A closed pipe containing liquid showed a pressure ' $\mathrm{P}_{1}$ ' by guage. When the valve is opened,…

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 $]$
  1. $\left[\frac{2\left(\mathrm{P}_{1}+\mathrm{P}_{2}\right)}{\rho}\right]^{1 / 2}$
  2. $\left[\frac{2\left(\mathrm{P}_{1}-\mathrm{P}_{2}\right)}{\rho}\right]^{1 / 2}$
  3. $\left[\frac{\rho}{2\left(\mathrm{P}_{1}-\mathrm{P}_{2}\right)}\right]^{1 / 2}$
  4. $\left[\frac{\rho}{2\left(\mathrm{P}_{1}+\mathrm{P}_{2}\right)}\right]^{1 / 2}$

Solution

By Bernoulli equation, $\begin{aligned} & P_{1}=P_{2}+\frac{1}{2} \rho v^{2} \\ \therefore & P_{1}-P_{2}=\frac{1}{2} \rho v^{2} \\ \therefore \quad v &=\sqrt{\frac{2\left(P_{1}-P_{2}\right)}{\rho}} \end{aligned}$ .

Asked in: MHT CET 2020 (19 Oct Shift 1)

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