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A river of width L is flowing with uniform velocity $V_{0}$ . A swimmer jumps in water at point A with speed of $V_{0}$ relative to water. Swimmer crosses the river in such a way that the resultant velocity of swimmer is always perpendicular to the line joining swimmer and point C. Point B is directly opposite to point A and point A, B and C are collinear. If swimmer crosses the river in time $T_{0};$ then the value of $T_{0}$ will be? ( $\int_{}^{} \sec θ d θ = \ln (\sec θ + \tan θ) + C$ )

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$T_{0} = \frac{L}{V_{0}} \ln (4 + \frac{1}{\sqrt{3}})$

$T_{0} = \frac{L}{V_{0}} \ln (2 + \frac{1}{\sqrt{3}})$

$T_{0} = \frac{L}{V_{0}} \ln (2 + \sqrt{3})$

$T_{0} = \frac{L}{V_{0}} \ln (4 + \sqrt{3})$

answer is B.

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Detailed Solution

$2 L . d θ = 2 V_{0} \cos θ . d t \frac{L}{V_{0}} \int_{0}^{π / 3} \frac{d θ}{\cos θ} = \int_{0}^{T} d t T = \frac{L}{V_{0}} . \ln (\sec θ + \tan θ) |_{0}^{π / 3} T = \frac{L}{V_{0}} \ln (2 + \sqrt{3})$

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