Question

A large open top container of negligible mass and uniform cross sectional area "A" has a small hole of cross sectional area "a" in its side wall near the bottom.The container is kept over a smooth horizontal floor and contains a liquid of density "ρ" and mass "m".Assuming that liquid starts flowing through the hole A,the acceleration of the container will be

Answer 1

volume of water level = m/ρ
so
volume of water level = Area × height
m/ρ = A × h
height of water level = m/ρA
velocity of water at bottom = root 2gh
= root 2gm/ρA
so force on container = ρav^2
= ρa×2gm/ ρA
=2agm/A
so
acceleration of container = force/mass
=2agm/Am
=2ag/A ..ans