A sphere of small size is at the bottom of a lake of depth 200 m. Due to pressure its fractional change in volume is α x 10^-7. What is value of α , if bulk modulus of sphere is 5 x 10¹² Pa? (Use g = 10 m/s²)

The correct answer is 4.

Given Data:
Depth of the lake, $h=200\text{m}$
Bulk modulus of the sphere, $B=5\times {10}^{12}\text{Pa}$
Acceleration due to gravity, $g=10{\text{m/s}}^2$
Density of water, $\rho ={10}^3{\text{kg/m}}^3$

As depicted in the provided schematic diagram, a small black sphere is located at the bottom of a water column representing a lake of depth 200 m.

The excess pressure (gauge pressure) acting on the sphere at the bottom of the lake due to the weight of the water column is calculated as:

$\Delta P=\rho gh$

Substituting the given values:

$\Delta P={10}^3\times 10\times 200=2\times {10}^6\text{Pa}$

The bulk modulus $B$ of a material is defined as the ratio of the change in pressure to the fractional change in volume:

$B=\frac{\Delta P}{\left(\frac{\Delta V}{V}\right)}$

Rearranging the equation to find the expression for the fractional change in volume $\frac{\Delta V}{V}$:

$\frac{\Delta V}{V}=\frac{\Delta P}{B}$

Substituting the values of $\Delta P$ and $B$ into this expression:

$\frac{\Delta V}{V}=\frac{2\times {10}^6}{5\times {10}^{12}}$

$\frac{\Delta V}{V}=0.4\times {10}^{-6}=4\times {10}^{-7}$

Comparing this result with the given fractional change in volume expression, which is $\alpha \times {10}^{-7}$:

$\alpha =4$