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`.)

Options

A. `sqrt((2muM)/(pirhoH))`
B. `sqrt((muM)/(2pirhoH))`
C. `sqrt((muM)/(rhoH))`
D. none

Answer

A

Explanation

Force on vessel = rate of change of linear momentum imparted to flow water through hole.
Force on vessel = mv (where m is coming out per second)
(v = volume coming out per second)
`F=mv" "F=rhoVvimplies F=rho.av.vimplies F=rhoav^(2)`
`muMglerho.(pid^(2))/(4)v^(2)["put v" = sqrt(2gh)]" "implies muMgle(rhopid^(2))/(4).2ghimpliesd_(min)=sqrt((2muM)/(pirhoH))`