question_answer

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-on2 units will be

A) \[\frac{3}{4}m{{l}^{2}}\]                               

B)  \[2m{{l}^{2}}\]

C) \[\frac{5}{4}m{{l}^{2}}\,\,\,\,\]                 

D) \[\frac{3}{2}m{{l}^{2}}\,\,\,\,\]

Correct Answer: C

Solution :

Moment of inertia c<the system about axis \[AX,\] \[={{I}_{A}}+{{I}_{B}}+{{I}_{C}}\] \[={{m}_{A}}{{({{r}_{A}})}^{2}}+{{m}_{B}}{{({{r}_{B}})}^{2}}+{{m}_{C}}{{({{r}_{C}})}^{2}}\] \[=m{{(0)}^{2}}+m{{(l)}^{2}}+m{{(l\cos 60)}^{2}}\] \[=m{{l}^{2}}+\frac{m{{l}^{2}}}{4}=\frac{5m{{l}^{2}}}{4}\]