MHT CET · Physics · Mechanical Properties of Fluids
A closed pipe containing a liquid showed a pressure \(P_1\) by gauge. When the valve was opened, pressure was reduced to \(P_2\). The speed of water flowing out of the pipe is ( \(\rho=\) density of water )
- A \(\left[\frac{4\left(P_1-P_2\right)}{\rho}\right]^{1 / 2}\)
- B \(\left[\frac{4\left(P_2-P_1\right)}{\rho}\right]^{1 / 2}\)
- C \(\left[\frac{2\left(P_1-P_2\right)}{\rho}\right]^{1 / 2}\)
- D \(\left[\frac{2\left(\mathrm{P}_2-\mathrm{P}_1\right)}{\rho}\right]^{1 / 2}\)
Answer & Solution
Correct Answer
(C) \(\left[\frac{2\left(P_1-P_2\right)}{\rho}\right]^{1 / 2}\)
Step-by-step Solution
Detailed explanation
According to Bernoulli's equation,
\(\begin{array}{ll}
& P_1+\frac{1}{2} \rho v_1^2=P_2+\frac{1}{2} \rho v_2^2 \\
\therefore \quad & v_2^2=\frac{2\left(P_1-P_2\right)}{\rho}, \quad\left[\because v_1=0\right] \\
& v_2=\sqrt{\frac{2\left(P_1-P_2\right)}{\rho}}
\end{array}\)
\(\begin{array}{ll}
& P_1+\frac{1}{2} \rho v_1^2=P_2+\frac{1}{2} \rho v_2^2 \\
\therefore \quad & v_2^2=\frac{2\left(P_1-P_2\right)}{\rho}, \quad\left[\because v_1=0\right] \\
& v_2=\sqrt{\frac{2\left(P_1-P_2\right)}{\rho}}
\end{array}\)
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