A) \[\frac{2D}{{{V}_{B}}\sqrt{3}}\]
B) \[\frac{\sqrt{3}D}{2{{V}_{B}}}\]
C) \[\frac{D}{{{V}_{B}}\sqrt{2}}\]
D) \[\frac{D\sqrt{2}}{{{V}_{B}}}\]
Correct Answer: A
Solution :
From figure, \[{{V}_{B}}\text{ }sin\,\theta \,\,=\,\,{{V}_{W}}\] \[\sin \,\theta \,=\,\frac{{{V}_{W}}}{{{V}_{B}}}\,=\,\frac{1}{2}\,\,\Rightarrow \,\,\theta =30{}^\circ \,\,\,\,\,\,\,\,\,\,[\because \,\,{{V}_{B}}=2{{V}_{W}}]\] Time taken to cross the river, \[t=\frac{D}{{{V}_{B}}\cos \,\theta }=\frac{D}{{{V}_{B}}\,\cos \,30{}^\circ }=\frac{2D}{{{V}_{B}}\,\sqrt{3}}\]You need to login to perform this action.
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