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Question

# The figure shows a light container kept on a horizontal rough surface of coefficient of friction μ=Shv ('v' is the volume of the container). A very small hole of area S is made at a depth h. Water of volume V is filled in the container. The friction is not sufficient to keep the container at rest. The initial acceleration of the container is,

A
VgSh
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B
g
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C
zero
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D
Shgv
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Solution

## The correct option is D ShgvVefflux=√2gh Force exerted by the liquid on the container ρ s(√2gh)2 Acceleration of the container, a=2ρSgh−μρvgρv = (2Shv−μ)g = (2Shv−Shv)g = Shgv

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