A) \[\frac{1}{6}\]
B) \[\frac{1}{3}\]
C) \[\frac{1}{2}\]
D) \[1\]
E) \[\frac{1}{10}\]
Correct Answer: A
Solution :
As mass\[m\]placed oy surface with vertical X-sectfon for which \[y=\frac{{{x}^{3}}}{6}\] Thus, \[\tan \theta =\frac{dy}{dx}=\frac{d\left( \frac{{{x}^{3}}}{6} \right)}{dx}=\frac{{{x}^{2}}}{2}\] At the condition of limiting equilibrium\[\mu =\tan \theta \] \[\Rightarrow \] \[0.5=\frac{{{x}^{2}}}{2}\Rightarrow x=\pm \,1\] Now when \[x=1\] then, \[y=\frac{{{(1)}^{3}}}{6}=\frac{1}{6}\] and when \[x=-1\] then, \[y=\frac{{{(-1)}^{3}}}{6}=\frac{-1}{6}\]
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