A large tank is filled with water to a height `H`. A small hole is made at the base of the tank. It takes `T_(1)` time to decrease the height of water to `(H)/(eta) (eta gt 1)`, and it takes `T_(2)` times to take out the rest of water. If `T_(1) = T_(2)`, then the value of `eta` is

Options

A. 2
B. 3
C. 4
D. `2 sqrt(2)`

Answer

C

Explanation

Time in which height of liquid falls from `H_1` to `H_2`
`t prop (sqrt(H_1) - sqrt(H_2))`
`(T_1)/(T_2) = (sqrt H - sqrt((H)/(eta)))/(sqrt((H)/(eta))- 0)`
`1 = (1 - (1)/(eta))/((1)/(sqrt(eta)))`
`(2)/(sqrt(eta)) = 1 rArr eta = 4`.