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JEE Main 2021 (24 Feb, Shift 1) · Friction on an Incline

Friction on an Incline question from JEE Main 2021 (24 Feb, Shift 1)

Q18JEE Main 2021 · 24 FEB · SHIFT 1Medium

Friction on an Incline

An inclined plane is bent in such a way that the vertical cross-section is given by y=x24y = \frac{x^2}{4} where y is in vertical and x in horizontal direction. If the upper surface of this curved plane is rough with coefficient of friction μ\mu = 0.5, the maximum height in cm at which a stationary block will not slip downward is _________ cm.

Numerical answer type. Work it out, then open the answer.

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Answer: 25

  1. Slope tan⁡θ=dydx=x2\tan\theta = \dfrac{dy}{dx} = \dfrac{x}{2}; limit of no slipping tan⁡θ=μ=0.5\tan\theta = \mu = 0.5.
  2. x=1 m⇒y=14 mx = 1\text{ m} \Rightarrow y = \dfrac{1}{4}\text{ m}
  3. Maximum height =0.25 m=25 cm= 0.25\text{ m} = 25\text{ cm}.
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