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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
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B
only a depends on h
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C
only v depends on h
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D
neither a nor v depends on h
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Solution

## The correct option is C only v depends on hThe velocity with which the liquid comes out is v0=√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=ρAv2F=ρAv2=ma, where m is the mass of the liquid inside the container.ρAv2=ρAh×a⇒a=2gDifferentiating 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.

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