Question

An insect is crawling up a hemispherical bowl of radius R. If the coefficient of fraction is $1/3,$ then insect will be able to go up to height h equal to $(take3/\sqrt{10}=0.95)$

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

Answer 1

The correct option is B $\sqrt{R}10$

An insect is crawling up a hemispherical bowl whose radius $=R$

If the coefficient of fraction $=\frac{1}{3}$

Then the insect will be able to go up to height

$h=?$

$\frac{3}{\sqrt{10}}=0.95$

$N=w\cos \theta f=w\sin \theta \frac{f}{N}=tan\theta =\frac{1}{3}\Rightarrow tan\theta =18.43\therefore \cos \theta =\frac{1}{\sqrt{(1/3{)}^2+1}}\frac{r-h}{r}=\frac{1}{\sqrt{(1/3{)}^2+1}}=0.986h=r(1-\frac{1}{{\mu }^2+1})=r(1-0.9486)h=r\left(0.9486\right)wherer=Rthen,h=R\left(0.9486\right)$