Question

An ant fell in a hemispherical bowl of radius 'r'. Now the ant is trying to climb out of it. Find the maximum height that the ant can reach above the bottom most point of the bowl. Given the coefficient of static friction between ant and bowl is '$\mu$' and the ant climbs very slowly.

• A. r (1 - cos ($ta{n}^{-1}\mu$)
• B. r - cos($ta{n}^{-1}\mu$)
• C. r (1 - sin ($ta{n}^{-1}\mu$)
• D. r - cos$\mu$

Answer 1

The correct option is A

r (1 - cos ($ta{n}^{-1}\mu$)

Let the mass of the ant be m,let us say ,at maximum height, at ant subtends on angle $\theta$ at the centre.

$\therefore$ As it is equillibrium

N=mg cos$\theta$ ........(i)

mg sin$\theta$ = $f_{max}$ ......(ii)

mg sin$\theta$ = $\mu$N

mg sin$\theta$ = $\mu$mg cos$\theta$(from (i))

$\Rightarrow \mu =tan\theta$

$\Rightarrow \theta =ta{n}^{-1}\mu$

From the diagram,

h = r - r cosθ

= r (1 - cosθ)

= r (1 - cos ($ta{n}^{-1}\mu$)