A ball falling in a lake of depth 200 m shows 0.1% decrease in its volume at the bottom. What is the bulk modulus of the material of the ball

To find the bulk modulus of the material of the ball, we can use the definition of bulk modulus ($B$). Bulk modulus is defined as the ratio of volumetric stress (change in pressure, $\Delta P$) to the volumetric strain ($\frac{\Delta V}{V}$).

The formula for bulk modulus is:
$B=\frac{\Delta P}{\left(\frac{\Delta V}{V}\right)}$

Let's first calculate the change in pressure ($\Delta P$) experienced by the ball at the bottom of the lake. The pressure exerted by a liquid column of height $h$ is given by:
$\Delta P=\rho gh$

Where:
- $\rho$ is the density of water, which is approximately ¹⁰³ kg/m³.
- $g$ is the acceleration due to gravity, which is 9.8 m/s².
- $h$ is the depth of the lake, which is given as 200 m.

Substituting these values into the pressure formula:
$\Delta P={10}^3\times 9.8\times 200$
$\Delta P=1.96\times {10}^6{\text{N/m}}^2$

Next, we look at the volumetric strain. The question states that there is a 0.1% decrease in the volume of the ball at the bottom of the lake. This fractional change in volume can be written as:
$\frac{\Delta V}{V}=\frac{0.1}{100}={10}^{-3}$

Now, substitute the values of $\Delta P$ and $\frac{\Delta V}{V}$ back into the Bulk Modulus equation:
$B=\frac{1.96\times {10}^6}{{10}^{-3}}$
$B=1.96\times {10}^9{\text{N/m}}^2$

To match the format of the options, we can rewrite $1.96\times {10}^9$ as:
$B=19.6\times {10}^8{\text{N/m}}^2$

Thus, the Bulk Modulus of the material of the ball is 19.6 x10⁸ N /m².