Four point masses,each of value m,are placed | vedclass

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(C) Let the square be $ABCD$ with side length $l$. The masses are at $A, B, C, D$.The axis passes through $A$ and is parallel to the diagonal $BD$.The

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$3ml^2$

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(C) Let the square be $ABCD$ with side length $l$. The masses are at $A, B, C, D$.The axis passes through $A$ and is parallel to the diagonal $BD$.The distance of point $A$ from the axis is $0$.The distance of point $B$ from the axis is $d_B = \frac{l}{\sqrt{2}}$.The distance of point $D$ from the axis is $d_D = \frac{l}{\sqrt{2}}$.The distance of point $C$ from the axis is $d_C = \sqrt{l^2 + (l/\sqrt{2})^2} = \sqrt{l^2 + l^2/2} = \sqrt{3l^2/2} = l\sqrt{3/2}$.Alternatively,the distance of $C$ from the diagonal $BD$ is $l\sqrt{2}$.The moment of inertia $I$ is given by $\sum m_i r_i^2$.$I = m(0)^2 + m(l/\sqrt{2})^2 + m(l\sqrt{2})^2 + m(l/\sqrt{2})^2$$I = 0 + m(l^2/2) + 2ml^2 + m(l^2/2)$$I = ml^2 + 2ml^2 = 3ml^2$.

Four point masses,each of value $m$,are placed at the corners of a square $ABCD$ of side $l$. The moment of inertia of this system about an axis passing through $A$ and parallel to $BD$ is

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