball using these given information.
As a first step, we could recall the expression for Bulk modulus of the material of the ball.
$B=\dfrac{\Delta P}{\left( \dfrac{\Delta V}{V} \right)}$ …………………………………………….. (1)
Where, ‘$\Delta P$’ is the pressure on the ball as it is falls down the lake and $\dfrac{\Delta V}{V}$ is the decrease in volume observed.
As the ball is falling down the lake the pressure on the ball will be the gauge pressure at the given depth.
The gauge pressure is given by,
$\Delta P=P-{{P}_{0}}=\rho gh$
Where, P is the pressure at depth h, ${{P}_{0}}$ is the atmospheric pressure, $\rho $ is the density of the water, g is the acceleration due to gravity and h is the depth at which the pressure is to be found.
$\Delta P=\rho gh=1000\times 9.8\times 200$
$\Rightarrow \Delta P=1.96\times {{10}^{6}}Pa$ ……………………………………….. (2)
Also, the ball is said to decrease in size by 0.1%, so,
$\dfrac{\Delta V}{V}=\dfrac{0.1}{100}$
$\Rightarrow \dfrac{\Delta V}{V}=0.001$ ………………………………………………….. (3)
Now we could directly substitute (2) and (3) in (1) to get,
$B=\dfrac{1.96\times {{10}^{6}}}{{{10}^{-3}}}$
$\therefore B=19.6\times {{10}^{8}}N{{m}^{-2}}$
Therefore, we found the bulk modulus of the material of the ball to be$19.6\times {{10}^{8}}N{{m}^{-2}}$.

Hence, option A is found to be the correct answer.

Note:
Basically, from the value of bulk modulus of a substance we could understand how resistant to compression that particular material is. We have other moduli that describe material stress to strain like the shear modulus and the Young’s modulus. But for the case of liquids only bulk moduli is meaningful.