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JEE Main 2024 (6 Apr, Shift 2) · Banking of Roads & Bridges

Banking of Roads & Bridges question from JEE Main 2024 (6 Apr, Shift 2)

Q9JEE Main 2024 · 6 APR · SHIFT 2Medium

Banking of Roads & Bridges

A car of 800 kg800\text{ kg} is taking turn on a banked road of radius 300 m300\text{ 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)(g = 10\text{ m/s}^2, \sqrt{3} = 1.73)

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Answer: (a)

  1. vmax2=rgμ+tan⁡θ1−μtan⁡θv_{max}^2 = rg\frac{\mu + \tan\theta}{1 - \mu\tan\theta}, with tan⁡30∘=11.73=0.578\tan 30^\circ = \frac{1}{1.73} = 0.578.
  2. 0.2+0.5781−0.2×0.578=0.7780.884≈0.880\frac{0.2 + 0.578}{1 - 0.2 \times 0.578} = \frac{0.778}{0.884} \approx 0.880
  3. vmax2=300×10×0.880≈2640v_{max}^2 = 300 \times 10 \times 0.880 \approx 2640, vmax≈51.4v_{max} \approx 51.4 m/s
  4. So the answer is (a) 51.4 m/s.
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