Question

An insect is at the bottom of a hemispherical 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 between the ground and the insect is $0.75$, then $h$ is: ($g=10m{s}^{-2}$)

• A. $0.45m$
• B. $0.60m$
• C. $0.20m$
• D. $0.80m$

Answer 1

The correct option is C

$0.20m$

Step 1: Given data.

Hemisphere of radius, $R=1m$

Coefficient friction, $\mu =0.75$

Acceleration due to gravity, $g=10m{s}^{-2}$

Height = $h$

Mass = $m$

Step 2: Find the height $h$ from the bottom.

Now, draw a free body diagram

From the free body diagram, net force down the plane.

$\mu N=mg\sin \left(\theta \right)$ ….$\left(i\right)$

Where $N$ is the normal reaction

From free body diagram, net force perpendicular to the plane

$N=mg\cos \left(\theta \right)$ ….$\left(ii\right)$

Substitute equation $\left(ii\right)$ in $\left(i\right)$ we get.

$\mu mg\cos \left(\theta \right)=mg\sin \left(\theta \right)$

$\Rightarrow$ $\mu =\frac{\sin \left(\theta \right)}{\cos \left(\theta \right)}$

$\Rightarrow$ $\mu =\tan \left(\theta \right)$

$\Rightarrow$ $0.75=\tan \left(\theta \right)$

$\Rightarrow$ $\theta ={37}^0$

Again:

About $Y-axis$,

$h=R-R\cos \left(\theta \right)$

$\Rightarrow$$h=1-1\times \cos \left({37}^0\right)$

$\Rightarrow$$h=1-\frac{4}{5}$

$\Rightarrow$$h=0.2m$

Hence, option C is correct.