In the figure shown, a light container is kept on a horizontal rough surface of coefficient of friction μ=ShV . A very small hole of area S is made at depth h. Water of volume V is filled in the container. The friction is not sufficient to keep the container at rest. The acceleration of the container initially, is
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a
VShg
b
G
c
Zero
d
ShVg
answer is D.
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Detailed Solution
Let the density of water be ρ , then the force by escaping liquid on container =F= ρS(2gh)2 .∴ Acceleration of container, a=F-fm=2ρSgh−μρVgρV=(2ShV−μ)gNow,μ=ShV ∴a=ShVg