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 `h/2` to zero is

Options

A. sqrt2
B. `sqrt(2)`
C. `1/sqrt(2)`
D. `1/(sqrt(2-1))`

Answer

C

Explanation

Time taken to reach half level
i.e., from h to `h//2` is,
`t_(1) = sqrt(2/g)[sqrt(h_(1)) -sqrt(h_(2))] = sqrt(2/g) [ sqrt(h) - sqrt(h//2))]`
`therefore t_(1) = sqrt((2h)/g) [ 1 - 1/sqrt(2)]`
Time taken to empty is from `h//2` to 0
`t_(2) = sqrt(2/g) [ sqrt(h/2 -0)]`
`therefore t_(2) = sqrt((2h)/g) 1/sqrt(2)`
`therefore t_(1)/t_(2) = sqrt(2) - 1`