The figure shows a light container kept on a horizontal rough surface of coefficient of friction $μ = \frac{S h}{v}$ ('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,

\frac{V g}{S h}

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\frac{S h g}{v}

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Solution

The correct option is D $\frac{S h g}{v}$
$V_{efflux} = \sqrt{2 g h}$

Force exerted by the liquid on the container $ρ s {(\sqrt{2 g h})}^{2}$

Acceleration of the container, $a = \frac{2 ρ S g h - μ ρ v g}{ρ v}$

= $(\frac{2 S h}{v} - μ) g$

= $(\frac{2 S h}{v} - \frac{S h}{v}) g$

= $\frac{S h g}{v}$

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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