Answer:
In
Fig. 3(HT) .11, suppose the insect can crawl up the bowl upto a pt. P from the
bottom B. In doing so, the insect rises through a height BA = h, above the
bottom B of the bowl of radius r.
As is clear from Fig. 3(HT). 11,
\[R=W\cos \alpha \]
\[f=W\,\sin \alpha \], where W
is weight of insect and f is force of friction.
\[\mu =\frac{f}{R}=\frac{W\,\sin \alpha }{W\,\cos \alpha
}=\tan \alpha =\frac{1}{\sqrt{3}}\]
In \[\Delta OPA\], \[\tan \alpha
=\frac{AP}{OA}=\frac{\sqrt{{{r}^{2}}-{{y}^{2}}}}{y}=\frac{1}{\sqrt{3}}\] or \[\frac{{{r}^{2}}-{{y}^{2}}}{{{y}^{2}}}=\frac{1}{3}\]
or \[{{y}^{2}}=\frac{3{{r}^{2}}}{4},y=\sqrt{3}r/2\]
\[h=BA=OB-OA=r-y\]
\[h=r-\frac{\sqrt{3}}{2}\] \[r=0.134\,r=13.4%\]of
r
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