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(B) The moment of inertia $I$ of a system of point masses about an axis is given by $I = \sum m_i r_i^2$,where $r_i$ is the perpendicular distance of

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$1.04$

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(B) The moment of inertia $I$ of a system of point masses about an axis is given by $I = \sum m_i r_i^2$,where $r_i$ is the perpendicular distance of the $i$-th mass from the axis of rotation $PQ$.From the figure,the distances of the masses from the axis $PQ$ are:- Two masses of $5 \ kg$ are at a distance of $0.2 \ m$ from the axis $PQ$.- Two masses of $2 \ kg$ are at a distance of $(0.2 + 0.2) = 0.4 \ m$ from the axis $PQ$.Calculating the moment of inertia:$I = 2 	imes (5 \ kg) 	imes (0.2 \ m)^2 + 2 	imes (2 \ kg) 	imes (0.4 \ m)^2$$I = 2 	imes 5 	imes 0.04 + 2 	imes 2 	imes 0.16$$I = 10 	imes 0.04 + 4 	imes 0.16$$I = 0.4 + 0.64$$I = 1.04 \ kg-m^2$

Four masses are fixed on a massless rod as shown in the figure. The moment of inertia about the axis $PQ$ is about ........ $kg-m^2.$

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