Question

Bulk modulus for rubber is $9.8\times {10}^8{\text{N/m}}^2$. To what depth should a rubber ball be taken in a lake, so that its volume is decreased by $0.1$%?

• A. $10\text{m}$
• B. $100\text{m}$
• C. $200\text{m}$
• D. $20\text{m}$

Answer 1

The correct option is B $100\text{m}$

Here, $\frac{\Delta V}{V}=\frac{0.1}{100},g=10{\text{m/s}}^2$
For rubber, $B=9.8\times {10}^8{\text{N/m}}^2$


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

Let $h$ be the required depth and the density of water be $\rho$ then,

$h\rho g=-\frac{B\Delta V}{V}$

Considering only magnitude of all quantities we have,

$h=\frac{B\Delta V}{\rho gV}=\frac{9.8\times {10}^8\times 0.1}{{10}^3\times 9.8\times 100}=100\text{m}$

Hence, $\left(B\right)$ is the correct answer.

Why this Q? Bulk Modulus: $B=\frac{\text{Volume stress}}{\text{Volume strain}}=\frac{\Delta P}{-\frac{\Delta V}{V}}\Rightarrow B=-\frac{V\Delta P}{\Delta V}$