Correct answer is (3)

Time to cross the River width $\omega =1000\mathrm{m}$ is $=\frac{1\mathrm{km}}{4\mathrm{km}/\mathrm{h}}$

Drift $\mathrm{x}=\mathrm{Vm}/\mathrm{g}\times \mathrm{t}$

Where $\mathrm{Vm}/\mathrm{g}$ is velocity of River w.r. to ground.

$\begin{array}{rl}& \mathrm{x}=\mathrm{Vm}/\mathrm{g}\times \frac{1}{4}=750\mathrm{m}=\frac{3}{4}\mathrm{km}\\ \\ & \mathrm{Vm}/\mathrm{g}=3\mathrm{km}/\mathrm{hr}\end{array}$

Correct Option is (C)

From graph,

$v_{RG}=15\sqrt{2}\tan {45}^{\circ }$

$=15\sqrt{2}$

$=\frac{30}{\sqrt{2}}$

Correct answer is (30)

Both velocity vectors are of same magnitude therefore resultant would pass exactly midway through them

$\theta$ = 30${}^{\circ }$

Correct answer is (5)

$V_R=10\sin {30}^{\circ }$

$V_R=\frac{10}{2}=5$ m/s

V_R = 5 m/s

Correct answer is (120)

The situation is depicted in the following figure.

where, V_MR = velocity of man = 12 km/h

and v_R = velocity of water flow in river = 6 km/h

As, v_MR should be along CD.

$\Rightarrow v_R-v_{MR}\sin \theta =0$

$\Rightarrow 6-12\sin \theta =0\Rightarrow \sin \theta =\frac{6}{12}$

$\Rightarrow \sin \theta =\frac{1}{2}$

$\Rightarrow \theta ={\sin }^{-1}\left(\frac{1}{2}\right)={\sin }^{-1}(\sin {30}^{\circ })$ [$\because$ $\sin {30}^{\circ }=\frac{1}{2}$]

$\Rightarrow \theta ={30}^{\circ }$

$\therefore$ $\alpha ={90}^{\circ }+\theta ={90}^{\circ }+{30}^{\circ }={120}^{\circ }$

$\Rightarrow \alpha ={120}^{\circ }$