A large tank filled with water to a height `h` is to be emptied through a small hole at the bottom. The ratio of times taken for the level of water to fall from h to `(h)/(2)` and from `(h)/(2)` to zero is

Explanation

Let R be the radius of the tank and r be the radius of the hole. Then velocity of water flowing out oh hole, `upsilon = sqrt(2 gh).` if dh is the fall in height of water in time dt, then `pi R^2 (dh)/(dt) = pi r^2 upsilon = pi r^2 sqrt(2gh) or - (dh)/(sqrth) = (r^2)/(R^2) sqrt(2g) dt`
Required ration, `(t_1)/(t_2) = (int_h^(h//2) (dh)/(sqrth))/(int_(h//2)^0 (dh)/(sqrth)) = ((sqrt(h)/(2) - sqrth))/(0 - sqrt(h)/(2)) = sqrt 2-1`