A tank full of water has a small hole at the bottom. If one-fourth of the tank is emptied in t1 seconds and the remaining three-fourth of the tank is empited in t2 seconds. Then the ratio t1/t2 is

velocity v₂ of water coming out of hole at the bottom of tank = √(2 g h)  from equation 1 above.  Here we neglect v₁ at the water surface as it is very small as compared to v₂. Let A be the area of cross section of the tank.
 
         Volume of water in tank = V = A h 
         water flow rate out of tank = decrease rate in volume inside tank
                        A v₂  =  d V/dt = - A dh/dt  
                 =>     v₂ = √(2 g h )  =  - dh/dt 

                =>       - dh/√h = √(2g) dt        ------- equation 3
Integrating the expressions on both sides , 

 

Time for water level to go from h₀ to 3 h₀/4,  is

 

Time for water level to go from 3 h₀ /4 to 0  is  

 

Ratio t1/t2 = 2 – sqrt(3) / sqrt(3)