A.

22.4 m/s

A car moves in a circular track so it perform circular motion. 

According to second law the force providing this acceleration is

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

But according to static friction

         f_s ≤ μ_s N

     f = $\frac{\mathrm{m} {\mathrm{v}}^2}{\mathrm{R}} ⩽ {\mathrm{\mu }}_{\mathrm{s}} \mathrm{N}$

     ${\mathrm{v}}_{\max } = \sqrt{\mathrm{\mu } \mathrm{R} \mathrm{g}}$

Given:-  μ = 1.28 

where μ is  the coefficient of friction 

  R = 40 m 

The maximum velocity

          v_max = $\sqrt{\mathrm{\mu } \mathrm{R} \mathrm{g}}$ 

                   = $\sqrt{1.28 \times 40 \times 9.8}$

             v_max = 22.4 m/s 

B.

56.0 km/s

Escape velocity is the minimum speed needed to a free object to escape speed from the gravitational influence of a massive body.

The escape velocity of the planet

        V_escape = $\sqrt{\frac{2\mathrm{G} \mathrm{m}}{\mathrm{R}}} \propto \sqrt{\frac{\mathrm{M}}{\mathrm{R}}}$

         V_escape = $\sqrt{\frac{{\displaystyle \frac{ 2{\mathrm{M}}_{\mathrm{P}}}{{\mathrm{R}}_{\mathrm{P}}}}}{{\displaystyle \frac{2 {\mathrm{M}}_{\mathrm{e}}}{{\mathrm{R}}_{\mathrm{e}}}}}}$

∴        $\frac{{\mathrm{V}}_{\mathrm{P} }}{{\mathrm{V}}_{\mathrm{e}}} = \sqrt{\frac{{\mathrm{M}}_{\mathrm{P}}}{{\mathrm{R}}_{\mathrm{P}}}\times \frac{{\mathrm{R}}_{\mathrm{e}}}{{\mathrm{R}}_{\mathrm{P}}}}$

where V_p - velocity of planet, V_e - velocity of earth

 ⇒      $\frac{{\mathrm{V}}_{\mathrm{P}}}{{\mathrm{V}}_{\mathrm{e}}} = \sqrt{100 \times \frac{1}{4}}$ 

  ⇒      $\frac{{\mathrm{V}}_{\mathrm{P}}}{{\mathrm{V}}_{\mathrm{e}}} = 5$

 ⇒        V_P = 5 × 11.2 

 ⇒        V_P = 56 km/s

B.

1 s

The time of flight = $\frac{2 \mathrm{u} \mathrm{sin\theta }}{\mathrm{g}}$

                         = $\frac{2 \times 9.8 \times \sin {30}^{\mathrm{o}}}{9.8}$

                          =  1 s                 ...( sin 30^o = $\frac{1}{2}$ ) 

C.

250 J

From the formula of work done

       W = P ΔV

Where W is the work done, P is the pressure and ΔV is the change in volume

    W = 10³ × 0.25

     W = 250 J

B.

1 km/h

The law of conservation of momentum states that for two objects colliding in an isolated system, the total momentum before and after collision is equal.

Applying law of conservation of momentum

             m₁ v₁ = MV

[ m₂ v₂ = 0 because v₂ = 0 and M = m₁ + m₂ , V = final velocity 

             m × 3 = ( m + 2m ) v₁

 So,          v₁ = 1 km/h