A) 1
B) 3
C) 4
D) \[\sqrt{41}\]
Correct Answer: B
Solution :
The shortest possible path is straight line AC. Let u be the velocity of river water and v that of boat making an angle \[\theta \] with line AC. Magnitude of velocity of boat along AC \[{{v}_{y}}=v\cos \theta \] Time \[=15\min =\frac{15}{60}h=0.25\,h\] \[{{v}_{y}}=\] Velocity along \[AC\] \[=\frac{Displacement}{Time}\] \[=\frac{1}{0.25}=4\,km/h\] but \[{{v}_{y}}=v\cos \theta \] \[\Rightarrow \cos \theta =\frac{{{v}_{y}}}{v}=\frac{4}{5}\] Velocity of river water \[u=v\sin \theta \] \[=v\sqrt{1-{{\cos }^{2}}\theta }\] \[=5\times \sqrt{1-{{\left( \frac{4}{5} \right)}^{2}}}\] \[=5\times \frac{3}{5}=3\,km/h\] Alternative: From above figure \[{{v}^{2}}={{u}^{2}}+v_{y}^{2}\] \[\Rightarrow \] \[u=\sqrt{{{v}^{2}}-v_{y}^{2}}=\sqrt{{{5}^{2}}-{{4}^{2}}}\] = 3 km/hYou need to login to perform this action.
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