Question

Three particles, each of mass $m$ are situated at the vertices of an equilateral triangle $ABC$ of side C 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. 2$m{\ell }^2$
• B. $\frac{5}{4}m{\ell }^2$
• C. $\frac{3}{2}m{t}^2$
• D. $\frac{3}{4}{m}^2$

Answer 1

The correct option is B $\frac{5}{4}m{\ell }^2$

$\begin{array}{c}Inthegivenfigure,\\ ThemomentofinertiaaboutthelineAX,I_{AX}=m{l}^2+m{d}^2\\ weknowthat,\\ \sin \theta =\frac{perpendicular}{hypotenuse}\\ \sin {30}^0=\frac{d}{1}\\ \frac{1}{2}=\frac{d}{1}\\ d=\frac{1}{2}\\ therefore,I_{AX}=m{l}^2+m{\left(\frac{1}{2}\right)}^2\\ I_{AX}=m{l}^2+\frac{m{l}^2}{4}\\ I_{AX}=\frac{5m{l}^2}{4}\end{array}$

Hence,

option $\left(B\right)$ is correct answer.