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 figure). The moment of inertia of the system about a line AX perpendicular to AB and in plane of ABC, in \[\operatorname{gram}-c{{m}^{2}}\] 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 :

The moment of inertia of the system \[=\,\,\,{{m}_{A}}{{r}_{A}}^{2}+\,\,{{m}_{B}}{{r}_{B}}^{2}+\,{{m}_{C}}{{r}_{C}}^{2}\] \[{{\operatorname{m}}_{A}},\,\,{{m}_{B}},{{m}_{C}}\] are masses at A,B,C respectively \[{{r}_{A}},\,\,{{r}_{B}},{{r}_{C}}\] are perpendicular distance from AX moment of inertia. \[=\,\,{{m}_{A}}{{(0)}^{2}}+m{{(l)}^{2}}+\,\,m{{(l\,sin\,\,30{}^\circ )}^{2}}\] \[=\,\,\,m{{l}^{2}}+\,\,m{{l}^{2}}\left( \frac{1}{4} \right)\,\,=\,\,\frac{5}{4}m{{l}^{2}}\]