Search any question & find its solution
Question:
Answered & Verified by Expert
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 $]$
Options:
opened, pressure was reduced to ' $\mathrm{P}_{2}$ '. The speed of water flowing out of the pipe
is $[\rho=$ density of water $]$
Solution:
2136 Upvotes
Verified Answer
The correct answer is:
$\left[\frac{2\left(\mathrm{P}_{1}-\mathrm{P}_{2}\right)}{\rho}\right]^{1 / 2}$
(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,
$\begin{aligned}
& P_{1}=P_{2}+\frac{1}{2} \rho v^{2} \\
\therefore & P_{1}-P_{2}=\frac{1}{2} \rho v^{2} \\
\therefore \quad v &=\sqrt{\frac{2\left(P_{1}-P_{2}\right)}{\rho}}
\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.