Three particles of each mass ' $m$ ' are kept at the three vertices of an equilateral triangle of side ' 1 '. The moment of inertia of system of the particles about any side of the triangle is

• A

  $\frac{m l^2}{4}$
• B

  $m l^2$
• Correct answer
  C

  $\frac{3}{4} m l^2$
• D

  $\frac{2}{3} m l^2$

Solution

The moment of inertia of the system of particle about $\mathrm{XX}^{\prime}$ is $\begin{aligned} & \mathrm{I}=\mathrm{mx}^2=\mathrm{m}\left(1 \sin 60^{\circ}\right)^2 \\ & =\frac{3}{4} \mathrm{ml}^2 \end{aligned}$

Hence, the correct answer is option: 'C'.