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(B) Let $	heta$ be the angle the radius makes with the vertical at the maximum height reached by the insect.At this point,the component of gravity al

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$r\left[1-\frac{1}{\sqrt{1+\mu^2}}\right]$

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(B) Let $	heta$ be the angle the radius makes with the vertical at the maximum height reached by the insect.At this point,the component of gravity along the tangent balances the limiting friction.The forces acting on the insect are: gravity $(mg)$ acting downwards,normal reaction $(N)$ perpendicular to the surface,and friction $(f)$ acting upwards along the tangent.Resolving gravity into components: $mg \sin 	heta$ along the tangent and $mg \cos 	heta$ perpendicular to the surface.Thus,$N = mg \cos 	heta$.The limiting friction is $f = \mu N = \mu mg \cos 	heta$.For the insect to be in equilibrium,$f = mg \sin 	heta$.Therefore,$\mu mg \cos 	heta = mg \sin 	heta$,which gives $	an 	heta = \mu$.From the geometry of the bowl,the height $h$ from the bottom is $h = r - r \cos 	heta = r(1 - \cos 	heta)$.Since $	an 	heta = \mu$,we have $\cos 	heta = \frac{1}{\sqrt{1+\mu^2}}$.Substituting this into the expression for $h$,we get $h = r\left[1 - \frac{1}{\sqrt{1+\mu^2}}ight]$.

If the coefficient of friction between an insect and a bowl is $\mu$ and the radius of the bowl is $r$,the maximum height to which the insect can crawl in the bowl is:

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