A boat can row with a speed of $10 k m / h r$ in still water. The river flows steadily at $5 k m / h r$ and the width of the river is $2 k m$ . If the boat man cross the river with reference to minimum distance of approach then time elapsed in rowing the boat will be

\frac{2 \sqrt{3}}{5} h r

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\frac{2}{5 \sqrt{3}} h r

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\frac{3 \sqrt{2}}{5} h r

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\frac{5 \sqrt{2}}{3} h r

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Solution

The correct option is B $\frac{2}{5 \sqrt{3}} h r$
$_{b r} =_{b} -_{r}$

Using pythagoras theorem $Vb=Vbr2−Vr2V_{b}=\sqrt{V_{\text{br}}^2 -V_r^2}$

Time taken,
$t = \frac{d}{\sqrt{V_{br}^{2} - V_{r}^{2}}} = \frac{2}{\sqrt{10^{2} - 5^{2}}} = \frac{2}{5 \sqrt{3}} hrs$

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A boat can row with a speed of 10 km/hr in still water. The river flows steadily at 5 km/hr and the width of the river is 2 km. If the boat man cross the river with reference to minimum distance of approach then time elapsed in rowing the boat will be

The correct option is B 25√3 hr→Vbr=→Vb−→Vr Using pythagoras theorem Vb=Vbr2−Vr2V_{b}=\sqrt{V_{\text{br}}^2 -V_r^2}Vb​=Vbr2​−Vr2​​ Time taken, t=d√V2br−V2r=2√102−52=25√3 hrs

A boat can row with a speed of $10 k m / h r$ in still water. The river flows steadily at $5 k m / h r$ and the width of the river is $2 k m$ . If the boat man cross the river with reference to minimum distance of approach then time elapsed in rowing the boat will be

The correct option is B $\frac{2}{5 \sqrt{3}} h r$
$_{b r} =_{b} -_{r}$

Using pythagoras theorem $Vb=Vbr2−Vr2V_{b}=\sqrt{V_{\text{br}}^2 -V_r^2}$

Time taken,
$t = \frac{d}{\sqrt{V_{br}^{2} - V_{r}^{2}}} = \frac{2}{\sqrt{10^{2} - 5^{2}}} = \frac{2}{5 \sqrt{3}} hrs$