A swimmer crosses a flowing stream of width `omega` to and fro in time `t_(1)`. The time taken to cover the same distance up and down the stream is `t_(2)`. If `t_(3)` is the time the swimmer would take to swim a distance `2omega` in still water, then

Options

A. `T_(1) = T_(2).T_(3)`
B. `T_(1)^(2) = T_(2).T_(3)`
C. `T_(2)^(2) = T_(1).T_(3)`
D. `T_(2)^(3) = T_(1).T_(2)`

Answer

B

Explanation

Let `v` and `u` be speeds of swimmer and river respectively.
`T_(1) = (2b)/(sqrt(v^(2) - u^(2))) , T_(2) = (b)/(v - u) + (b)/(v + u) = (2bv)/(v^(2) - u^(2)) , T_(3) = (2b)/(v)`
`:. T_(1)^(2) = t_(2) . T_(3)`