A cylindrical tank of height `H` is open at the top end and it has a radius `r`. Water is filled in it up to a height of `h`. The time taken to empty the tank through a hole of radius `r'` in its bottom is

Options

A. `sqrt((2h)/(g))(R^(2))/(r )`
B. `sqrt((2H)/(g))(R^(2))/(r^(2)) `
C. `sqrt(h)/(H)`
D. None of these

Answer

D

Explanation

`piR^(2)(dh)/(dt)=pir^(2)sqrt(2gh)implies(dh)/(dt)=-(r^(2))/(R^(2))sqrt(2gh)`
On integrating properly
`t=sqrt((2h)/(g))(R^(2))/(r^(2))`