A) 2 % of R
B) 3 % of R
C) 4 % of R
D) 5 % of R
Correct Answer: D
Solution :
Referring figure, \[F=\omega \sin \alpha \] and \[N=\omega \cos \alpha \] \[m=\tan a\frac{F}{N}=\frac{1}{3}\] But \[\tan \left( \frac{\pi }{2}-\alpha \right)=\cot \alpha =\frac{x}{\sqrt{{{R}^{2}}-{{x}^{2}}}}\] \[\therefore \] \[x=\frac{3R}{\sqrt{10}}=0.949\,\,R\] Now, \[h=R-x\] \[=0.051\,\,R\] \[=\mathbf{5%}\,\,\mathbf{of}\,\,\mathbf{R}\].You need to login to perform this action.
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