JEE Physics 2025 Question with Solution

Given:

• Initial pressure of the liquid $P_i = 1 \, \text{atm}$
• Final pressure of the liquid $P_f = 5 \, \text{atm}$
• Change in volume $\Delta V = -0.8 \, \text{cm}^3 = -0.8 \times 10^{-6} \, \text{m}^3$
• Bulk modulus $B = 2 \times 10^9 \, \text{Pa}$

Find: Initial volume $V$ of the liquid.

The change in pressure is

$$\Delta P = P_f - P_i = 5 \, \text{atm} - 1 \, \text{atm} = 4 \, \text{atm} = 4 \times 10^5 \, \text{Pa}$$

Using the formula for bulk modulus,

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

So,

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

Substituting the given values,

$$V = -2 \times 10^9 \left( \frac{-0.8 \times 10^{-6}}{4 \times 10^5} \right)$$

Thus,

$$V = 4 \times 10^{-3} \, \text{m}^3 = 4 \, \text{litre}$$

Therefore, the initial volume of the liquid is $4 \, \text{litre}$.

Given:

• $\Delta P = 4 \times 10^5 \, \text{Pa}$
• $K = 2 \times 10^9 \, \text{Pa}$
• $\Delta V = -0.8 \times 10^{-6} \, \text{m}^3$

Find: Initial volume $V_0$.

Use the bulk modulus relation

$$K = -\frac{\Delta P}{\Delta V/V_0}$$

Substitute the known values:

$$2 \times 10^9 = -\frac{4 \times 10^5}{-0.8 \times 10^{-6}/V_0}$$

This gives

$$2 \times 10^9 = \frac{4 \times 10^5 \times V_0}{0.8 \times 10^{-6}}$$

Now solve for $V_0$:

$$V_0 = \frac{2 \times 10^9 \times 0.8 \times 10^{-6}}{4 \times 10^5}$$$$V_0 = \frac{1.6 \times 10^3}{4 \times 10^5} = 4 \times 10^{-3} \, \text{m}^3$$

Convert to litres using $1 \, \text{m}^3 = 1000 \, \text{litre}$:

$$V_0 = 4 \times 10^{-3} \times 1000 = 4 \, \text{litre}$$

Therefore, the required numerical answer is 4.