C.

$\frac{\mathrm{v}}{3}$

According to conservation of momentum

change in moment is eual to final momentum

m₁v₁+m₂v₂= (m₁+m₂)v'

mv+2m×0= (m+2m)v'

so,   v'=$\frac{\mathrm{v}}{3}$

D.

$\left[{\mathrm{M}}^{-1}{\mathrm{L}}^3{\mathrm{T}}^{-2}\right]$

$$\begin{gathered} \mathrm{From} \mathrm{the} \mathrm{formula} \mathrm{G}=\frac{{\mathrm{FR}}^2}{{\mathrm{m}}_1{\mathrm{m}}_2} \\ \mathrm{Dimensions} \mathrm{of} \\ \mathrm{G}=\frac{\mathrm{dimensions} \mathrm{of} \mathrm{force} \mathrm{F}\times \mathrm{dimension} \mathrm{of} {\left(\mathrm{R}\right)}^2}{\mathrm{dimension} \mathrm{of} \mathrm{mass} \mathrm{M}\times \mathrm{dimension} \mathrm{of} \mathrm{mass} \mathrm{M}} \\ =\frac{\left[{\mathrm{MLT}}^{-2}\right]\left[{\mathrm{L}}^2\right]}{\left[\mathrm{M}\right]\left[\mathrm{M}\right]}=\left[{\mathrm{M}}^{-1}{\mathrm{L}}^3{\mathrm{T}}^{-2}\right] \end{gathered}$$

C.

9.8 m

Here :- Initial velocity u=-4.9m/s (-ve is due to vertically upward motion)

Total time t = 2sec

The height of the bridge is given by

s =ut+$\frac{1}{2}{\mathrm{at}}^2$

Here we consider a=g$\Rightarrow$acceleration due to gravity

$$\begin{gathered} \mathrm{s}=-4.9\times 2+\frac{1}{2}\times 9.8\times {\left(2\right)}^2 \\ =-9.8+19.6 =9.8\mathrm{m} \end{gathered}$$

B.

0.3Ns

Here:- Mass of the ball m= 10 g ; time dt =0.1 sec ;

acceleration of the ball = 20m/s²

Force F =ma =0.15×20 =3N

Impulsive forces are forces that act over short times producing rapid changes in motion

$\Rightarrow$ Impulsive force = F×dt = 3×0.1 =0.3 Ns

C.

36×10²¹ N

Here :- Mass of earth M_e =6×10²⁴ kg

Angular velocity of earth ω =2×10^-7 rad/sec

Radius of circular orbit R =1.5×10⁸ km

                                     =1.5×10¹¹m

 Force exerted on earth by the sun is given by

F= M_eRω²

   =6×10²⁴×1.5×10¹¹×(2×10^-7)²

   =3.6×10²²

   =36×10²¹ N