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Question: Answered & Verified by Expert
A car of $800 \mathrm{~kg}$ is taking turn on a banked road of radius $300 \mathrm{~m}$ and angle of banking $30^{\circ}$. If coefficient of static friction is 0.2 then the maximum speed with which car can negotiate the turn safely: $\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2, \sqrt{3}=1.73\right)$
PhysicsLaws of MotionJEE MainJEE Main 2024 (06 Apr Shift 2)
Options:
  • A $264 \mathrm{~m} / \mathrm{s}$
  • B $51.4 \mathrm{~m} / \mathrm{s}$
  • C $70.4 \mathrm{~m} / \mathrm{s}$
  • D $102.8 \mathrm{~m} / \mathrm{s}$
Solution:
2988 Upvotes Verified Answer
The correct answer is: $51.4 \mathrm{~m} / \mathrm{s}$
$\begin{aligned} & \mathrm{m}=800 \mathrm{~kg} \\ & \mathrm{r}=300 \mathrm{~m} \\ & \theta=30^{\circ} \\ & \mu_{\mathrm{s}}=0.2 \\ & \mathrm{~V}_{\max }=\sqrt{\operatorname{Rg}\left[\frac{\tan \theta+\mu}{1-\mu \tan \theta}\right]} \\ & =\sqrt{300 \times \mathrm{g} \times\left[\frac{\tan 30^{\circ}+0.2}{1-0.2 \times \tan 30^{\circ}}\right]} \\ & =\sqrt{300 \times 10 \times\left[\frac{0.57+0.2}{1-0.2 \times 0.57}\right]} \\ & \mathrm{V}_{\max }=51.4 \mathrm{~m} / \mathrm{s}\end{aligned}$

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