Three particles of each mass ' $m$ ' are kept at the three vertices of an equilateral triangle. If $\mathrm{I}_1$ is the moment of inertia of the system of the particles about an axis along one side of the triangle and $\mathrm{I}_2$ is the moment of inertia of the system of the particles about an axis along the perpendicular bisector of a side, then $\mathrm{I}_1: \mathrm{I}_2$ is

• A

  $\sqrt{3}: 2$
• B

  $\sqrt{3}: 4$
• C

  $3: 4$
• Correct answer
  D

  $3: 2$

Solution

$I_1=$ Moment of Inertia along axis of one side of triangle
$\begin{aligned} & d=\sqrt{\frac{3}{4} a^2} \\ & I_1=m d^2 \\ & I_1=m\left(\frac{3 a^2}{4}\right)\end{aligned}$ $I_2=$ Moment of Inertia about axis along perpendicular bisector of side.
$\begin{aligned} & I_2=m\left(\frac{a}{2}\right)^2+m\left(\frac{a}{2}\right)^2 \\ I_2 & =\frac{m a^2}{2} \\ \therefore \quad & I_1: I_2 \text { is } m\left(\frac{3 a^2}{4}\right): \frac{m a^2}{2} \\ \therefore I_1: I_2 & =3: 2\end{aligned}$

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