A swimmer crosses a river of width d flowing | vedclass

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(C) Let $v_r$ be the velocity of the swimmer with respect to the river.The swimmer heads at an angle of $120^{\circ}$ with the river flow $v$. The com

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$4(2-\sqrt{3})v$

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(C) Let $v_r$ be the velocity of the swimmer with respect to the river.The swimmer heads at an angle of $120^{\circ}$ with the river flow $v$. The component of $v_r$ perpendicular to the river flow is $v_r \sin(120^{\circ}-90^{\circ}) = v_r \sin 30^{\circ} = v_r/2$ is incorrect based on the geometry. Actually,the angle with the normal to the river flow is $120^{\circ}-90^{\circ} = 30^{\circ}$.The time taken to cross the river is $t = \frac{d}{v_r \cos 30^{\circ}} = \frac{d}{v_r (\sqrt{3}/2)} = \frac{2d}{\sqrt{3} v_r}$.The drift in the direction of the river flow is given by $x = (v - v_r \sin 30^{\circ})t$.Given $x = d/2$,we have $d/2 = (v - v_r/2) \left( \frac{2d}{\sqrt{3} v_r} \right)$.Simplifying,$1/2 = \frac{2v - v_r}{\sqrt{3} v_r} \implies \sqrt{3} v_r = 4v - 2v_r$.$v_r(2 + \sqrt{3}) = 4v \implies v_r = \frac{4v}{2 + \sqrt{3}}$.Rationalizing the denominator: $v_r = \frac{4v(2 - \sqrt{3})}{(2 + \sqrt{3})(2 - \sqrt{3})} = \frac{4v(2 - \sqrt{3})}{4 - 3} = 4(2 - \sqrt{3})v$.

$A$ swimmer crosses a river of width $d$ flowing at velocity $v$. While swimming,he heads himself always at an angle of $120^{\circ}$ with the river flow and on reaching the other end he finds a drift of $d/2$ in the direction of flow of river. The speed of the swimmer with respect to the river is

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