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4-2.Friction
hard
If the coefficient of friction between an insect and bowl is $\mu$ and the radius of the bowl, is $r$, the maximum height to which the insect can crawl in the bowl is :
A$\frac{r}{\sqrt{1+\mu^2}}$
B$r\left[1-\frac{1}{\sqrt{1+\mu^2}}\right]$
C$r \sqrt{1+\mu^2}$
D$r \sqrt{1+\mu^2}-1$
Solution
$h=r-r \cos \theta$$\mu mg \cos \theta=m g \sin \theta$
$\tan \theta=u$
$\cos \theta=\frac{1}{\sqrt{1+\mu^2}}$
$h=r(1-\cos \theta)=r\left[1-\frac{1}{\sqrt{1+\mu^2}}\right]$
Standard 11
Physics
