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(A) According to Bernoulli's principle for streamline flow in a horizontal tube, the sum of static pressure and dynamic pressure remains constant at a

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$\sqrt{v^{2}+\frac{p}{\rho}}$

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(A) According to Bernoulli's principle for streamline flow in a horizontal tube, the sum of static pressure and dynamic pressure remains constant at all points along a streamline.$p + \frac{1}{2} \rho v^{2} = p_{1} + \frac{1}{2} \rho v_{1}^{2}$Given that at the second point, $p_{1} = p/2$.Substituting this into the equation:$p + \frac{1}{2} \rho v^{2} = \frac{p}{2} + \frac{1}{2} \rho v_{1}^{2}$Rearranging the terms to solve for $v_{1}^{2}$:$\frac{1}{2} \rho v_{1}^{2} = p - \frac{p}{2} + \frac{1}{2} \rho v^{2}$$\frac{1}{2} \rho v_{1}^{2} = \frac{p}{2} + \frac{1}{2} \rho v^{2}$Multiplying both sides by $2/\rho$:$v_{1}^{2} = \frac{p}{\rho} + v^{2}$$v_{1} = \sqrt{v^{2} + \frac{p}{\rho}}$

Water is flowing in streamline motion through a horizontal tube. The pressure at a point in the tube is $p$ where the velocity of flow is $v$. At another point, where the pressure is $p/2$, the velocity of flow is (density of water $= \rho$)

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