A container filled with liquid up to height `h` is placed on a smooth horizontal surface. The 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, it acquires a velocity `v`. Neglect mass of the container. In this case

Options

A. both `a` and `v` depend on `h`
B. only `a` depends on `h`
C. only `v` depends on `h`
D. neither `a` nor `v` depends on `h`

Answer

C

Explanation

The velocity with which the liquid comes out is `v_(0)=sqrt(2gh)`
Let cross sectional area of the hole be `A`. Then the force exerted by the ejecting find due to change in momentum on the container is `F=rhoAv^(2)`.
`F=rhoAv^(2)=ma`, where `m` is the mass of the liquid inside the container
`rho Av^(2)=rhoAhxxa`
`a=2g`
Differentiating the above equation, we can have `v=2"gt"` where `t` is the time in which the liquid comes out completely which is dependent on `h`.