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

Options

A. `sqrt(2)`
B. `(1)/sqrt(2)`
C. `sqrt(2)-1`
D. `(1)/(sqrt(2)-1)`

Answer

C

Explanation

Time taken for the level to fall from H to H' `t=(A)/(A_(0)) sqrt((2)/(g))[sqrt(H)-sqrt(H^('))]`
According to problem- the time taken for the level to fall from h to `(h)/(2) t_(1)=(A)/(A_(0))sqrt((2)/(g))[sqrt(h)-sqrt((h)/(2))]`
and similarly time taken for the level to fall from `(h)/(2)` to zero
`t_(2) = (A)/(A_(0))sqrt((2)/(g))[sqrt((h)/(2))-0]`
`therefore (t_(1))/(t_(2)) = (1-(1)/sqrt(2))/((1)/(sqrt(2))-0)=sqrt(2)-1`.