$A$ man walking briskly in rain with speed $v$ must slant his umbrella forward making an angle $\theta$ with the vertical. $A$ student derives the following relation between $\theta$ and $v: \tan \theta = v$ and checks that the relation has a correct limit: as $v \rightarrow 0, \theta \rightarrow 0,$ as expected. (We are assuming there is no strong wind and that the rain falls vertically for a stationary man). Do you think this relation can be correct? If not,guess the correct relation.

On a calm day,a boat can go across a lake and return in time $T_0$ at a speed $V$. On a rough day,there is a uniform current at speed $v$ that helps the onward journey and impedes the return journey. If the time taken to go across and return on the rough day is $T$,then $T / T_0$ is:

Two particles having position vectors $r_1 = (3 \hat{i} + 5 \hat{j}) \text{ m}$ and $r_2 = (-5 \hat{i} - 3 \hat{j}) \text{ m}$ are moving with velocities $v_1 = (4 \hat{i} + 3 \hat{j}) \text{ m/s}$ and $v_2 = (a \hat{i} + 7 \hat{j}) \text{ m/s}$. If they collide after $2 \text{ s}$,then the value of $a$ is:

Two particles are moving along two long straight lines,in the same plane,with the same speed $= 20 \, cm/s$. The angle between the two lines is $60^{\circ}$,and their intersection point is $O$. At a certain moment,the two particles are located at distances $3 \, m$ and $4 \, m$ from $O$,and are moving towards $O$. Subsequently,the shortest distance between them will be

$A$ girl standing on a road holds her umbrella at $45^{\circ}$ with the vertical to keep the rain away. If she starts running without an umbrella with a speed of $15 \sqrt{2} \; kmh^{-1}$,the rain drops hit her head vertically. The speed of rain drops with respect to the moving girl is ........... $kmh^{-1}$

$A$ river is flowing due east with a speed $3\, ms^{-1}$. $A$ swimmer can swim in still water at a speed of $4\, ms^{-1}$ (figure).
$(a)$ If the swimmer starts swimming due north,what will be his resultant velocity (magnitude and direction)?
$(b)$ If he wants to start from point $A$ on the south bank and reach the opposite point $B$ on the north bank,
$(i)$ In which direction should he swim?
$(ii)$ What will be his resultant speed?
$(c)$ From the two different cases as mentioned in $(a)$ and $(b)$ above,in which case will he reach the opposite bank in a shorter time?