Four point masses m, 2m, 3m and 4m are kept a | vedclass

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(C) 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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(C) 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.Here, the axis passes through point $D$ and is perpendicular to the plane of the square.$1$. Mass $m$ at $A$: Distance $r_A = b\sqrt{2}$. $I_A = m(b\sqrt{2})^2 = 2mb^2$.$2$. Mass $2m$ at $B$: Distance $r_B = b$. $I_B = 2m(b)^2 = 2mb^2$.$3$. Mass $3m$ at $C$: Distance $r_C = b$. $I_C = 3m(b)^2 = 3mb^2$.$4$. Mass $4m$ at $D$: Distance $r_D = 0$. $I_D = 4m(0)^2 = 0$.Total moment of inertia $I = I_A + I_B + I_C + I_D = 2mb^2 + 2mb^2 + 3mb^2 + 0 = 7mb^2$.Wait, re-evaluating the distances: $A$ is at $(0, b)$, $B$ is at $(b, b)$, $C$ is at $(b, 0)$, $D$ is at $(0, 0)$.Distance from $D(0,0)$:$A(0,b) \rightarrow r = b$, $I_A = m(b)^2 = mb^2$.$B(b,b) \rightarrow r = b\sqrt{2}$, $I_B = 2m(b\sqrt{2})^2 = 4mb^2$.$C(b,0) \rightarrow r = b$, $I_C = 3m(b)^2 = 3mb^2$.$D(0,0) \rightarrow r = 0$, $I_D = 0$.Total $I = mb^2 + 4mb^2 + 3mb^2 + 0 = 8mb^2$.

Four point masses $m, 2m, 3m$ and $4m$ are kept at the corners $A, B, C$ and $D$ respectively of a square $ABCD$ of side '$b$'. The moment of inertia of the system about an axis perpendicular to the plane of the square and passing through the point $D$ is (in $mb^2$)

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