Question

For shotust time is ovand the othec shoue ( perp )
( t=frac{d}{v_{s}} )
( therefore ) weque ( v_{s}=frac{d}{10} )
and ( x=t x v_{text {rive }}=10 mathrm{v}_{text {rever }} )
now
( therefore quad sin theta=frac{10}{15}=frac{2}{3} )
Asoo ( sin theta=frac{d}{sqrt{x^{2}+d^{2}}} )
( therefore frac{4}{a}-frac{V_{s}^{2}}{v_{s}^{2}+v_{r i w}^{2}} quad frac{2}{3}=frac{operatorname{tov}_{s}}{10 sqrt{v_{r i v e}^{2}+v_{s}^{2}}} )
( 4 v_{text {rive } L}^{2}=S v_{s}^{2} )
( frac{V_{S}}{V_{text {fine }}}=frac{2}{sqrt{5}} quad therefore ) Ans ( =(C) )

# when he reache tarted swimmin d of swimme A swimmer crosses a river with minimum possi. ses a river with minimum possible time 10 second. And when he arts Swimming in the direction towards the point the point from where he started swir direction fixed the swimmer crosses the ne swimmer crosses the river in 15 sec. The ratio of speed of Sun water and the speed of river flow is (Assume speed of river flow is (Assume constant speed of river & swimmer) (A) 3 (B) Ā

Solution