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JEE Main 2018 (15 Apr, Shift 2) · Level Road & Rotating Platform

Level Road & Rotating Platform question from JEE Main 2018 (15 Apr, Shift 2)

Q13JEE Main 2018 · 15 APR · SHIFT 2Medium

Level Road & Rotating Platform

A disc rotates about its axis of symmetry in a horizontal plane at a steady rate of 3.53.5 revolutions per second. A coin placed at a distance of 1.25 cm from the axis of rotation remains at rest on the disc. The coefficient of friction between the coin and the disc is: (g = 10 m/s2)

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Answer: (d)

  1. ω=2πf=2π×3.5=7π≈22\omega = 2\pi f = 2\pi \times 3.5 = 7\pi \approx 22 rad/s, ω2≈483.6\omega^2 \approx 483.6.
  2. Coin stays: μg≥ω2r⇒μ≥ω2rg\mu g \ge \omega^2 r \Rightarrow \mu \ge \frac{\omega^2 r}{g}.
  3. μ=483.6×0.012510≈0.60\mu = \frac{483.6 \times 0.0125}{10} \approx 0.60
  4. So the answer is (d) 0.6.
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