A cylindrical vessel filled with water up to a height h weighs w and is resting on a horizontal plane. The side of wall of the vessel has a plugged circular hole touching the bottom. The coefficient of friction between the bottom of the vessel and the plane is $μ$ . The minimum diameter of the hole so that the vessel just begins to move on the floor after the removal of the plug is

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$\sqrt{\frac{μw}{πρhg}}$

$\sqrt{\frac{2 μw}{πρhg}}$

$\sqrt{\frac{πμw}{ρhg}}$

$\sqrt{\frac{μw}{2 πρhg}}$

answer is D.

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Detailed Solution

velocity of water through the hole: $u = \sqrt{2 gh}$
rate of mass flow: $R = ρAu$ [A = area of the hole]
reaction force on the wall: $F = R (u - 0) = {ρAu}^{2}$
For cylinder to just move,

$μw = {ρAu}^{2} = \frac{{ρπd}^{2}}{4} {(\sqrt{2 gh})}^{2}$
$∴ d = \sqrt{\frac{2 μw}{πρhg}}$

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