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(B) Let $v_s$ be the speed of the swimmer in still water and $v_r$ be the speed of the river.When the swimmer moves downstream,their effective speed i

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$3$

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(B) Let $v_s$ be the speed of the swimmer in still water and $v_r$ be the speed of the river.When the swimmer moves downstream,their effective speed is $(v_s + v_r)$.When the swimmer moves upstream,their effective speed is $(v_s - v_r)$.The float moves with the speed of the river,$v_r$.Let $t_1$ be the time taken to travel distance $D$ downstream,and $t_2$ be the time taken to travel back to the float.Time $t_1 = D / (v_s + v_r)$.At this point,the float has moved a distance $x = v_r t_1 = v_r D / (v_s + v_r)$.After turning back,the swimmer meets the float at a distance $D/2$ from $M$. This means the swimmer covers a distance $D - D/2 = D/2$ upstream.The float covers a distance $D/2 - x$ in the same time $t_2$.$t_2 = (D/2) / (v_s - v_r) = (D/2 - x) / v_r$.Substituting $x$,we get $t_2 = (D/2) / (v_s - v_r) = (D/2 - v_r D / (v_s + v_r)) / v_r$.Dividing by $D$ and simplifying: $1 / (2(v_s - v_r)) = (1/2 - v_r / (v_s + v_r)) / v_r$.$1 / (2(v_s - v_r)) = (v_s + v_r - 2v_r) / (2v_r(v_s + v_r)) = (v_s - v_r) / (2v_r(v_s + v_r))$.$(v_s + v_r) = (v_s - v_r)^2 / v_r$.Let $k = v_s / v_r$. Then $k + 1 = (k - 1)^2 = k^2 - 2k + 1$.$k^2 - 3k = 0$. Since $k 
eq 0$,$k = 3$.

$A$ man swimming downstream overtakes a float at a point $M$. After travelling a distance $D$,he turns back and passes the float at a distance of $D/2$ from the point $M$. The ratio of the speed of the swimmer with respect to still water $(v_s)$ to the speed of the river $(v_r)$ is:

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