An inclined plane is bent in such a way that the vertical cross-section is given by

Solution:

given

$y=\frac{x^{2}}{4}$

$\mu=0.5$

condition for block will not slip downward

$\mathrm{mg} \sin \theta=\mu \mathrm{mg} \cos \theta$

$\Rightarrow \tan \theta=\mu$

and we know that

$\Rightarrow \tan \theta=\frac{d y}{d x}$

$\Rightarrow \frac{d y}{d x}=\mu \Rightarrow \frac{x}{2}=0.5$

$\left[\begin{array}{l}y=\frac{x^{2}}{4} \\ \frac{d y}{d x}=\frac{x}{2}\end{array}\right]$

put $x=1$ in equation $y=x^{2} / 4$

$\Rightarrow y=\frac{(1)^{2}}{4} \Rightarrow y=\frac{1}{4} \Rightarrow y=0.25$

$y=25 \mathrm{~cm}$