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Question

There is small hole in the bottom of a fixed container containing a liquid upto ā€²hā€². The top of the liquid as well as the hole at the bottom are exposed to atmosphere. As the liquid comes out of the hole. Then,
(Area of the hole is ā€²aā€² and that of the top surface is ā€²Aā€²) :

A
the top surface of the liquid accelerates with acceleration =g
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B
the top surface of the liquid accelerates with acceleration =ga2A2
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C
the top surface of the liquid retards with retardation =gaA
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D
the top surface of the liquid retards with retardation =ga2A2
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Solution

The correct option is D the top surface of the liquid retards with retardation =ga2A2
The velocity of fluid at the hole is
V2=2gh1+(a2/A2)
Using continuity equation, at the two cross-sections A and B :


V1A=V2a
V1=aAV2
Acceleration of top surface
a1=V1dV1dh=(aA)V2ddh(aAV2)
a1=a2A2V2dV2dh
a1=a2A22gh2g.12ha1=ga2A2

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