D.

[M L^7/2 T^-2 ]

Given;- v = $\frac{\mathrm{A} \sqrt{\mathrm{x}}}{\mathrm{x} + \mathrm{B}}$             .... (i)

Dimensions of v = dimensions of potential energy

                         = [ M¹ L² T^-2 ]

From equqtion (i) 

Dimensions of B = Dimensions of x = [M^o L¹ T^o ]

∴ Dimensions of A =$\frac{\mathrm{dimension} \mathrm{of} \mathrm{v} \times \mathrm{dimensions} \mathrm{of} \left(\mathrm{x} + \mathrm{B}\right)}{\mathrm{dimensions} \mathrm{of} \sqrt{\mathrm{x}}}$

                            $= \frac{\left[{\mathrm{M}}^1 {\mathrm{L}}^2 {\mathrm{T}}^{-2}\right] \left[{\mathrm{M}}^{\mathrm{o}} {\mathrm{L}}^1 {\mathrm{T}}^{\mathrm{o}}\right]}{\left[{\mathrm{M}}^1 {\mathrm{L}}^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}} {\mathrm{T}}^{\mathrm{o}}\right]}$

                             = [ M¹ L^5/2 T^-2 ]

Hence dimensions of AB 

                            = [ M¹ L^5/2 T^-2 ] [ M^o L¹ T^o ]

                            = [ M L^7/2 T^-2 ]

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

D.

7.5 R

Let at O there will be a collision. If smaller sphere moves x distance to reach at O, then bigger sphere will move a distance of ( 9R - x )

          $\mathrm{F} = \frac{\mathrm{GM} \times 5\mathrm{M}}{{\left(12 \mathrm{R} - \mathrm{x}\right)}^2}$

          a_small  = $\frac{\mathrm{F}}{\mathrm{M}}$

          a_small = $\frac{\mathrm{G} \times 5\mathrm{M}}{{\left(12 \mathrm{R} - \mathrm{x}\right)}^2}$

           a_{big = $\frac{\mathrm{F}}{5\mathrm{M}}$}

           a_big = $\frac{\mathrm{GM}}{{\left(12\mathrm{R}\hspace{0.17em}-\mathrm{x} \right)}^2}$

            $\mathrm{x} = \frac{1}{2}$a_small t²                     

             $\mathrm{x} =\frac{1}{2}\frac{\mathrm{G} \times 5\mathrm{M}}{{\left(12 \mathrm{R} - \mathrm{x}\right)}^2} {\mathrm{t}}^2$             .....(i)

    ( 9R - x ) = $\frac{1}{2}$a_big t²

     (9R - x)  = $\frac{1}{2}\frac{\mathrm{GM}}{{\left(12 \mathrm{R} - \mathrm{x}\right)}^2}{\mathrm{t}}^2$           ....(ii)

Thus dividing equation (i) by equation (ii), we get

                $\frac{\mathrm{x}}{9\mathrm{R}- \mathrm{x}} = 5$

⇒            x = 45 R - 5x

⇒             6x = 45 R

∴               x = 7.5 R    

C.

$\frac{3}{2} {\mathrm{MR}}^2$

The moment of inertia of circular disc about an axis through its centre and normal to its plane

             I_CG $= \frac{1}{2} {\mathrm{MR}}^2$

Where M is the mass of the disc and R its radius.

According to the theorem of parallel axis, the moment of inertia of the uniform circular disc about an axis passing from the edge of the disc and normal to the disc,

         I = I_CG + MR² 

          = $\frac{1}{2} {\mathrm{MR}}^2 + {\mathrm{MR}}^2$

         I = $\frac{3}{2} {\mathrm{MR}}^2$

B.

9000 km

Time period of satellite

     T² = $\frac{4 {\mathrm{\pi }}^2}{\mathrm{GM}} {\mathrm{r}}^3$

     T² = $\frac{4{\mathrm{\pi }}^2}{\mathrm{g} {\mathrm{R}}^2} {\mathrm{r}}^3$                      [ since GM = gR² ]

⇒    T = $\frac{2\mathrm{\pi }}{\mathrm{R}} \sqrt{\frac{{\mathrm{r}}^3}{\mathrm{g}}}$

⇒     T ∝ r^3/2 

∴    $\frac{{\mathrm{T}}_1}{{\mathrm{T}}_2} = {\left(\frac{{\mathrm{r}}_1}{{\mathrm{r}}_2}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$       

Given   T₁ = 3h , T₂ = 24 h 

 (geostationary satellite) r₁ = R,  r₂ = 36000 km

⇒        $\frac{3}{24} = {\left(\frac{\mathrm{R}}{36000}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

⇒         R = ${\left(\frac{1}{8}\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.} \times 36000$

⇒         R = $\frac{1}{4} \times 36000$

⇒         R = 9000 km