A man wishes to swim across a river $0.5 k m$ wide. If he can swim at the rate of $2 k m / h$ in still water and the river flows at the rate of $1 k m / h$ . The angle (w.r.t the flow of the river) along which he should swim so as to reach a point exactly opposite his starting point, will be

60^{\circ}

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120^{\circ}

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145^{\circ}

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90^{\circ}

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Solution

The correct option is A $120^{\circ}$
$v_{r} = v_{m r} s i n θ$
$\Rightarrow 1 = 2 s i n θ$
so $θ = 30^{\circ}$
$Φ = 90^{\circ} + 30^{\circ} = 120^{\circ}$

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A man wishes to swim across a river 0.5 km wide. If he can swim at the rate of 2 km/h in still water and the river flows at the rate of 1 km/h. The angle (w.r.t the flow of the river) along which he should swim so as to reach a point exactly opposite his starting point, will be

The correct option is A 120∘vr=vmrsinθ⇒1=2sinθso θ=30∘Φ=90∘+30∘=120∘

A man wishes to swim across a river $0.5 k m$ wide. If he can swim at the rate of $2 k m / h$ in still water and the river flows at the rate of $1 k m / h$ . The angle (w.r.t the flow of the river) along which he should swim so as to reach a point exactly opposite his starting point, will be

The correct option is A $120^{\circ}$
$v_{r} = v_{m r} s i n θ$
$\Rightarrow 1 = 2 s i n θ$
so $θ = 30^{\circ}$
$Φ = 90^{\circ} + 30^{\circ} = 120^{\circ}$