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

Water flows in a horizontal pipe. When valve is closed, pressure is $$P_1$$. When opened, pressure falls to $$P_2$$. We need to find the proportionality of speed.

Apply Bernoulli's equation:

When the valve is closed, velocity = 0, pressure = $$P_1$$.

When the valve is opened, velocity = v, pressure = $$P_2$$.

$$P_1 + \frac{1}{2}\rho(0)^2 = P_2 + \frac{1}{2}\rho v^2$$

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

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

Therefore, $$v \propto \sqrt{P_1 - P_2}$$.

The correct answer is Option 4: $$\sqrt{P_1 - P_2}$$.