The correct answer is A. 437.5 cm4.
The moment of inertia of a hollow circular section about its centroidal axis is given by the following formula:
$I = \frac{\pi}{4}(d^4 – d_o^4)$
where $d$ is the outer diameter of the section, $d_o$ is the inner diameter of the section, and $\pi$ is the mathematical constant pi.
In this case, $d = 8$ cm and $d_o = 6$ cm. Substituting these values into the formula, we get:
$I = \frac{\pi}{4}(8^4 – 6^4) = \frac{\pi}{4}(256 – 144) = \frac{\pi}{4}(112) = 437.5$ cm4
Therefore, the moment of inertia of the hollow circular section is 437.5 cm4.
Option B is incorrect because it is the moment of inertia of a solid circular section. Option C is incorrect because it is the moment of inertia of a hollow circular section with an outer diameter of 6 cm and an inner diameter of 4 cm. Option D is incorrect because it is the moment of inertia of a hollow circular section with an outer diameter of 4 cm and an inner diameter of 2 cm.