    Question

# A river is flowing with a speed of 1 km/hr. A swimmer wants to go to point 'C' starting from 'A'. He swims with a speed of 5 km/hr at an angle θ w.r.t the river flow. If AB = BC = 400 m. At what angle with river bank should swimmer swim? A

θ=37

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B

θ=53

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C

θ=45

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D

θ=60

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Solution

## The correct option is B θ=53∘ vR=1^i vMR=5cosθ^i+5sinθ^j We know that velocity of man w.r.t. ground should point in the direction 45∘ to the bank as AB = BC vMR=v√2^i+v√2^j vMR=vM−vR 5 cosθ^i+5 sinθ^j=v√2^i+v√2^j−1^i 5 cosθ^i+5 sinθ^j=(v√2−1)^i+v√2^j ⇒5 cosθ=v√2−1 --------------(i) 5 sinθ=v√2 ----------------(ii) Substituting (ii) in (i) 5 cosθ=5 sinθ−1 On squaring both sides we get 25 cos2θ=25 sin2θ+1−10 sinθ 25−25 sin2θ=25 sin2θ+1−10 sinθ (∵cos2θ=1−sin2θ) 50 sin2θ−10 sinθ−24=0 sinθ=10±√100+4800100 = 10±√4900100 sinθ=45 θ=53∘  Suggest Corrections  0    Join BYJU'S Learning Program
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