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JEE Main 2003 (AIEEE) · Friction Basics

Friction Basics question from JEE Main 2003 (AIEEE)

Q35JEE Main 2003 · AIEEEEasy

Friction Basics

A marble block of mass 2 kg2\text{ kg} lying on ice when given a velocity of 6 m/s6\text{ m/s} is stopped by friction in 10 s.10\text{ s.} Then the coefficient of friction is

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Answer: (d)

  1. Retardation a=610=0.6 m/s2a = \dfrac{6}{10} = 0.6\text{ m/s}^2
  2. a=μg⇒μ=0.610a = \mu g \Rightarrow \mu = \dfrac{0.6}{10}
  3. μ=0.06\mu = 0.06, so the answer is (d).
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