JEE Physics 2024 Question with Solution

Given: The volume of the rubber ball decreases by $0.02\%$.

Find: The depth $h$ below the sea surface.

Use the bulk modulus relation:

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

Given fractional decrease in volume:

$$\frac{\Delta V}{V} = 0.02\% = \frac{0.02}{100} = 0.0002$$

Rearranging,

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

Substitute $B = 9 \times 10^8 \, \text{N m}^{-2}$:

$$\Delta P = -9 \times 10^8 \times 0.0002 = -1.8 \times 10^5 \, \text{N m}^{-2}$$

The required pressure increase in magnitude is:

$$\Delta P = 1.8 \times 10^5 \, \text{N m}^{-2}$$

Hydrostatic pressure at depth $h$ is:

$$\Delta P = \rho g h$$

where $\rho = 10^3 \, \text{kg m}^{-3}$ and $g = 10 \, \text{m s}^{-2}$.

Equating,

$$10^3 \times 10 \times h = 1.8 \times 10^5$$ $$10^4 h = 1.8 \times 10^5$$ $$h = \frac{1.8 \times 10^5}{10^4} = 18 \, \text{m}$$

Therefore, the depth below the surface of the sea is $18 \, \text{m}$. The correct option is A.

Given: $\beta = -\frac{\Delta P}{\Delta V/V}$ and the decrease in volume is $0.02\%$.

Find: Depth $h$.

Write directly:

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

Substitute the values from the solution:

$$10^3 \times 10 \times h = -9 \times 10^8 \times \left(-\frac{0.02}{100}\right)$$ $$h = 18 \, \text{m}$$

This works because the pressure increase due to depth is exactly the pressure required to produce the stated fractional compression. Hence, the correct option is A.