An insect crawling up a fixed hemispherical bowl of radius .R. If the coefficient of friction is $\frac{1}{3},$ then the insect can only crawl upto a height of

See fig.

The insect will crawl up the bowl till the component of its weight down the plane just equals the force of limiting friction

$\begin{array}{l}\mathrm{F}=\mathrm{Wsin}\mathrm{\alpha }\text{ and }\mathrm{N}=\mathrm{Wcos}\mathrm{\alpha }\\ \mathrm{\mu }=\tan \mathrm{\alpha }=\mathrm{F}/\mathrm{N}=1/3\end{array}$

$\text{ But }\tan \left(\frac{\mathrm{\pi }}{2}-\mathrm{\alpha }\right)=\cot \mathrm{\alpha }=\frac{\mathrm{x}}{\sqrt{\left({\mathrm{R}}^2-{\mathrm{x}}^2\right)}}$

$\text{ so that }\mathrm{X}=3/\sqrt{10}=0\cdot 949\mathrm{R}$

$\text{Now }\mathrm{h}=\mathrm{R}-\mathrm{x}=0\cdot 051\mathrm{R}\text{. }$