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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
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An insect crawls up a hemispherical surface very slowly. The coefficient of friction between insect and surface is $\large\frac{1}{3}$. If the line joining the center of hemispherical surface to the insect makes an angle $\alpha$ with the vertical the maximum possible angle of $\alpha$ is

$ a) \cot \alpha=3 \qquad b)\sec \alpha=3 \qquad c) cosec \alpha=3 \qquad d) none $

Can you answer this question?
 
 

1 Answer

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Suppose the insect can crawl upto a point P
$N=w \cos \alpha$-----(1)
$f=frictional\; force= w \sin \alpha$-----(2)
from (1) and (2)
$\tan \alpha=\large\frac{f}{N}$
In limiting case $\tan \alpha =\mu$
Therefore $ \tan \alpha=\large\frac{1}{3}$
$\cot \alpha=3$
Hence a is the correct answer. 

 

answered Jul 16, 2013 by meena.p
edited Feb 10, 2014 by meena.p
 

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