A closed pipe containing a liquid showed a pressure $P_1$ by gauge. When the valve was opened, pressure was…

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 )
  1. $\left[\frac{4\left(P_1-P_2\right)}{\rho}\right]^{1 / 2}$
  2. $\left[\frac{4\left(P_2-P_1\right)}{\rho}\right]^{1 / 2}$
  3. $\left[\frac{2\left(P_1-P_2\right)}{\rho}\right]^{1 / 2}$
  4. $\left[\frac{2\left(\mathrm{P}_2-\mathrm{P}_1\right)}{\rho}\right]^{1 / 2}$

Solution

According to Bernoulli's equation, $\begin{array}{ll} & P_1+\frac{1}{2} \rho v_1^2=P_2+\frac{1}{2} \rho v_2^2 \\ \therefore \quad & v_2^2=\frac{2\left(P_1-P_2\right)}{\rho}, \quad\left[\because v_1=0\right] \\ & v_2=\sqrt{\frac{2\left(P_1-P_2\right)}{\rho}} \end{array}$ .

Asked in: MHT CET 2024 (10 May Shift 1)

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