1. Set up the condition for slipping

At a point on the inner surface of the hemispherical ditch, let the insect be at an angle $\theta$ from the bottommost point.

For the insect to remain at rest while crawling slowly, friction must balance the component of weight along the surface.

• Component of weight along the tangent: $mg\sin\theta$
• Normal reaction: $N = mg\cos\theta$

Maximum static friction is $f_{\max} = \mu N = \mu mg\cos\theta$

The insect starts slipping when $mg\sin\theta = \mu mg\cos\theta$

So, $\tan\theta = \mu = 0.75 = \frac{3}{4}$

2. Find $\cos\theta$

Given $\tan\theta = \frac{3}{4}$

Using the $3$-$4$-$5$ triangle, $\cos\theta = \frac{4}{5}$

3. Relate height $h$ to $\theta$

The radius of the hemispherical ditch is $R = 1\text{ m}$

If $\theta$ is measured from the bottom, the vertical rise from the bottom is $h = R(1-\cos\theta)$

Therefore, $h = 1\left(1-\frac{4}{5}\right) = \frac{1}{5} = 0.20\text{ m}$

4. Match with the options

$h = 0.20\text{ m}$

So the correct option is C.

5. Comparison with stored answer

Stored correct answer: C

Our derived answer: C

Hence, they agree.