JEE Physics 2025 Question with Solution

Given: Area of window $A = 100 \, \text{cm}^2 = 10^{-2} \, \text{m}^2$, depth $h = 3 \, \text{m}$, density of water $\rho_w = 1000 \, \text{kg/m}^3$, density of other liquid $\rho_l = 1500 \, \text{kg/m}^3$, and $g \approx 10 \, \text{m/s}^2$.

Find: The external force required to keep the hinged door closed.

For a small door, pressure at depth $h$ may be treated as uniform. Pressure on each side is:

$$P = \rho g h$$

So the force due to water is:

$$F_w = \rho_w g h A = 1000 \times 10 \times 3 \times 10^{-2} = 300 \, \text{N}$$

The force due to the denser liquid is:

$$F_l = \rho_l g h A = 1500 \times 10 \times 3 \times 10^{-2} = 450 \, \text{N}$$

These forces act in opposite directions, so the net force tending to open the door is:

$$F_{\text{net}} = F_l - F_w = 450 - 300 = 150 \, \text{N}$$

Therefore, the external force required to keep the door closed is $150 \, \text{N}$. The correct option is A.

Given: The door separates two liquids at the same depth $3 \, \text{m}$. The area is $10^{-2} \, \text{m}^2$.

Find: The force needed so that the door does not open.

The pressure difference across the door is:

$$\Delta P = (\rho_l - \rho_w)gh$$

Substituting the values:

$$\Delta P = (1500 - 1000) \times 10 \times 3 = 15000 \, \text{Pa}$$

Now force equals pressure difference times area:

$$F = \Delta P \times A = 15000 \times 10^{-2} = 150 \, \text{N}$$

Thus, the required applied force is $150 \, \text{N}$, opposite to the force exerted by the denser liquid. The correct option is A.