Laws of Motion

JEE Main level, single correct, medium

142 of 210

Question 142 of 210

A block of mass mm is placed on a surface having vertical cross section given by y=x24y=\dfrac{{x}^{2}}{4}. If coefficient of friction is 0.50.5, the maximum height above the ground at which block can be placed without slipping is:

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Answer: (A) 14m\dfrac{1}{4}m

At equilibrium, tan⁡θ=μ\tan \theta =\mu.
Now, tan⁡θ\tan \theta is slope of the curve. Therefore,
tan⁡θ=dydx=x2\tan \theta =\dfrac{dy}{dx}=\dfrac{x}{2}
Now,
x2=0.5⇒x=1\dfrac{x}{2}=0.5\Rightarrow x=1
Required height
y=x24=14my=\dfrac{{x}^{2}}{4}=\dfrac{1}{4}m.

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