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If the coefficient of friction between an insect and bowl is `mu` and the radius of the bowl is r, find the maximum height to which the insect can crawl in the bowl.

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The insect will crawl up the bowl till the component of its weight along the bowl is balanced by limiting frictional force. So, resolving weight perpendicular to the bowl and along the bowl,

` R= mg cos theta `
` F_L = mg sin theta `
Dividing (ii) by (i)
` tan theta = (f_L)/(R) or tan theta = mu [ :. F_L = mu R ] `
` or ( sqrt( r^2 - y^2))/(y ) = mu or y=(r)/( sqrt((1+ mu ^2))`
` so ,h= r -y=r [1- (1)/( sqrt(+mu^2))]`
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