A man can swim at a speed of $3$ km/h in still water. A river is flowing with a speed of $2$ km/h . If man always keeps his direction of swimming $120^{0}$ from the direction of flow of water the time taken to cross the river of breadth = $500$ m, is

3 \sqrt{3}

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4 \sqrt{3}

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\frac{1}{3 \sqrt{3}} h

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2 \sqrt{3}

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Solution

The correct option is C $\frac{1}{3 \sqrt{3}} h$
$\begin{matrix} H o r i z o n t a l c o m p o n e n t o f h i s v e l o c i t y i s v_{h} = 3 \times cos ({(120 - 90)}^{0}) = 3 \times cos (30^{0}) = 3 \times \frac{\sqrt{3}}{2} k m / h r s o, t h e t i m e t a k e n i s t = \frac{0.5}{3 \frac{\sqrt{3}}{2}} = \frac{1}{3 \sqrt{3}} h . H e n c e, o p t i o n C i s c o r r e c t a n s w e r . \end{matrix}$

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A man can swim at a speed of 3km/h in still water. A river is flowing with a speed of 2 km/h . If man always keeps his direction of swimming 1200 from the direction of flow of water the time taken to cross the river of breadth =500m, is

The correct option is C 13√3hHorizontalcomponentofhisvelocityisvh=3×cos((120−90)0)=3×cos(300)=3×√32km/hrso,thetimetakenist=0.53√32=13√3h.Hence,optionCiscorrectanswer.

A man can swim at a speed of $3$ km/h in still water. A river is flowing with a speed of $2$ km/h . If man always keeps his direction of swimming $120^{0}$ from the direction of flow of water the time taken to cross the river of breadth = $500$ m, is