D.

the linear momentum is conserved

In an inelastic collision, the particles do not regain their shape and size completely after collision. Some fraction of mechanical energy is retained by the colliding particles in the form of deformation potential energy. Thus, the kinetic energy of particles no longer remains conserved. However in the absence of external forces, law of conservation of linear momentum still holds good.

B.

$\frac{1}{2}$

Given:- Kinetic energy of satellite is half of its potential energy. 
Potential energy of satelite

 $\mathrm{U} = \frac{{\mathrm{GM}}_{\mathrm{e}}\mathrm{m}}{{\mathrm{R}}_{\mathrm{e}}}$

Kinetic energy of satellite

          $\mathrm{K} = \frac{1}{2}\frac{{\mathrm{GM}}_{\mathrm{e}} \mathrm{m}}{{\mathrm{R}}_{\mathrm{e}}}$

Thus

    $$\begin{gathered} \frac{\mathrm{K}}{\mathrm{U}} = \frac{1}{2}\frac{{\mathrm{GM}}_{\mathrm{e}} \mathrm{m}}{{\mathrm{R}}_{\mathrm{e}}} \times \frac{{\mathrm{R}}_{\mathrm{e}}}{{\mathrm{GM}}_{\mathrm{e}} \mathrm{m}} \\ \frac{\mathrm{K}}{\mathrm{U}} = \frac{1}{2} \end{gathered}$$

B.

100 J

Net work done in sliding a body up to a height h on inclined plane

        = Work done against gravitational force + Work done against frictional force

⇒             W = W_g  +  W_f

                W = 300 J 

   but ,      W_g = mgh = 2 × 10 × 10

                      = 200 J

  Putting in Eq. (i), we get
                300 = 200 + W_f

⇒               W_f = 300 - 200

                  W_f = 100 J

C.

56 m

Speed is rate of change of distance.

Distance travelled by the particle is

               x = 40 + 12 t - t³

We know the speed is rate of change of distance

       i.e       $\mathrm{\nu } = \frac{\mathrm{dx}}{\mathrm{dt}}$

                  $\mathrm{\nu } = \frac{\mathrm{d}}{\mathrm{dt}}\left(40 + 12\mathrm{t} - {\mathrm{t}}^3\right)$

                     = 0 + 12 - 3t²

but final velocity ν = 0

∴               12 - 3t² = 0

⇒               $$\begin{gathered} {\mathrm{t}}^2 = \frac{12}{3} \\ {\mathrm{t}}^2 = 4 \end{gathered}$$

 ⇒              t = 2s

Hence, distance travelled by the particle before coming to rest is given by
             x = 40 + 12(2)-(2)³

             x = 64 - 8

              x = 56 m

A.

0.4 cm

The situation can be shown as: Let radius of complete disc is a and that of small disc is b. Also let centre of mass now shifts to 0₂ at a distance X₂ from original centre

The position of new centre of mass is given by

                      ${\mathrm{X}}_{\mathrm{CM}} = \frac{-\mathrm{\sigma }. {\mathrm{\pi b}}^2. {\mathrm{x}}_1}{\mathrm{\sigma }.\mathrm{\pi } {\mathrm{a}}^2 - \mathrm{\sigma }.\mathrm{\pi } {\mathrm{b}}^2}$

Here, a = 6 cm, b = 2 cm, x₁ = 3.2 cm

$$\begin{gathered} \therefore {\mathrm{X}}_{\mathrm{CM}} = \frac{-\mathrm{\sigma } \times \mathrm{\pi } {\left(2\right)}^2 \times 3.2}{\mathrm{\sigma } \times \mathrm{\pi } \times {\left(6\right)}^2 - \mathrm{\sigma } \times \mathrm{\pi } \times {\left(2\right)}^2} \\ = \frac{12.8}{32\mathrm{\pi }} \end{gathered}$$

⇒    X_CM = 0.4 cm