C.

6 m s^-1

If the man did not run, the bus would be at a distance s₁ at time t given by

   s₁ = 6 + $\frac{1}{2}\mathrm{\alpha }$ t²

       = 6 + $\frac{1}{2}$ × 3 × t²

   s₁ = 6 + $\frac{3}{2}$t²

If v is the speed of man, he would cover a distance s₂ = vt in time t.

To catch the bus,

           s₁ = s₂

     6 + $\frac{3}{2}$ t² = v t

⇒     t² $-$ $\frac{2 \mathrm{v}}{3}$t + 4 = 0

which gives

         t = $\frac{2 \mathrm{v}}{6} \pm {\left[ \frac{4 {\mathrm{v}}^2}{9} - 16 \right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$  

Now t will be real if  $\left(\frac{4 {\mathrm{v}}^2}{9} - 16\right)$ is positive or zero.

Minimum v corresponds to $\frac{4 {\mathrm{v}}^2}{16} - 16$= 0 which gives v = 6 m s^-1        

C.

90^o

      $\left|\overrightarrow{\mathrm{A}} + \overrightarrow{\mathrm{B}}\right| = \left|\overrightarrow{\mathrm{A}} - \overrightarrow{\mathrm{B}}\right|$

⇒ ${\left|\overrightarrow{\mathrm{A}}\right|}^2 + {\left|\overrightarrow{\mathrm{B}}\right|}^2 + 2 \left|\overrightarrow{\mathrm{A}}\right| \left|\overrightarrow{\mathrm{B}}\right| \mathrm{cos\theta } = {\left|\overrightarrow{\mathrm{A}}\right|}^2 + {\left|\overrightarrow{\mathrm{B}}\right|}^2 - 2 \left|\overrightarrow{\mathrm{A}}\right| \left|\overrightarrow{\mathrm{B}}\right| \mathrm{cos\theta }$   

⇒        cosθ = 0

⇒           θ = 90^o   

D.

infinity

On an artificial satellite orbiting the earth the acceleration is given by $\frac{\mathrm{GM}}{{\mathrm{R}}^2}$ towards the centre of the earth.

Now for a body of mass m on the satellite the graviational force due to earth is  $\frac{\mathrm{G} \mathrm{Mm}}{{\mathrm{R}}^2}$ towards the centre of the earth.

Let the reaction force on the surface of the satellite be N, then

          $\frac{\mathrm{G} \mathrm{Mm}}{{\mathrm{R}}^2} - \mathrm{N} = \mathrm{m} \left(\frac{\mathrm{G} \mathrm{M}}{{\mathrm{R}}^2}\right)$

⇒           N = 0

That is on the satellite there is a state of weightlessness or  g = 0

           T = 2 $\mathrm{\pi } \sqrt{\frac{\mathrm{l}}{\mathrm{g}}}$ = ∞

Hence     T = ∞

C.

24 N

Momentum of one bullet

             p = mv

                 = 20 × 10^-3 × 300 

             p = 6 kg m s^-1

N = number of bullet per sec = 4

∴   $\frac{\mathrm{dp}}{\mathrm{dt}}$ = change of momentum per sec or force

            = N ( p $-$ 0 )

     $\frac{\mathrm{dp}}{\mathrm{dt}}$ = 4 × 6

       $\frac{\mathrm{dp}}{\mathrm{dt}}$= 24 N

C.

7.2 N

Here,    mass of the stone m = 0.3 kg

           Length of a string = 1.5 m 

                        Speed v = 6 m/s

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

Where R is the length of the string.

                   = $\frac{\left(0.3\right) {\left(6\right)}^2}{1.5}$

             T = 7.2 N