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JEE Main 20246 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
6 April
Shift
Evening Shift
Subject
Physics
Chapter
Laws of Motion

JEE Main 2024 · 6 April · Evening Shift · Physics · Laws of Motion

A car of 800 ~kg is taking turn on a banked road of radius 300 ~m...

A car of 800 kg800 \mathrm{~kg} is taking turn on a banked road of radius 300 m300 \mathrm{~m} and angle of banking 30∘30^{\circ}. If coefficient of static friction is 0.2 then the maximum speed with which car can negotiate the turn safely: (g=10 m/s2,3=1.73)\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2, \sqrt{3}=1.73\right)

Options

  1. A264 m/s264 \mathrm{~m} / \mathrm{s}
  2. B51.4 m/s51.4 \mathrm{~m} / \mathrm{s}
  3. C70.4 m/s70.4 \mathrm{~m} / \mathrm{s}
  4. D102.8 m/s102.8 \mathrm{~m} / \mathrm{s}

Correct answer

B. 51.4 m/s51.4 \mathrm{~m} / \mathrm{s}

Step-by-step solution

m=800 kgr=300 mθ=30∘μs=0.2 Vmax⁡=Rg⁡[tan⁡θ+μ1−μtan⁡θ]=300×g×[tan⁡30∘+0.21−0.2×tan⁡30∘]=300×10×[0.57+0.21−0.2×0.57]Vmax⁡=51.4 m/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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