Question

A ball falling-in a lake of depth $400\text{ }m$ has a decrease of $0.2$ in its volume at the bottom. The bulk modulus of the material of the ball is (in $N{{m}^{-2}}$)
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 : The bulk modulus is an evaluation of the potential of an object to resist any volumetric changes when under contraction from on all sides. Hence, we will use the bulk modulus formula where we will put the height or depth of the lake $\left( d \right)$ and the volumetric decrease $\dfrac{\Delta v}{v}$of the object in relationship along with the density $\left( \rho \right)$ of the object and $g$ is the gravity as:
$B=\dfrac{h\rho g}{\left( \dfrac{\Delta v}{v} \right)}$

Complete step by step solution:
First, let us write the depth at which the ball has fallen off the lake as $400m$.
Now, the volumetric decreases at the bottom as $\dfrac{\Delta v}{v}=\dfrac{0.2}{100}$ (percentage value to fraction).
And the density of the water is taken as $1000\text{ }kg/{{m}^{3}}$ and the gravity is $10\text{ }m/{{s}^{2}}$.
Now placing all the values in the bulk modulus formula, we get the result as:
$\Rightarrow B=\dfrac{400\times 1000\times 9.8}{\left( \dfrac{0.2}{100} \right)}$
$\Rightarrow B=\dfrac{400\times 1000\times 9.8\times 100}{0.2}$
$\Rightarrow B=\dfrac{400\times 1000\times 9.8\times 100\times 10}{2}$
$\Rightarrow B=19.6\times {{10}^{8}}N/{{m}^{2}}$
Therefore, the bulk modulus on the ball is given as $B=19.6\times {{10}^{8}}N/{{m}^{2}}$.

So, option A is correct

**Note :** Another method to find the bulk modulus is to find the change in pressure generated due to the ball reaching to the bottom from the top and after that divide the pressure by the volumetric change in the ball as:
Pressure change from top to bottom of the lake: ${{P}_{bottom}}-{{P}_{top}}=\rho gh$
The volumetric change in the ball is $\Delta V=0.2\times {{10}^{-2}}$
Hence, the bulk modulus of the ball is $\dfrac{\rho gh}{\Delta V}=\dfrac{400\times 1000\times 9.8}{0.2\times {{10}^{-2}}}$
$\dfrac{\rho gh}{\Delta V}=19.6\times {{10}^{8}}N/{{m}^{2}}$.