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 :
Let the velocity of the boat, if it has to cross the river directly on water the line AB be \[{{v}_{A}}\]and the angle between and \[{{v}_{B}}\] be \[\theta \]. Then, from the figure \[\sin \theta =\frac{{{v}_{w}}}{{{v}_{B}}}\] Given, \[{{v}_{B}}=2{{v}_{w}}\] \[\therefore \] \[\sin \theta =\frac{{{v}_{w}}}{2{{v}_{w}}}=\frac{1}{2}\] \[\Rightarrow \] \[\theta ={{30}^{o}}\] Now, time taken by the boat to cross the river directly from A to B \[t=\frac{D}{{{v}_{A}}}=\frac{D}{{{v}_{B}}\cos \theta }\] \[=\frac{D}{{{v}_{B}}\times \cos {{30}^{o}}}\] or \[t=\frac{2D}{{{v}_{B}}\sqrt{3}}\]
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