Question

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 -

• A. Both $a$ and $v$ depends 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 CSA of hole is $A$ then the force exerted by the ejecting fluid due to change in momentum on container is $F=\rho A{v}^2$

$\Rightarrow F=\rho A{v}^2=ma$ where $m$ is the mass of liquid inside the container

$\Rightarrow \rho A{v}^2=\rho Ah\times a$

$\Rightarrow a=2g$

Differentiating above equation we can have $v=2gt$ where $t$ is the time in which liquid comes out
completely which is depending on $h$.