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)` time 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. `2sqrt2`

Answer

C

Explanation

`t=(A)/(A_0)sqrt(2/g)[sqrt(H_1)-sqrt(H_2)]`
`(T_1)/(T_2)=((A)/(A_0)sqrt(2/g)[sqrt(H)-sqrt(H/eta)])/((A)/(A_0)sqrt(2/g)[sqrt(H/eta)-0])`
`implies sqrt(H/eta)=sqrt(H)-sqrt(H/eta)`
`implies 4 H/eta = H implies eta=4`.