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A light cylindrical vessel is kept on a horizontal surface. Area of base is A. A hole of cross sectional area ‘a’ is made just at its bottom side. The minimum coefficient of friction necessary to prevent sliding the vessel due to the impact force of the emerging liquid is (a<<A)

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$\frac{A}{2 a}$

$\frac{2 a}{A}$

$\frac{2 A}{a}$

$\frac{a}{A}$

answer is C.

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Detailed Solution

$\frac{d v}{d t} = a v$ , $\frac{d m}{d t} = a ρ v$

Force on container $F = \frac{d p}{d t} = v \frac{d m}{d t}$

$F = v a ρ v = a ρ v^{2}$

This force must be equal to frictional force for the container not moving

$μ m g = a ρ v^{2} (v = \sqrt{2 g h})$

$μ p A h g = a ρ 2 g h$

$μ = \frac{2 a}{A}$

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