What work should be done in order to sqeeze all water from a horizontally located cylinder (figure) during the time t by means of a constant force acting on the poston? The volume of water in the cylinder is equal to V, the cross-sectional area of the orifice is s, with s being considerablu less than the piston area. The friction and viscosity are negligibly small. Density of water is `rho`

Explanation

`(F)/(A)=(1)/(2)rhov^(2)`
or, `F=(1)/(2) rho Av^(2)` …(i)
Here, v is the velocity of liquid, with which it comes out of the hole.

Futher `V=Ax` …(ii)
`t=(V)/(sv)` …(iii)
and `w=F:x` …(iv)
From the above four equations,
`w=((1)/(2) Arhov^(2))((V)/(A))`
`=(1)/(2) rho (V^(2))/(s^(2)t^(2)).V=(1)/(2) (rho v^(3))/(s^(2)t^(2))`