An inclined plane is bent in such a way that the vertical cross-section is given by $y =\frac{ x ^{2}}{4}$ where $y$ is in vertical and $x$ in horizontal direction. If the upper surface of this curved plane is rough with coefficient of friction $\mu=0.5,$ the maximum height in $cm$ at which a stationary block will not slip downward is............$cm$
JEE MAIN 2021, Difficult
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At maximum ht. block will experience maximum friction force. Therefore if at this height slope of the tangent is $\tan \theta,$ then $\theta=$ Angle of repose.
$\therefore \tan \theta=\frac{d y}{d x}=\frac{2 x}{4}=\frac{x}{2}=0.5$
$\Rightarrow x=1$ and therefore $y=\frac{x^{2}}{4}=0.25 m$
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