A) \[\frac{1}{\sqrt{2}}\]
B) \[\frac{1}{3}\]
C) \[\frac{1}{3\sqrt{2}}\]
D) None of these
Correct Answer: C
Solution :
[c] \[\sin \theta = \frac{r}{3r}=\frac{1}{3},cos\,\,\theta =\sqrt{8/9}\] If mass of smaller cylinder is m, then mass of bigger one will be 4m. For the equilibrium of upper cylinder, \[2\,{{N}_{1}} sin \theta = 4 mg\] \[\therefore {{N}_{1}} = \frac{2 mg}{sin \theta }\] Now for the equilibrium of lower cylinder, \[{{\operatorname{N}}_{2}} =~{{N}_{1}}\,\,sin\,\theta + mg~~~...\left( ii \right)\] and \[f={{N}_{1}} \,\,cos\,\,\theta \] or \[\mu {{N}_{2}}={{N}_{1}}\cos \,\,\theta \] ?.(iii) After solving above equations, we get \[\mu =\frac{1}{3\sqrt{2}}\].You need to login to perform this action.
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