An insect is at the bottom of a hemisphere ditch of radius 1m. It crawls up the ditch but starts slipping after it is at height 'h' from the bottom. If the coefficient of friction b/w the ground and insect is 0.75. Then h is : (g=10m/s^{2})  
Option: 1 0.6 m
 
Option: 2 0.45 m
Option: 3 0.2 m  
Option: 4 0.8 m

Answers (1)

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

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

h=r\left[1-\frac{1}{\sqrt{1+\mu^{2}}}\right]=1\left[1-\frac{1}{\sqrt{1+ (\frac{3}{4})^{2}}}\right] =1(1-\frac{4}{5})=\frac{1}{5}=0.2 \ m

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