question_answer

An insect is crawling up a fixed hemispherical bowl of radius R. If the coefficient of friction is 1/3, then the height upto which insect can crawl is

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,     \[h=R-x\]             \[=0.051\,\,R\]             \[=\mathbf{5%}\,\,\mathbf{of}\,\,\mathbf{R}\].