A problem from fluids

Let " S " denote the cross - sectional area of the basin and " s " denote the area of the orifice , through which the plug at the bottom drains the sink . Also , let " H " be the height of the basin .

Now , the time " T₂ " can be given in terms of these quantities , as -

T₂ = Ss √ 2 H√ g ............ ( 1 )

Again , the basin will overflow if and only if the volume rate of flow of water into the basin becomes greater than the volume rate of flow of water out of the basin .

So , our condition will be fulfilled if -

dV₁dT > dV₂dT

Or , ΔV₁ΔT > ΔV₂ΔT

Or , S HT₁ > s √ 2 g H

Or , Ss √ 2 H√ g > 2 T₁

Or , From ( 1 ) , ......... T₂ > 2 T₁

Hence the said ratio must be strictly less than " 12 " .
Note : - Obviously , the area of the " orifice " must be very small ( negligible ) as compared to the area of cross section .

How did you get T² with S and s terms?
I got it as T₂=(√2H)/√g

Getting same answer ..t1/t2 < 1/2