A light cylindrical vessel of radius r is kept on a rough horizontal surface with sufficient friction so that it cannot slide but can topple. It is filled with water up to a height 2h and a small hole of area a is punched in it so that the water coming out of it falls at the maximum distance from its wall along horizontal surface. Water comes out horizontally from the hole. The value of h for which the cylinder topples is (are)
The Attempt at a Solution
Using Bernoulli's principle ,velocity of the fluid coming out of the hole v= √(2gy) where y be the distance of the hole from the top .
Now since water coming out of it falls at the maximum distance from the cylinder wall along horizontal surface ,we need to maximize the distance S = (√(2gy))(√(2(2h-y)/g) .
From this we get y=h
Water coming out of the hole gives an impulse Fdt = (dm)v to the rest of the fluid ,where dm = ρA(dx)
Fdt = ρA(dx)v
or F = ρAv2
If the block has to topple about the rightmost point P , then there has to be a net torque about that point.
i.e ρAv2h - ρπr2(2h)gr ≥0
But this is none of the given options .
I am not sure if my reasoning and approach is correct .
I would be grateful if somebody could help me with the problem .