Explanation

(a) In figure $O$ is the centre of the bowl of radius $r$. The insect will crawl (from $B$ to $A$) till component of its weight ($mg$) along the bowl is balanced by the force of limiting friction ($f$)
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$ie,\ mg\sin{\theta = f = \mu R = \mu\ mg\cos\theta}$
or $\mu = \tan{\theta = \frac{AC}{OC}}$
or $= \frac{\sqrt{OA^{2} - OC^{2}}}{OC} = \frac{\sqrt{r^{2} - y^{2}}}{y}$
or $\mu^{2} = \frac{r^{2} - y^{2}}{y^{2}}$
$\mu^{2}y^{2} + y^{2} = r^{2}$
$y = \frac{r}{\sqrt{\mu^{2} + 1}}$
$h = BC = OB - OC = r - y$
$= r - \frac{r}{\sqrt{\mu^{2} + 1}} = r\left\lbrack 1 - \frac{1}{\sqrt{\mu^{2} + 1}} \right\rbrack$