A cylindrical vessel filled with water upto a height of `2m` stands on horizontal plane. The side wall of the vessel has a pugged circular hole touching the bottom. If the minimum diameter of the hole so that the vessel begins to move on the floor if the plug is removed is `(x)/(10sqrt(pi))` meter then `x` will be (if the coefficient of frication between the bottom of the vessel and the plane is `0.4` and total mass of water plus vessel is `100 kg`.)

Explanation

`a` : area of hole
Reaction force `F = rho av^2 = rhoa. 2gh`
`f_(max) = mu N = mu mg`
`F ge f_(max)`
`2 rho a g h ge mu mg`
`a ge (mu m)/(2 rho h) = (0.4 xx 100)/(2 xx 1000 xx 2) = 0.01`
`pi r^2 ge 0.01`
`r ge sqrt((0.01)/(pi))`
`r_(min) = (0.01)/(sqrt(pi)) = 0.0564 m`
`d_(min) = 2 r_(min) = 2 xx 0.0564 m`
=`0.113 m`.
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