JEE Physics 2025 Question with Solution

Given: Water flows in a horizontal pipe. When the valve is closed, the pressure gauge reads $P_1$ and the water is at rest, so velocity is $0$. When the valve is opened, the pressure becomes $P_2$ and the water flows with speed $v$.

Find: How $v$ depends on $P_1 - P_2$.

Use Bernoulli's principle for horizontal flow:

$$P + \frac{1}{2}\rho v^2 = \text{constant}$$

When the valve is closed:

$$P_1 = \text{constant}$$

When the valve is opened:

$$P_2 + \frac{1}{2}\rho v^2 = \text{constant}$$

Equating these:

$$P_1 = P_2 + \frac{1}{2}\rho v^2$$

So,

$$\frac{1}{2}\rho v^2 = P_1 - P_2$$

Therefore,

$$v^2 = \frac{2(P_1 - P_2)}{\rho}$$

Hence,

$$v = \sqrt{\frac{2(P_1 - P_2)}{\rho}}$$

So the speed is proportional to $\sqrt{P_1 - P_2}$. Therefore, the correct option is A.

Given: The pipe is horizontal, so height terms do not change. Pressure falls from $P_1$ to $P_2$ when water starts flowing.

Find: The proportional relation for speed $v$.

From Bernoulli's equation for horizontal flow,

$$P_1 - P_2 = \frac{1}{2}\rho v^2$$

Thus,

$$v \propto \sqrt{P_1 - P_2}$$

Therefore, the correct option is A. This shortcut works because the pressure drop is converted into kinetic energy per unit volume.