Three particles, each of mass $m$ gram, are situated at the vertices of an equilateral triangle…
Three particles, each of mass $m$ gram, are situated at the vertices of an equilateral triangle $\mathrm{ABC}$ side $l \mathrm{~cm}$ (as shown in the figure). The moment of inertia of the system about a line $\mathrm{AX}$ perpendicular to $A B$ and in the plane of $A B C$, in gram $\mathrm{cm}^2$ units will be:
$\frac{3}{4} m l^2$
$2 \mathrm{ml}$
$\frac{5}{4} m l^2$
$\frac{3}{2} m l^2$
Solution
According to the question
$\begin{aligned}
I_{A X} & =m l^2+m\left(\frac{l}{2}\right)^2 \\
& =m l^2+\frac{5}{4} \\
& =\frac{5}{4} m l^2
\end{aligned}$
.
According to the question
$\begin{aligned}
I_{A X} & =m l^2+m\left(\frac{l}{2}\right)^2 \\
& =m l^2+\frac{5}{4} \\
& =\frac{5}{4} m l^2
\end{aligned}$
.