Q 11-04-180JEE MainJEE Main 2020 (6 Sep, Shift 1)Medium
An insect is at the bottom of a hemispherical ditch of radius $1$ m. It crawls up the ditch but starts slipping after it is at height $h$ from the bottom. If the coefficient of friction between the ground and the insect is $0.75$, then $h$ is: ($g = 10\ \text{m s}^{-2}$)
Answer: (A) $0.20$ m
Let the radius to the insect make angle $\theta$ with the vertical. The surface there is inclined at $\theta$, and the insect is about to slip when $\tan\theta = \mu = 0.75$.
Then $\cos\theta = 0.8$, and the height above the bottom is
$$h = R(1 - \cos\theta) = 1\times(1 - 0.8) = 0.20\ \text{m}$$
Solution by Sreeraj P, M.Sc Physics