1Again , as I later noticed , the " viscosity of the liquid " has to be neglected as well if my solution is correct .
Consider a rectangular basin with a very wide mouth. When a tap above the basin is opened, it fills in time T1. With the basin filled completely and the tap closed, opening the plug at the bottom drains the sink in time T2. With the tap and the plug both opened and starting with an empty basin, for what ratios T1T2 will the basin ultimately overflow?
(From 200 Challenging Problems in Physics)
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4 Answers
1Let " 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 .
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