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 _____.
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 _____.
Correct Answer: 50
Approach Solution - 1
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).
Approach Solution -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
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