A tank full of water has a small hole at the bottom. If one-fourth of the tank is emptied in `t_(1)` seconds and the remaining three-fourths of the tank is emptied in `t_(2)` seconds. Then the ratio `(t_(1))/(t_(2))` is

Options

A. `sqrt3`
B. `sqrt2`
C. `(2- sqrt2)/(sqrt2)`
D. `(2 - sqrt3)/(sqrt3)`

Answer

D

Explanation

Let A and a be cross-sectional areas of the tank and hole respectively. Let h be height of water in the tank at time t. If `(-(dh)/(dt))` represent the rate of fall of level, then `-A (dh)/(dt) = a sqrt(2gh) or dt = -(A)/(a sqrt(2g)) (dh)/(sqrth)`
`:.` Required ratio `(t_(1))/(t_(2)) = (underset(h)overset(3h//4)int (dh)/(sqrth))/(underset(3h//4)overset(0)int (dh)/(sqrth)) = ([sqrt((3h)/(4)) - sqrth])/([0- sqrt((3h)/(4))]) = (2- sqrt3)/(sqrt3)`