Correct Answer: 24

Solution : Given: A telegraph post top touches the ground at a distance of $8\sqrt{3}$ m from its foot and makes an angle of $30^{\circ}$ with the horizontal.
We know that $\tan\theta=\frac{\text{Perpendicular}}{\text{Base}}$ and $\sin\theta=\frac{\text{Perpendicular}}{\text{Hypotenuse}}$.

From the given figure,
In $ \triangle CDB$, $\tan30^{\circ}=\frac{CB}{BD}$
⇒ $\frac{1}{ \sqrt{3}}=\frac{BC}{8\sqrt{3}}$
⇒ $BC= 8 \ m$
Again, in $ \triangle CDB$, $\sin30^{\circ}=\frac{CB}{DC}$
⇒ $\frac{1}{ 2}=\frac{8}{CD}$
⇒ $CD=AC=16 \ m$
The height of the post is AB i.e. BC + CD $=8+16=24$ m
Hence, the correct answer is 24 m.