A cylindrical vessel contains a liquid of density $ρ$ up to a height $h$ . The liquid is closed by a piston of mass m and area of cross-section $A$ . There is a small hole at the bottom of the vessel. The speed $v$ with which the liquid comes out of the hole is

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TestHub · Accepted Answer

Applying Bernoulli's theorem at 1 and 2 : difference in pressure energy between 1 and $2 =$ difference in kinetic energy between 1 and 2

or $ρ g h + \frac{m g}{A} = \frac{1}{2} ρ v^{2}$

or $v = \sqrt{2 g h + \frac{2 m g}{ρ A}} = \sqrt{2 (g h + \frac{m g}{ρ A})}$

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