A liquid of density 800 kg/m³ is filled in a cylindrical vessel upto a height of 3 m. This cylindrical vessel stands on a horizontal plane. There is a circular hole on the side of the vessel. What should be the minimum diameter of the hole to move the vessel on the
floor, if plug is removed. Take the coefficient of friction between the bottom of the vessel and the plane as 0.5 and total mass of vessel plus vessel as 95 kg.

0.107 m

0.053 m

0.206 m

0.535 m

Solution

0.107 m

Given:-

ρ = 800 kg/m³

h = 3 m

μ = 0.5

m = 95 kg

d_min = ?

Let area of hole be a

∴ Reaction force F = ρ av²

= ρa. 2gh [ $∵$ v = $\sqrt{2 g h}$ ]

and f_max = μN

= μ mg

μ= coefficient of friction

⇒ F ≥ f_max

⇒ 2 ρ agh ≥ μ mg

⇒ a ≥ $\frac{μ m}{2 ρ h}$

= $\frac{0.5 \times 95}{2 \times 800 \times 3}$

a = 0.009

$π$ r² ≥ 0.009

r ≥ $\sqrt{\frac{0.009}{π}}$

r_min ≥ 0.0535 m

d_min = 2 r_min

= 2 × 0.0535

d_min = 0.107 m

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