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If the coefficient of friction between on insect and bowl is \mu and the radius of the bowl is r. Find the maximum height to which insect can crawl up in the bowl.

Option: 1

\frac{r}{\sqrt{1+\mu ^{2}}}

 

 

 


Option: 2

\frac{r\mu }{\sqrt{1+\mu ^{2}}}


Option: 3

r\left [ 1-\frac{1}{\sqrt{1+\mu ^{2}}} \right ]


Option: 4

\frac{ru}{\sqrt{1-\mu ^{2}}}


Answers (1)

best_answer

As we learned

Motion an insect in the rough bowl -

Till the component of its weight along with the bowl is balanced by limiting frictional force, then only the insect crawl up the bowl, up to a certain height h  

        

         Let m=mass of the insect, r=radius of the bowl, μ= coefficient of friction for limiting condition at point A

          R=m g \cos \theta \quad \ldots \ldots\left(\text { i) } \quad \text { and } \quad F_{l}=m g \sin \theta \quad \ldots \ldots \text { (ii) }\right.

 \begin{array}{l}{\text { Dividing (ii) by (i) }} \\ {\qquad \tan \theta=\frac{F_{l}}{R}=\mu \quad\left[\text { using } F_{l}=\mu R\right]}\end{array}

\begin{array}{l}{\therefore \quad \frac{\sqrt{r^{2}-y^{2}}}{y}=\mu \quad \text { or } y=\frac{r}{\sqrt{1+\mu^{2}}}} \\ {\text { So } \quad h=r-y=r\left[1-\frac{1}{\sqrt{1+\mu^{2}}}\right], \therefore h=r\left[1-\frac{1}{\sqrt{1+\mu^{2}}}\right]}\end{array}

where 

h = height up to which insect can climb

m = mass of insect

r = radius of the bowl

\mu= coefficient of friction

 

Posted by

Kuldeep Maurya

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