A man can swim with a certain speed, in still water to cross a river to a point directly opposite to the starting point in time $t_{0}$ . When the river flows, he crosses the river directly along the same path in a time $t_{1}$ . If ‘d’ be the river width, then the velocity of water current is

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$\frac{d (t_{1} - t_{0})}{t_{1} t_{0}}$

$\frac{d \sqrt{t_{1}^{2} - t_{0}^{2}}}{t_{1} t_{0}}$

$\frac{d \sqrt{t_{1}^{2} + t_{0}^{2}}}{t_{1} t_{0}}$

$\frac{d (t_{1} + t_{0})}{t_{1} t_{0}}$

answer is B.

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Detailed Solution

$t_{0} = \frac{d}{V m} a n d t_{1} = \frac{d}{\sqrt{V m^{2} - V r^{2}}} \frac{t_{0}}{t_{1}} = \frac{\sqrt{V m^{2} - V r^{2}}}{V m} \Rightarrow \frac{t_{0}^{2}}{t_{1}^{2}} = 1 - \frac{V r^{2}}{V m^{2}} \frac{V r^{2}}{V m^{2}} = \frac{t_{1}^{2} - t_{0}^{2}}{t_{1}^{2}}; V r = \frac{V m}{t_{1}} \sqrt{t_{1}^{2} - t_{0}^{2}} = \frac{d}{t_{0} t_{1}} \sqrt{t_{1}^{2} - t_{0}^{2}}$