Step 1: Understanding projectile motion
The range \( R \) of a projectile launched at an angle \( \theta \) is given by:
\[
R = \frac{u^2 \sin 2\theta}{g}
\]
The maximum range \( R_{\max} \) occurs when the projectile is launched at \( 45^\circ \) or \( \frac{\pi}{4} \):
\[
R_{\max} = \frac{u^2}{g}
\]
Given that the maximum range is 80 m, we set:
\[
\frac{u^2}{g} = 80
\]
Step 2: Finding the range at \( \frac{\pi}{12} \)
For an angle \( \theta = \frac{\pi}{12} \), the range is:
\[
R' = \frac{u^2 \sin 2\theta}{g}
\]
Substituting \( 2\theta = \frac{\pi}{6} \):
\[
R' = 80 \times \sin \frac{\pi}{6}
\]
Since:
\[
\sin \frac{\pi}{6} = \frac{1}{2}
\]
\[
R' = 80 \times \frac{1}{2} = 40 \text{ m}
\]
Step 3: Verifying the correct option
Comparing with given options, the correct answer is:
\[
\mathbf{40 \text{ m}}
\]