An insect crawls up a hemispherical surface very slowly. The coefficient of friction between the insect and the surface is 1/3. If the line joining the centre of the hemispherical surface to the insect makes an angle a with the vertical, the maximum possible value of a so that the insect does not slip is given by [JEE ONLINE 12-05-2012] |
A) \[\cot \alpha =3\]
B) \[\sec \alpha =3\]
C) \[\cos ec\,\alpha =3\]
D) \[\cos \,\alpha =3\]
Correct Answer: A
Solution :
[a] |
The insect crawls up the bowl up to a certain height h only till the component of its weight along the bowl is balanced by limiting frictional force. |
For limiting condition at point A |
\[R=mg\cos \alpha \] ...(i) |
\[{{F}_{1}}=ma\sin \alpha \] ...(ii) |
Dividing eq.(ii) by (i) |
\[\tan \alpha =\frac{1}{\cot \alpha }=\frac{{{F}_{1}}}{R}=\mu \left[ As\,{{F}_{1}}=\mu R \right]\] |
\[\Rightarrow \]\[\tan \alpha =\mu =\frac{1}{3}\left[ \because \mu =\frac{1}{3}(\text{Given}) \right]\] |
\[\therefore \]\[\cot \alpha =3\] |
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