A container filled with liquid upto height h is placed on a smooth horizontal surface. Container is having a small hole at the bottom. As the liquid comes out from the hole, the container moves in a backward direction with acceleration 'a' and finally, when all the liquid is drained out acquires a velocity v . Neglect mass of container. In this case -
Only v depends on h
The velocity with which the liquid comes out is v0=√2gh
Let CSA of hole is A then the force exerted by the ejecting fluid due to change in momentum on container is F=ρAv2
⇒F=ρAv2=ma where m is the mass of liquid inside the container
Differentiating above equation we can have v=2gt where t is the time in which liquid comes out
completely which is depending on h.