JEE MAIN - Physics (2025 - 23rd January Evening Shift - No. 15)

Water flows in a horizontal pipe whose one end is closed with a valve. The reading of the pressure gauge attached to the pipe is $P_1$. The reading of the pressure gauge falls to $P_2$ when the valve is opened. The speed of water flowing in the pipe is proportional to
$\left(P_1-P_2\right)^2$
$\sqrt{\mathrm{P}_1-\mathrm{P}_2}$
$P_1-P_2$
$\left(P_1-P_2\right)^4$

Explanation

When water flows from a region of higher pressure to a region of lower pressure, the difference in pressure is converted into kinetic energy of the flowing water. According to Bernoulli's principle for horizontal flow (where gravitational effects can be ignored), this conversion can be expressed as:

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

Here,

$$\rho$$ is the density of water,

$$v$$ is the speed of the water,

$$P_1$$ is the initial pressure when the valve is closed,

$$P_2$$ is the pressure after the valve is opened.

By solving for $$v$$, we have:

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

This equation shows that the speed of the water, $$v$$, is proportional to the square root of the pressure difference:

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

Thus, the correct relationship is given by the square root of the pressure difference.

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