Question

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 :- Hint : [P = pgh]

• A. 19.6 × 10⁸ N/m²
• B. 19.6 × 10¹⁰ N/m²
• C. 19.6 × 10⁻¹⁰ N/m²
• D. 19.6 × 10⁻⁸ N/m²

Answer 1

19.6 × 10⁸ N/m²

Answer 2

The pressure at the bottom of the lake is given by $P = \rho g h$. Given: Depth, $h = 200$ m Percentage decrease in volume, $\frac{\Delta V}{V} = -0.1\% = -0.001$ Density of water, $\rho \approx 1000 \, \text{kg/m}^3$ Acceleration due to gravity, $g \approx 9.8 \, \text{m/s}^2$

Calculate the pressure: $P = (1000 \, \text{kg/m}^3) \times (9.8 \, \text{m/s}^2) \times (200 \, \text{m})$ $P = 196000 \, \text{N/m}^2 = 1.96 \times 10^5 \, \text{N/m}^2$

The bulk modulus ($B$) is defined as: $B = -\frac{V \Delta P}{\Delta V} = -\frac{\Delta P}{\Delta V/V}$

Substituting the values: $B = -\frac{1.96 \times 10^5 \, \text{N/m}^2}{-0.001}$ $B = \frac{1.96 \times 10^5}{10^{-3}} \, \text{N/m}^2$ $B = 1.96 \times 10^8 \, \text{N/m}^2$

This value matches option (A) when written as $19.6 \times 10^8 \, \text{N/m}^2$. There appears to be a typo in the provided options or the question's percentage value, as $1.96 \times 10^8 \, \text{N/m}^2$ is the direct calculation. However, if we assume the percentage decrease was $0.01\%$, then $B = 1.96 \times 10^9 \, \text{N/m}^2$, which is $19.6 \times 10^8 \, \text{N/m}^2$. Assuming this intended calculation, option (A) is the closest fit.