A block of mass $m$ is placed on a surface having vertical cross section given by $y=x^2 / 4$. If coefficient of friction is $0.5$ , the maximum height above the ground at which block can be placed without slipping is:
JEE MAIN 2024, Difficult
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$\frac{d y}{d x}=\tan \theta=\frac{x}{2}=\mu=\frac{1}{2}$
$x=1, y=1 / 4$
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