The ratio of the distance carried away by the water current, downstream, in crossing a river, by a person, making same angle with downstream and upstream is `2 : 1`. The ratio of the speed of person to the water current cannot be less than.

Options

A. `1//3`
B. `4//5`
C. `2//5`
D. `4//3`

Answer

A

Explanation

(a) Motion of the person making an angle (say prop) with the downstream.
The time taken to cross the river `= (d)/(v sin prop)`
The distance carried away downstream in the same time = speed xx time.
`x_1 = (u + v cos prop) (d)/(v sin prop)` ....(i)
Motion of the person making `prop` angle with upstream.
The time taken to cross the river is equal to `(d)/(v sin prop)`
Distance carried away downstream in the same time
`x_2 = [u + v cos(180^@ - prop)] (d)/(v sin prop)`
`rArr x_(2) = (u - vcosalpha)(d)/(vsinalpha)` ...(ii)
given `((u + v cos prop) (d)/(v sin prop))/((u - v cos prop)(d)/(v sin prop)) = (2)/(1)`
`((u + v cos prop))/((u - v cos prop))= (2)/(1) rArr 3v cos prop = u`
`rArr (v)/(u) = (sec prop)/(3)`
`sec prop ge 1 rArr (sec prop)/(3) ge (1)/(3)`
From Eq. (iii),`(v)/(u) ge (1)/(3)`
So, `v//u` cannot be less than `1//3`,