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JEE Main 2002 (AIEEE) · Level Road & Rotating Platform

Level Road & Rotating Platform question from JEE Main 2002 (AIEEE)

Q15JEE Main 2002 · AIEEEEasy

Level Road & Rotating Platform

The maximum velocity (in m s−1\text{m s}^{-1}) with which a car driver must traverse a flat curve of radius 150 m and coefficient of friction 0.60.6 to avoid skidding is

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Answer: (b)

  1. vmax=μrgv_{max} = \sqrt{\mu r g}
  2. vmax=0.6×150×10=900=30v_{max} = \sqrt{0.6 \times 150 \times 10} = \sqrt{900} = 30 m/s (taking g=10g = 10 m/s2^2)
  3. So the answer is (b) 30.
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