A man crosses a river in a boat. If he cross the river in minimum time he takes $10 min$ with a drift $120 m$. If he crosses the river taking shortest path, he takes $12.5 min$. Find Width of the river?

Let $v_r$ = velocity of river

$v_{br}$ = velocity of river in still water and

$\omega$ = width of river

Given, $t_{min}$ = 10 min or $\frac{\omega }{v_{rr}}=10 -----\left(i\right)$

Drift in this case will be,

$$\begin{gathered} x=v_{r}t \\ 120=10v_r -----\left(ii\right) \end{gathered}$$

Shortest ath is taken when $v_b$ is along AB. In this case, 

$$\begin{gathered} v_b=\sqrt{v_{tr}^2-v_r^2} \\ Now, 12.5=\frac{a}{v_b}=\frac{6}{\sqrt{v_{hp}^2-v_p^2}}----\left(iii\right) \end{gathered}$$

Solving these three equations, we get $\omega$ = width of river = 200 m.