A car is moving along a circular path having coefficient of friction 0.5 and radius of curvature $16.2 \mathrm{~m}$. Then the maximum velocity of the car that can travel without falling outwards is (Acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )

• A

  $18 \mathrm{~ms}^{-1}$
• Correct answer
  B

  $32.4 \mathrm{kmh}^{-1}$
• C

  $18 \mathrm{kmh}^{-1}$
• D

  $32.4 \mathrm{~ms}^{-1}$

Solution

Coefficient of friction, $\mu=0.5$ Radius of curvature, $r=16.2 \mathrm{~m}$ Along the radial direction, we have
friction provides the centripetal force, $\mathrm{f}=\mu \mathrm{N}=\mu \mathrm{mg}$ $\begin{aligned} & \frac{\mathrm{m}^2}{\mathrm{r}}=\mu \mathrm{mg} \\ & v=\sqrt{\mu \mathrm{rg}}=\sqrt{0.5 \times 16.2 \times 10} \\ & =9 \mathrm{~m} / \mathrm{s} \\ & =32.4 \mathrm{~km} / \mathrm{h}\end{aligned}$

Hence, the correct answer is option: 'B'.