Three identical uniform thin metal rods form the three sides of an equilateral triangle. If the moment of inertia of the system of these three rods about an axis passing through the centroid of the triangle and perpendicular  to the plane of the triangle is ‘n’ times the moment of inertia of one rod separately about an axis passing through the centre of the rod and perpendicular to its length , the value of ‘n’ is

Let l be the length of the rod, and O be the centroid of the triangle.

${\mathrm{AB}}^2={\mathrm{BD}}^2+{\left(\frac{l}{2}\right)}^2$

$l^2={\mathrm{BD}}^2+\frac{l^2}{4}$

${\mathrm{BD}}^2=\frac{3}{4}{l}^2\Rightarrow \mathrm{BD}=\frac{\sqrt{3}}{2}.l$

$\mathrm{OD}=\frac{\sqrt{3}}{2}.l\times \frac{1}{3}=\frac{l}{2\sqrt{3}}$

The moment of inertia about an axis that passes through the rod's middle and is perpendicular to it,

$\mathrm{l}=\frac{\mathrm{m}{l}^2}{12}$

We now get the moment of inertia about the centroid by using the parallel axes theorem.

$$\begin{gathered} =\mathrm{m}{\left(\frac{l}{2\sqrt{3}}\right)}^2+\mathrm{I} \\ =\frac{\mathrm{m}{l}^2}{4\times 3}+\frac{\mathrm{m}{l}^2}{12} \\ =\frac{\mathrm{m}{l}^2}{12}+\frac{\mathrm{m}{l}^2}{12} \\ =\frac{2\mathrm{m}{l}^2}{12} \end{gathered}$$

Moment of inertia for the three-rod system

$3\times \frac{2\mathrm{m}{l}^2}{12}=6\times \frac{\mathrm{m}{l}^2}{12}=6\mathrm{l}$

According to the question,

$$\begin{gathered} \frac{6m{l}^2}{12}=\mathrm{n}\left(\frac{m{l}^2}{12}\right) \\ \Rightarrow \mathrm{n}=6 \end{gathered}$$

Hence the correct answer is 6.