Q. Three particles, each of mass $m$ gram, are situated at the vertices of an equilateral triangle $ABC$ of side $l\ cm$ (as shown in the figure). The moment of inertia of the system about a line $AX$ perpendicular to $AB$ and in the plane of $ABC$, in $gram$-$cm^{2}$ units will be
A) $\frac{3}{4}ml^{2}$
B) $2ml^{2}$
C) $\frac{5}{4}ml^{2}$
D) $\frac{3}{2}ml^{2}$
Explanation
(c) Moment of inertia of the system about axis $AX$,
$= I_{A} + I_{B} + I_{C}$
$= m_{A}\left( r_{A} \right)^{2} + mg\left( r_{B} \right)^{2} + mc\left( r_{C} \right)^{2}$
$= m(0)^{2} + m(l)^{2} + m\left( l\cos 60 \right)^{2}$
$= ml^{2} + \frac{ml^{2}}{4} = \frac{5ml^{2}}{4}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.