1. Interpret the pressure meter readings

The pressure meter gives the gauge pressure in the pipe.

• Before opening valve: water is at rest, so pressure is $P_1 = 4.5 \times 10^4\ \text{Pa}$
• After opening valve: water flows, and pressure becomes $P_2 = 2.0 \times 10^4\ \text{Pa}$

So the pressure drop is $\Delta P = P_1 - P_2 = (4.5 - 2.0)\times 10^4 = 2.5\times 10^4\ \text{Pa}$

2. Apply Bernoulli’s principle

Initially, the water is at rest, so initial speed is zero.

Assuming the height remains the same, Bernoulli’s equation gives: $P_1 + \frac{1}{2}\rho (0)^2 = P_2 + \frac{1}{2}\rho v^2$

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

For water, $\rho = 1000\ \text{kg/m}^3$

Substitute values: $2.5\times 10^4 = \frac{1}{2}(1000)v^2$

$2.5\times 10^4 = 500v^2$

$v^2 = \frac{2.5\times 10^4}{500} = 50$

$v = \sqrt{50}\ \text{m/s}$

3. Compare with given form

The speed is given as $\sqrt{V}\ \text{m/s}$.

So, $V = 50$

4. Comparison with stored answer

Stored correct answer = $50$

Our derived answer also = $50$.

Hence, the answer agrees.