JEE Main · 2020 · 06 Sep 2020 · Shift 1 · Laws of Motion

As the insect climbs the inside of the bowl, the radius to it makes an angle $\theta$ with the vertical, the normal force is $N=mg\cos \theta$, and the component of gravity along the surface is $mg\sin \theta$. Slipping just begins when this driving component equals the maximum friction $\mu N$, i.e. $mg\sin \theta =\mu mg\cos \theta$, giving $\tan \theta =\mu =0.75=\frac{3}{4}$, so $\theta =37^{\circ }$ and $\cos \theta =\frac{4}{5}$. The height risen above the lowest point is $h=R(1-\cos \theta )=1\left(1-\frac{4}{5}\right)=0.2\text{m}$.