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

Correct Answer: 50

To solve this problem, we use Bernoulli's equation for fluid flow. The principle states that in a steady flow, the sum of all forms of mechanical energy in a fluid along a streamline is constant. The equation in terms of pressure \( P \), fluid density \( \rho \), velocity \( v \), and gravitational potential energy is given by: 

\( P_1 + \frac{1}{2}\rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho gh_2 \)

In our scenario, when the valve is closed, the water is at rest so \( v_1 = 0 \). Also, assuming the heights \( h_1 \) and \( h_2 \) are the same, they cancel out. Therefore, the equation simplifies to:

\( P_1 = P_2 + \frac{1}{2}\rho v_2^2 \)

We are given:

• \( P_1 = 4.5 \times 10^4 \, N/m^2 \)
• \( P_2 = 2.0 \times 10^4 \, N/m^2 \)

Substituting these into the simplified equation:

\( 4.5 \times 10^4 = 2.0 \times 10^4 + \frac{1}{2}\rho v_2^2 \)

Solving for \( v_2^2 \):

\( \frac{1}{2}\rho v_2^2 = 4.5 \times 10^4 - 2.0 \times 10^4 = 2.5 \times 10^4 \)

Assuming \( \rho = 1000 \, kg/m^3 \) (density of water), the equation becomes:

\( \frac{1}{2} \times 1000 \times v_2^2 = 2.5 \times 10^4 \)

Simplifying gives:

\( 500 v_2^2 = 2.5 \times 10^4 \)

\( v_2^2 = \frac{2.5 \times 10^4}{500} = 50 \)

Therefore, \( v_2 = \sqrt{50} \, m/s \). The value of \( V \) is 50, which fits the expected range (50, 50).

Answer 2

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