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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

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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 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 ⇒ρAv2=ρAh×a ⇒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.

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