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(B) Let $V_{SR} = 5 	ext{ km/hr}$ be the velocity of the swimmer with respect to the river.Let $V_R = 4 	ext{ km/hr}$ be the velocity of the river w

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$4$

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(B) Let $V_{SR} = 5 	ext{ km/hr}$ be the velocity of the swimmer with respect to the river.Let $V_R = 4 	ext{ km/hr}$ be the velocity of the river with respect to the ground.The angle with the river flow is $127^{\circ}$. The velocity components of the swimmer relative to the ground are:$V_{Sx} = V_R + V_{SR} \cos(127^{\circ}) = 4 + 5(-\cos 53^{\circ}) = 4 - 5(0.6) = 4 - 3 = 1 	ext{ km/hr}$.$V_{Sy} = V_{SR} \sin(127^{\circ}) = 5 \sin(53^{\circ}) = 5(0.8) = 4 	ext{ km/hr}$.The time taken to cross the river of width $d = 0.2 	ext{ km}$ is $t_1 = \frac{d}{V_{Sy}} = \frac{0.2}{4} = 0.05 	ext{ hr} = 3 	ext{ minutes}$.The horizontal distance covered downstream is $CB = V_{Sx} 	imes t_1 = 1 	ext{ km/hr} 	imes 0.05 	ext{ hr} = 0.05 	ext{ km} = 50 	ext{ m}$.Time taken to walk distance $CB$ at $3 	ext{ km/hr}$ is $t_2 = \frac{0.05 	ext{ km}}{3 	ext{ km/hr}} = \frac{1}{60} 	ext{ hr} = 1 	ext{ minute}$.Total time $= t_1 + t_2 = 3 	ext{ min} + 1 	ext{ min} = 4 	ext{ minutes}$.

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