QJEEM25Ph2f68: JEE Main 2025 Physics - Class 11 JEEM25Ph2f68 Solution: Water flows in a horizontal pipe whose one end is closed with a …

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.