MHT CET · Physics · Mechanical Properties of Fluids
A closed pipe containing liquid showed a pressure ' \(\mathrm{P}_{1}\) ' by guage. When the valve is
opened, pressure was reduced to ' \(\mathrm{P}_{2}\) '. The speed of water flowing out of the pipe
is \([\rho=\) density of water \(]\)
- A \(\left[\frac{2\left(\mathrm{P}_{1}+\mathrm{P}_{2}\right)}{\rho}\right]^{1 / 2}\)
- B \(\left[\frac{2\left(\mathrm{P}_{1}-\mathrm{P}_{2}\right)}{\rho}\right]^{1 / 2}\)
- C \(\left[\frac{\rho}{2\left(\mathrm{P}_{1}-\mathrm{P}_{2}\right)}\right]^{1 / 2}\)
- D \(\left[\frac{\rho}{2\left(\mathrm{P}_{1}+\mathrm{P}_{2}\right)}\right]^{1 / 2}\)
Answer & Solution
Correct Answer
(B) \(\left[\frac{2\left(\mathrm{P}_{1}-\mathrm{P}_{2}\right)}{\rho}\right]^{1 / 2}\)
Step-by-step Solution
Detailed explanation
(B)
By Bernoulli equation,
\(\begin{aligned}
& P_{1}=P_{2}+\frac{1}{2} \rho v^{2} \\
\therefore & P_{1}-P_{2}=\frac{1}{2} \rho v^{2} \\
\therefore \quad v &=\sqrt{\frac{2\left(P_{1}-P_{2}\right)}{\rho}}
\end{aligned}\)
By Bernoulli equation,
\(\begin{aligned}
& P_{1}=P_{2}+\frac{1}{2} \rho v^{2} \\
\therefore & P_{1}-P_{2}=\frac{1}{2} \rho v^{2} \\
\therefore \quad v &=\sqrt{\frac{2\left(P_{1}-P_{2}\right)}{\rho}}
\end{aligned}\)
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