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(B) According to Bernoulli's principle for a fluid in motion,the sum of pressure energy,kinetic energy,and potential energy per unit volume remains co

Vedclass · Accepted Answer

$\left[\frac{2(P_{1}-P_{2})}{\rho}\right]^{1/2}$

Vedclass · Answer

(B) According to Bernoulli's principle for a fluid in motion,the sum of pressure energy,kinetic energy,and potential energy per unit volume remains constant.For the fluid inside the pipe,the initial pressure is $P_{1}$ when the velocity is zero.When the valve is opened,the pressure drops to $P_{2}$ and the fluid gains a velocity $v$.Applying Bernoulli's equation between the inside of the pipe and the outlet:$P_{1} + 0 = P_{2} + \frac{1}{2} \rho v^{2}$Rearranging the terms to solve for $v$:$P_{1} - P_{2} = \frac{1}{2} \rho v^{2}$$v^{2} = \frac{2(P_{1} - P_{2})}{\rho}$$v = \sqrt{\frac{2(P_{1} - P_{2})}{\rho}}$Thus,the correct option is $B$.

$A$ closed pipe containing liquid showed a pressure $P_{1}$ by gauge. When the valve is opened,the pressure was reduced to $P_{2}$. The speed of water flowing out of the pipe is $[\rho = \text{density of water}]$

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