The figure shows two ships moving in the $x-y$ plane with velocities $V_A$ and $V_B$. The ships move such that $B$ always remains north of $A$. The ratio $\frac{V_A}{V_B}$ is equal to ........

$A$ ship is travelling due east at $10 \, km/hr$. $A$ second ship,heading $30^o$ east of north,is always due north of the first ship. The speed of the second ship in $km/hr$ is:

$A$ river $200 \,m$ wide is flowing at a rate of $3.0 \,m/s$. $A$ boat is sailing at a velocity of $15 \,m/s$ with respect to the water in a direction perpendicular to the river. How far from the point directly opposite to the starting point does the boat reach on the opposite bank (in $\,m$)?

$A$ swimmer wants to cross a river which is flowing at a speed $v$. If the swimmer can swim in still water at speed $V$,the direction he should swim to cross the river in the least time is

$A$ man can swim with velocity $v$ relative to water. He has to cross a river of width $d$ flowing with a velocity $u$ $(u > v)$. The distance through which he is carried downstream by the river is $x$. Which of the following statements is correct?

$A$ boy runs on a horizontal road with a speed of $4 \ m \ s^{-1}$ while it is raining. He sees that the rain is making an angle $\theta$ with the vertical while running from west to east. However, when he runs from east to west, the angle is $\alpha$. The rain is pouring down at an angle $45^{\circ}$ with the vertical and at a speed of $8 \ m \ s^{-1}$ as shown in the figure. The ratio $\frac{\tan \theta}{\tan \alpha}$ is,