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. `2sqrt(2)`

Answer

C

Explanation

Apply result from Q. 16 of Ex-II
`t=(A)/(A_(0))sqrt((2)/(g))lfloor(sqrt(H)-sqrt(x))rfloorimpliesT_(1)=(A)/(A_(0))sqrt((2)/(g))lfloor(sqrt(H)-sqrt(H//eta))rfloor" "…(1)`
`T_(2)=(A)/(A_(0))sqrt((2)/(g))lfloorsqrt(H//eta)-sqrt(0)rfloor" "...(2)`
From (1) and (2) `eta=4`