What work should be done in order to squeeze all water from a horizontally located cylinder (figure) 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 to s, with s being considerably less than the piston area. The friction and viscosity are negligibly small.

Explanation

Let the force acting on the piston be F and the length of the cylinder be l.
Then, work done `=Fl` (1)
Applying Bernoulli's theorem for points A and B, `p=1/2rhov^2` where `rho` is the density and v is the velocity at point B. Now, force on the piston,
`F=pA=1/2rhov^2A` (2)
where A is the cross section area of piston. Also, discharge through the orifice during time interval `t=Svt` and this is equal to the volume of the cylinder, i.e.,
`V=Svt` or `v=(V)/(St)` (3)
From Eq. (1), (2) and (3) work done
`=1/2rhov^2Al=1/2rhoA(V^2)/((St)^2)l=1/2rhoV^3//S^3t^2` (as `Al=V`)