A container of cross-section area ‘S’ and height ‘h’ is filled with mercury upto the brim. Then the container is sealed airtight and a hole of small cross-section area ‘S/n’ (where ‘n’ is a positive constant) is punched in its bottom. Find out the time interval upto which the mercury will come out from the bottom hole. [Take the atmospheric [pressure to be equal to $h_0$  height of mercury column :  $h>h_0$ ]

Let the velocity of efflux of mercury coming out of hole be v at an instant mercury level in container is y. At same instant the speed of top surface of fluid is v.
 
From the equation of continuity
 $$\begin{gathered} \frac{S}{n}v=Sv \\ \because \frac{S}{n}<<S\left(i\right) \\ \therefore v>>\upsilon \end{gathered}$$
 
Applying Bernoullis theorem between A and B
 $\therefore v=\sqrt{2g(Y-h_0)}\left(ii\right)$
Hence mercury flows out of both till  $y=h_0$.
From eq. (i)
$\upsilon =-\frac{dy}{dt}=\frac{1}{h}v=\frac{1}{h}\sqrt{2g(Y-h_0)}$    
 Or   $\frac{-dy}{\sqrt{2gY-h_0}}=\frac{1}{h}dt$   
Integrating between limits

At     l = 0  y=n
And  t=T,       Y=  
 $$\begin{gathered} -\underset{n}{\overset{h_0}{\int }}\frac{dy}{\sqrt{2g(Y-h_0)}}=\frac{1}{n}\underset{0}{\overset{T}{\int }}dt \\ t=n\sqrt{\frac{2}{g}(h-h_0)} \end{gathered}$$