A cylindrical vessel contains a liquid of density `rho` up to 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

Explanation

Applying Bernoulli's theorem at 1 and 2, difference in pressure energy between 1 and 2 `=` difference in kinetic energy between 1 and 2
`(p_(0)+rhogh+(mg)/(A))+0=(p_(0))+(1)/(2)rhov^(2)`
`p_(0)+rhogh+(mg)/(A)-p_(0)=(1)/(2)rhov^(2)`
`rhogh+(mg)/(A)=(1)/(2)rhov^(2)implies v=sqrt(2gh+(2mg)/(rhoA))=sqrt(2(gh+(mg)/(rhoA)))`