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(C) The moment of inertia of the system about the axis $AX$ is given by the sum of the moments of inertia of individual particles:$I = I_A + I_B + I_C

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$ \frac{5}{4} m l^2 $

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(C) The moment of inertia of the system about the axis $AX$ is given by the sum of the moments of inertia of individual particles:$I = I_A + I_B + I_C$Since particle $A$ lies on the axis $AX$,its perpendicular distance $r_A = 0$. Thus,$I_A = m(0)^2 = 0$.Particle $B$ is at a distance $l$ from $A$ along the line $AB$. Since $AX$ is perpendicular to $AB$,the perpendicular distance of $B$ from $AX$ is $r_B = l$. Thus,$I_B = m(l)^2 = m l^2$.Particle $C$ forms an equilateral triangle with $A$ and $B$. The perpendicular distance of $C$ from the axis $AX$ is the horizontal projection of side $AC$ onto the line perpendicular to $AX$ (which is parallel to $AB$). This distance is $r_C = l \cos(60^{\circ}) = l \cdot \frac{1}{2} = \frac{l}{2}$.Thus,$I_C = m(r_C)^2 = m(\frac{l}{2})^2 = \frac{m l^2}{4}$.Total moment of inertia $I = 0 + m l^2 + \frac{m l^2}{4} = \frac{5}{4} m l^2$.

Three particles,each of mass $m \; g$,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 $g \cdot cm^2$ units will be:

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