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JEE Main 2024 (30 Jan, Shift 2) · Friction on an Incline

Friction on an Incline question from JEE Main 2024 (30 Jan, Shift 2)

Q8JEE Main 2024 · 30 JAN · SHIFT 2Medium

Friction on an Incline

A block of mass mm is placed on a surface having vertical cross-section given by y=x2/4y = x^2/4. If coefficient of friction is 0.5, the maximum height above the ground at which block can be placed without slipping is:

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Answer: (d)

  1. Slope of the surface tan⁡θ=dydx=x2\tan\theta = \dfrac{dy}{dx} = \dfrac{x}{2}.
  2. Limit of no slipping: tan⁡θ=μ⇒x2=0.5⇒x=1 m\tan\theta = \mu \Rightarrow \dfrac{x}{2} = 0.5 \Rightarrow x = 1\text{ m}.
  3. y=124=14 my = \dfrac{1^2}{4} = \dfrac{1}{4}\text{ m}
  4. So the answer is (d) 1/4 m.
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