Water flows in a horizontal pipe whose one end is closed with a valve. The reading of the pressure gauge attached to the pipe is \( P_1 \). The reading of the pressure gauge falls to \( P_2 \) when the valve is opened. The speed of water flowing in the pipe is proportional to:
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- \( \sqrt{P_1 - P_2} \)
- \( (P_1 - P_2)^2 \)
- \( (P_1 - P_2)^4 \)
- \( P_1 - P_2 \)
The Correct Option is A
Approach Solution - 1
To solve the problem of finding the relationship between the speed of water flowing in a pipe and the pressure difference, we can apply Bernoulli's principle. Bernoulli's principle states that in a streamline flow of an ideal fluid, the sum of the pressure energy, kinetic energy, and potential energy per unit volume is a constant.
The general form of Bernoulli's equation for a horizontal flow (neglecting height-related potential energy difference) is:
\(P + \frac{1}{2} \rho v^2 = \text{constant}\)
Where:
- \(P\) is the pressure exerted by the fluid.
- \(\rho\) is the density of the fluid.
- \(v\) is the velocity of the fluid.
Now, when the valve is closed, the pressure is \(P_1\) and the velocity is zero because the fluid is at rest. Thus, the Bernoulli equation becomes:
\(P_1 = \text{constant}\)
When the valve is opened, the pressure drops to \(P_2\) and the water starts to flow with velocity \(v\). Applying Bernoulli's equation, we get:
\(P_2 + \frac{1}{2} \rho v^2 = \text{constant}\)
Equating the two equations, we have:
\(P_1 = P_2 + \frac{1}{2} \rho v^2\)
Simplifying for the velocity \(v\), we get:
\(\frac{1}{2} \rho v^2 = P_1 - P_2\)
\(v^2 = \frac{2 (P_1 - P_2)}{\rho}\)
\(v = \sqrt{\frac{2 (P_1 - P_2)}{\rho}}\)
This shows that the velocity \(v\) is proportional to \(\sqrt{P_1 - P_2}\). Therefore, the correct answer is \(\sqrt{P_1 - P_2}\).
Approach Solution -2
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