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What work should be done in order to squeeze all water from a horizontally located cylinder during the time $t$ by means of a constant force acting on the piston ? The volume of water in the cylinder is equal to $V$ , the cross–sectional area of the orifice is $s$ , with $s$ being considerably less than the piston area. The friction and viscosity are negligibly small.

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$\frac{ρ V^{2}}{2 s^{2} t^{2}}$

$\frac{ρ V^{3}}{2 s^{2} t^{2}}$

$\frac{3 ρ V^{3}}{4 s^{2} t^{2}}$

$\frac{2 ρ V^{2}}{3 s^{2} t^{2}}$

answer is B.

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Detailed Solution

Using Bernoulli's equation between $A$ and $B$ , we have

$P = P_{a} + \frac{1}{2} ρ v^{2} ∴ P - P_{a} = \frac{1}{2} ρ v^{2}$

The net force on the piston, $F = (P - P_{a}) A$
Thus work done is given by,

$W = (P - P_{0}) ∆ V = \frac{1}{2} ρ v^{2} \times (V - 0) = \frac{1}{2} ρ v^{2} V G i v e n V = v s t ∴ W = \frac{1}{2} ρ (\frac{V^{3}}{s^{2} t^{2}})$

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