Question

The reading of pressure metre attached with a closed pipe is $4.5 \times 10^4 \, N/m^2$. On opening the valve, water starts flowing and the reading of pressure metre falls to $2.0 \times 10^4 \, N/m^2$. The velocity of water is found to be $\sqrt{V} \, m/s$. The value of $V$ is \\\\\\\_.

Answer 1

Using Bernoulli’s theorem for the flow of an ideal fluid:

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

Given:
- $P_1 = 4.5 \times 10^4 \, \text{N/m}^2$,
- $P_2 = 2.0 \times 10^4 \, \text{N/m}^2$,
- $\rho = 1000 \, \text{kg/m}^3$.

Substituting values:

$4.5 \times 10^4 + \frac{1}{2} \cdot 1000 \cdot v^2 = 2.0 \times 10^4$

Rearranging:

$\frac{1}{2} \cdot 1000 \cdot v^2 = 4.5 \times 10^4 - 2.0 \times 10^4$

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

$v^2 = 50 \implies v = \sqrt{50} \, \text{m/s}.$

Therefore: $v = 50 \, \text{m/s}.$

The Correct answer is: 50 m/s