Interpret the two pressure readings

• When the valve is closed, water is at rest in the horizontal pipe.

• So the pressure gauge reads the static pressure: $P_1$

• When the valve is opened, water flows with speed $v$.

• The pressure gauge now reads the dynamic flowing pressure: $P_2$

Apply Bernoulli’s principle

For a horizontal pipe, the height term is the same everywhere, so Bernoulli’s equation gives $P + \frac{1}{2}\rho v^2 = \text{constant}$

Comparing the condition when fluid is at rest and when it is flowing, $P_1 = P_2 + \frac{1}{2}\rho v^2$

Find the relation for speed

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

Hence, $v^2 \propto P_1 - P_2$ $v \propto \sqrt{P_1 - P_2}$

Match with the options

The speed is proportional to $\sqrt{P_1-P_2}$

So the correct option is B.