C.

60^o

Draw the situation as Y shown. OA represents the path of the particle starting from origin O (0 , 0 ).  Draw a perpendicular from point A to x-axis. Let path of the particle makes an angle 0 with the x-axis, then 

tan θ = slope of line OA 

$$\begin{gathered} \tan \mathrm{\theta } = \frac{\mathrm{AB}}{\mathrm{OB}} \\ =\frac{3}{\sqrt{3}} \\ \tan \mathrm{\theta } = \sqrt{3} \end{gathered}$$

⇒ θ  = 60^o

B.

10

According to Kinematic equations for uniformly accelerated motion

$\mathrm{\theta } = {\mathrm{\omega }}_{\mathrm{o}} \mathrm{t} + \frac{1}{2}\mathrm{\alpha } {\mathrm{t}}^2$

Where ,The angular acceleration is α = 3 rad/s²

the initial angular speed is ω_o =  2 rad/s

The time is t = 2 s

Given:- α = 3.0 rad/s² , ω_o = 2.0 rad/s, t = 2 s

Hence

$\mathrm{\theta } = 2 \times 2 + \frac{1}{2} \times 3 \times {\left(2\right)}^2$

θ = 4 + 6

θ = 10 rad

C.

is independent of both R and v

When magnetic field is perpendicular to motion of charged particle, then

Centripetal force = magnetic force

$$\begin{gathered} \mathrm{i}.\mathrm{e} \frac{\mathrm{m} {\mathrm{v}}^2}{\mathrm{R}} = \mathrm{BqV} \\ \Rightarrow \mathrm{R} = \frac{\mathrm{mv}}{\mathrm{Bq}} \end{gathered}$$

Wherw B is the magnetic field,

q-  charge on particle

Further, time period of the motion

$$\begin{gathered} \mathrm{T} = \frac{2 \mathrm{\pi R}}{\mathrm{v}} \\ = \frac{2\mathrm{\pi } \left({\displaystyle \frac{\mathrm{mv}}{\mathrm{Bq}}}\right)}{\mathrm{v}} \\ \Rightarrow \mathrm{T} = \frac{2 \mathrm{\pi m}}{\mathrm{Bq}} \end{gathered}$$

It is independent of both R and v.

A.

$\frac{2 {\mathrm{\nu }}_{\mathrm{d} }{\mathrm{\nu }}_{\mathrm{u}}}{{\mathrm{\nu }}_{\mathrm{d}} + {\mathrm{\nu }}_{\mathrm{u}}}$

Average velocity is defined as the change in position or displacement divided by the time intervals ( Δt ) in which displacement occurs.

$\overrightarrow{\mathrm{v}} = \frac{∆\mathrm{x}}{∆\mathrm{t}}$

Let t₁ and t₂ times taken by the car to go from X to Y and then from Y to X respectively.

$$\begin{gathered} {\mathrm{t}}_1 + {\mathrm{t}}_2 = \frac{\mathrm{XY}}{{\mathrm{v}}_{\mathrm{u}}} + \frac{\mathrm{XY}}{{\mathrm{v}}_{\mathrm{d}}} \\ {\mathrm{t}}_1 + {\mathrm{t}}_2 = \mathrm{XY} \left(\frac{{\mathrm{v}}_{\mathrm{u}} + {\mathrm{v}}_{\mathrm{d}}}{{\mathrm{v}}_{\mathrm{u}} {\mathrm{v}}_{\mathrm{d}}}\right) \end{gathered}$$

Total distance travelled = XY + XY

                                 = 2XY

Therefore, average speed of the car for this round trip is

$$\begin{gathered} {\mathrm{v}}_{\mathrm{av}} = \frac{2 \mathrm{XY}}{\mathrm{XY} \left({\displaystyle \frac{{\mathrm{v}}_{\mathrm{u}} + {\mathrm{v}}_{\mathrm{d}}}{{\mathrm{v}}_{\mathrm{u}} {\mathrm{v}}_{\mathrm{d}}}}\right)} \\ \Rightarrow {\mathrm{v}}_{\mathrm{av}} = \frac{2 {\mathrm{v}}_{\mathrm{u}} {\mathrm{v}}_{\mathrm{d}}}{{\mathrm{v}}_{\mathrm{u}} + {\mathrm{v}}_{\mathrm{d}}} \end{gathered}$$

A.

$\frac{\mathrm{\nu }}{\mathrm{g\mu }}$

Block B will come to rest, if force applied to it will vanish due to frictional force acting between block B and surface, ie,
  force applied  =  frictional force

   μmg = m$\mathrm{\alpha }$

where μ - constant of proportionality 

$$\begin{gathered} \Rightarrow \mathrm{\mu mg} = \mathrm{m\alpha } \\ \Rightarrow \mathrm{\mu mg} = \mathrm{m} \left(\frac{\mathrm{\nu }}{\mathrm{t}}\right) \\ \Rightarrow \mathrm{t} = \frac{\mathrm{\nu }}{\mathrm{\mu g}} \end{gathered}$$