A swimmer crosses a river with minimum possible time 10 second. And when he reaches the other end, he starts swimming in the direction towards the point from where he started swimming. Keeping the direction fixed the swimmer crosses the river in 15sec. The ratio of speed of swimmer with respect to water and the speed of river flow is (Assume constant speed of river and swimmer)

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$\frac{3}{2}$

$\frac{2}{\sqrt{5}}$

$\frac{9}{4}$

$\frac{\sqrt{5}}{2}$

answer is A.

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Detailed Solution

$v =$ velocity of man w.r.t. river

$u =$ velocity of river

$A \overset{t}{\to} B = \frac{d}{v} \Rightarrow 10 = \frac{d}{v} \Rightarrow d = 10 v$ ………..(1)

$B \overset{t}{\to} C = \frac{d}{v \cos θ}$

$\Rightarrow 15 = \frac{d}{v \cos θ} \Rightarrow d = 15 v \cos θ$ …………..(2)

Eq. (1) and (2) $\Rightarrow \cos θ = \frac{2}{3} \Rightarrow \sec θ = \frac{3}{2}$

$∵ \tan θ = \frac{u}{v} ∴ \sqrt{\sec^{2} θ - 1} = \frac{u}{v}$

$\Rightarrow \frac{u}{v} = \sqrt{\frac{9}{4} - 1} = \frac{\sqrt{5}}{2} \Rightarrow \frac{v}{u} = \frac{2}{\sqrt{5}}$

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