Question

A uniform rod of mass 'M' and length 'L' is bent In the form of a regular hexagon Moment of inertia of the hexagon so formed about an axis passing through its centre and perpendicular to its plane is

• A. $\frac{7M{L}^2}{343}$
• B. $\frac{9M{L}^2}{343}$
• C. $\frac{5M{L}^2}{216}$
• D. $\frac{3M{L}^2}{109}$

Answer 1

The correct option is C $\frac{5M{L}^2}{216}$

$\begin{array}{c}I=\left[\left(\frac{M}{6}{\left(\frac{L}{6}\right)}^2\cdot \frac{1}{12}\right)+\left\{\frac{M}{6}{\left(\frac{\sqrt{3}}{2}\cdot \frac{L}{6}\right)}^2\right\}\right]\times 6\\ =\left[\frac{M{L}^2}{6\times 36\times 12}+\frac{3M{L}^2}{6\times 4\times 36}\right]\times 6\\ =\frac{M{L}^2}{12}\left[\frac{1}{4}+\frac{1}{36}\right]\\ =\frac{M{L}^2}{12}\times \frac{10}{36}\\ =\frac{5M{L}^2}{216}\end{array}$

Hence,

option $\left(C\right)$ is correct answer.