Question

An insect is crawling up a hemispherical bowl of radius R. If the coefficient of friction is $\frac{1}{3}$, what will be height h to which the insect will be able to crawl? Take $(\frac{3}{\sqrt{10}}=0.95)$

• A. $\frac{R}{5}$
• B. $\frac{R}{10}$
• C. $\frac{R}{20}$
• D. $\frac{R}{30}$

Answer 1

The correct option is C

$\frac{R}{20}$

The insect will crawl up the bowl until the component $mgsin\theta$ of his weight down the plane equals the limiting frictional force $F=\mu mgcos\theta$.
Thus, $mgsin\theta =\mu mgcos\theta \Rightarrow tan\theta =\mu =\frac{1}{3}\Rightarrow cos\theta =\frac{3}{\sqrt{10}}$

Now, $h=R(1-cos\theta )=R(1-\frac{3}{\sqrt{10}})=R(1-0.95)=0.05R=\frac{R}{20}$
Hence, the correct choice is (c).