Three point masses each of mass m are placed at the corners of an equilateral triangle of side a. Then the moment of inertia of this system about an axis passing along one side of the triangle is

The correct option is 3/4 ma².

Let us find the moment of inertia of the system step-by-step.

Consider an equilateral triangle of side $a$ with three point masses, each of mass $m$, placed at its three vertices. Let the vertices of the triangle be denoted as $A$, $B$, and $C$.

We are required to find the moment of inertia of this system about an axis passing along one side of the triangle. Let us choose this axis to be along the side $BC$.

The moment of inertia $I$ of a system of discrete point masses is given by the formula:
$I=\sum m_i{r}_i^2$
where $m_i$ is the mass of the $i$-th particle and $r_i$ is the perpendicular distance of the $i$-th particle from the axis of rotation.

Let us calculate the perpendicular distance of each mass from the axis of rotation (the line $BC$):

1. The mass at vertex $B$ lies directly on the axis of rotation. Therefore, its perpendicular distance is:
$r_B=0$

2. The mass at vertex $C$ also lies directly on the axis of rotation. Therefore, its perpendicular distance is:
$r_C=0$

3. The mass at vertex $A$ is at a perpendicular distance equal to the altitude (height) $h$ of the equilateral triangle from the side $BC$.

Using the properties of an equilateral triangle of side $a$, the altitude $h$ is given by:
$h=asin(60°)=a\frac{\sqrt{3}}{2}$
Thus, the perpendicular distance for the mass at $A$ is:
$r_A=\frac{\sqrt{3}}{2}a$

Now, we substitute these distances into the moment of inertia formula:
$I=m_A{r}_A^2+m_B{r}_B^2+m_C{r}_C^2$
$I=m{\left(\frac{\sqrt{3}}{2}a\right)}^2+m(0{)}^2+m(0{)}^2$
$I=m\left(\frac{3}{4}{a}^2\right)$
$I=\frac{3}{4}m{a}^2$

Therefore, the moment of inertia of the system about the axis passing along one side of the triangle is $\frac{3}{4}m{a}^2$.