Q64 / 90 · JEE Main 2014 (06 Apr) · Physics

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

Correct answer
A
$\frac{1}{6} m$
B
$\frac{2}{3} m$
C
$\frac{1}{3} m$
D
$\frac{1}{2} m$

Solution

For block remain stationary.

μ = \tan θ (θ

is angle at that point)

∴ tan θ = μ = \frac{1}{2} (slope)

y = \frac{x^{3}}{6}

∴ \frac{d y}{d x} = \frac{3 x^{2}}{6} = \frac{x^{2}}{2} \Rightarrow slope = \frac{1}{2} \Rightarrow x = 1

∴ y = \frac{{(1)}^{3}}{6}

= \frac{1}{6}

m

Hence, the correct answer is option: 'A'.

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