Question

A container filled with liquid up to height $h$ is place 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

• 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 1

The correct option is C only $v$ depends on $h$

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 fluid due to change in momentum on the container is $F=\rho A{v}^2$
$F=\rho A{v}^2=ma$, where $m$ is the mass of the liquid inside the container.
$\rho A{v}^2=\rho Ah\times a$
$\Rightarrow a=2g$
Differentiating the above equation, we can have $v=2gt$ where $t$ is the time in which the liquid comes out completely which is dependent on $h$.