JEE Main 2014 · 6 Apr · Shift 1 · Physics Q4

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PhysicsSingle correct+4 / -1Question 4 of 90

A block of mass m is placed on a surface with a vertical cross section given by $y=\dfrac{x^3}{6}$ . If the coefficient of friction is 0.5, the maximum height above the ground at which the block can be placed without slipping is :

\dfrac16

\dfrac23

\dfrac13

\dfrac12

Answer and worked solution

Official answer: A

The block does not slip while the slope is at most $\tan\theta=\mu=0.5$ . Since $\dfrac{dy}{dx}=\dfrac{x^2}{2}\le\dfrac12$ , we need $x\le1$ , so
$y_{\max}=\frac{1^3}{6}=\frac16\text{ m}.$

Topics:Friction

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JEE Main 2014 · 6 Apr · Shift 1, Physics question 4