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(B) Let the width of the river be $d$.For the shortest time,the swimmer swims perpendicular to the river flow. The time taken is $t = \frac{d}{v}$.For

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$\sin \,	heta$

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(B) Let the width of the river be $d$.For the shortest time,the swimmer swims perpendicular to the river flow. The time taken is $t = \frac{d}{v}$.For the shortest distance,the swimmer must swim at an angle $	heta$ with the upstream such that the resultant velocity is perpendicular to the river bank. The component of velocity perpendicular to the bank is $v \sin 	heta$ (where $	heta$ is the angle with the upstream). Thus,the time taken is $t' = \frac{d}{v \sin 	heta}$.The ratio of the time taken for the shortest time to the time taken for the shortest distance is $\frac{t}{t'} = \frac{d/v}{d/(v \sin 	heta)} = \sin 	heta$.

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