A.

converts translational energy to rotational energy

When a body rolls down without slipping along an inclined plane of inclination θ, it rotates about a horizontal axis through its centre of mass and also its centre of mass moves. Therefore, rolling motion may be regarded as a rotational motion about an axis through its centre of mass plus a translational motion of the centre of mass. As it rolls down, it suffers loss in gravitational potential energy provided translational energy due to frictional force is converted into rotational energy.

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 

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

B.

20 N

The accelerating force on the rocket

     = upward thrust = $\frac{∆\mathrm{m}}{∆\mathrm{t}}.\mathrm{u}$

Given:-

$\frac{∆\mathrm{m}}{∆\mathrm{t}} = 50 \times {10}^{-3} \mathrm{kg}/\mathrm{s}$ , 

ν  = 400 m/s

Accelerating force = 50 × 10 ×  400

                           = 20 N

B.

4 rad s^-1

According to second law,  the force f provding the acceleration is 

            F = $\frac{{\mathrm{mv}}^2}{\mathrm{R}}$

where m is the mass of the body. This force directed towards the centre is called the centripetal force. For a stone rotated in a circle by a string, the centripetal force is provided by the string.

Maximum tension in the thread is given by 

           T_max = mg + $\frac{{\mathrm{mv}}^2}{\mathrm{r}}$

⇒         T_max = mg + mrω²               ( $\because \mathrm{v} = \mathrm{r} \mathrm{\omega }$ )

              ω² = $\frac{{\mathrm{T}}_{\max } - \mathrm{mg}}{\mathrm{mr}}$

Given

         T_max = 37 N,   m = 500 g = 0.5 kg,

          g = 10 ms^-2,    r = 4 m

∴              ${\mathrm{\omega }}^2 = \frac{37 - 0.5 \times 10}{0.5 \times 4}$

                ${\mathrm{\omega }}^2 = \frac{37 - 5}{2}$

⇒              ω² = 16

⇒              ω = 4 rad s^-1