A horizontal cylinder of length `l` contains water. On the left a constant force `F` is applied on the piston and on the right there is a small hole of area a. What is the velocity of the piston? Hence, calculate the work done in emptying the vessel in time `t`. brgt given area of cylinder is A

Explanation

Let `A` be the area of the piston and `v_(1)` be its velocity. Then
`(F/A+P_(0))+1/2rhov_(1)^(2)=P_(0)+1/2rhov_(2)^(2), Av_(1)=av_(2)`
`impliesF/A=1/2rhov_(1)^()2[(A^(2))/(a^(2))-1]=(rhov_(1)^(2)A^(2))/(a^(2))((A^(2))/(a^(2))-1)`

Net work done is `Fxx` distance `=Fl=(2v_(1)^(2)A^(2)t)/(2a^(2))`