A boat whose speed in still water is $v$ crosses a river of width $d$ flowing with speed $u$ $(u < v)$. The time taken to cross along the shortest path is $\dfrac54$ times the minimum possible time of crossing. The ratio $\dfrac uv$ is
Answer: (C) $\dfrac35$
Concept. Minimum time: head straight across, $t_1 = \dfrac dv$. Shortest path (straight across, possible since $u < v$): head upstream so that the resultant is perpendicular to the banks, with crossing speed $\sqrt{v^2 - u^2}$.
Step 1. Shortest-path time.
$$t_2 = \frac{d}{\sqrt{v^2 - u^2}}$$
Step 2. Ratio.
$$\frac{t_2}{t_1} = \frac{v}{\sqrt{v^2 - u^2}} = \frac54 \;\Rightarrow\; 1 - \frac{u^2}{v^2} = \frac{16}{25}$$
Step 3. Solve. $\dfrac uv = \dfrac35$.
Answer. $\dfrac35$ (Option C).
Solution by Sreeraj P, M.Sc Physics