As shown in the figure, a pump is designed as | vedclass

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(C) According to the equation of continuity, the volume flow rate is constant:$AV = av \implies V = \frac{a}{A}v$where $V$ is the velocity of the pist

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$\sqrt{\frac{2 F}{\rho A}}$

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(C) According to the equation of continuity, the volume flow rate is constant:$AV = av \implies V = \frac{a}{A}v$where $V$ is the velocity of the piston and $v$ is the velocity of the liquid emerging from the orifice.Applying Bernoulli's principle between the inside of the cylinder (near the piston) and the orifice:$P_{in} + \frac{1}{2} \rho V^2 = P_{out} + \frac{1}{2} \rho v^2$The pressure inside the cylinder is $P_{in} = P_0 + \frac{F}{A}$, where $P_0$ is the atmospheric pressure. The pressure at the orifice is $P_{out} = P_0$.Substituting these into the Bernoulli equation:$(P_0 + \frac{F}{A}) + \frac{1}{2} \rho V^2 = P_0 + \frac{1}{2} \rho v^2$$\frac{F}{A} = \frac{1}{2} \rho (v^2 - V^2)$Substituting $V = \frac{a}{A}v$:$\frac{F}{A} = \frac{1}{2} \rho (v^2 - (\frac{a}{A}v)^2) = \frac{1}{2} \rho v^2 (1 - \frac{a^2}{A^2})$Since $A \gg a$, we have $\frac{a^2}{A^2} \approx 0$, so:$\frac{F}{A} = \frac{1}{2} \rho v^2 \implies v = \sqrt{\frac{2F}{\rho A}}$

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