The vessel of area of cross-section A has liquid to a height H. There is a hole at the bottom of vessel having area of cross-section a. The time taken to decrease the level from  H1 to H2 will be

The average velocity of effluxv=2gH1+2gH22 let t be the time taken to empty the tank from level H1 to H2. Then2gH1+2gH22×a×t=AH1−H2 or t=Aa×2g×H1−H2H1+H2 =Aa×2g×H1−H2×H1−H2H1+H2H1−H2 =Aa×2g×H1−H2 Alternate method. Let I be the height of the water level at any instant. Now, the rate of decrease of water level is-dh/ dt. Therefore,−Adhdt=av=a(2gh) or −dhdt=aA(2gh) Integrating this expression, we get −∫H1H2 dhh=aA(2g)∫0t dt −2[h]H1H2=aA(2g)⋅t 2H1−H2=aA(2g)⋅t ∴ t=Aa2gH1−H2