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Question

# 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

A
VShg
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B
g
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C
zero
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D
SghV
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Solution

## The correct option is D SghVGiven, Friction coefficient, μ=ShV Area of hole is S Velocity of liquid, v=√2gh Force, F=dPdt=rateofflow×velocity=(ρSvdtdt)v=ρSv2=ρS(2gh) Friction force, Fr=μMg=ShV(ρV)g=ρshg Net force = Fluid force – friction force Fnet=2ρSgh−ρSgh=ρSgh Acceleration of container, a′=FnetM=ρSghρV=SghV heance acceleration of container is SghV.

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