JEE Physics 2025 Question with Solution

Given: Depth of water is $h = 2.5 \, \text{km} = 2500 \, \text{m}$. Density of water is $\rho = 10^3 \, \text{kg m}^{-3}$, acceleration due to gravity is $g = 10 \, \text{m s}^{-2}$, and bulk modulus is $K = 2 \times 10^9 \, \text{N m}^{-2}$.

Find: The fractional compression $\frac{\Delta V}{V}$ of water at this depth.

Use the relation

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

where the pressure at depth is

$$P = \rho g h$$

Substituting the given values,

$$P = 10^3 \times 10 \times 2500 = 2.5 \times 10^7 \, \text{N m}^{-2}$$

Now,

$$\frac{\Delta V}{V} = \frac{2.5 \times 10^7}{2 \times 10^9} = 0.0125$$

Expressing this as percentage,

$$0.0125 \times 100 = 1.25\%$$

Therefore, the fractional compression of water is $1.25\%$, so the correct option is D. The solution states option A, but the shown working gives $1.25\%$.

Given: Water is at a depth $2.5 \, \text{km}$ below sea level.

Find: Fractional compression.

First calculate hydrostatic pressure:

$$P = \rho g h$$

With

$$\rho = 10^3 \, \text{kg m}^{-3}, \quad g = 10 \, \text{m s}^{-2}, \quad h = 2500 \, \text{m}$$

we get

$$P = 10^3 \times 10 \times 2500 = 25 \times 10^6 = 2.5 \times 10^7 \, \text{N m}^{-2}$$

Now apply the definition of bulk modulus:

$$K = \frac{P}{\Delta V / V}$$

So,

$$\frac{\Delta V}{V} = \frac{P}{K} = \frac{2.5 \times 10^7}{2 \times 10^9}$$ $$\frac{\Delta V}{V} = 1.25 \times 10^{-2} = 0.0125$$

As a percentage,

$$0.0125 = 1.25\%$$

Hence the fractional compression is $1.25\%$, matching option D.