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

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(D) Let the axis be $nn'$. The axis passes through $A$ and is parallel to the diagonal $BD$.The distance of point $A$ from the axis is $0$. Thus,the m

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$3ml^2$

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(D) Let the axis be $nn'$. The axis passes through $A$ and is parallel to the diagonal $BD$.The distance of point $A$ from the axis is $0$. Thus,the moment of inertia of the mass at $A$ 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 $d_C = \sqrt{2}l$ (since $C$ is at the opposite corner of the square from $A$).The total moment of inertia $I$ is given by:$I = m(d_A)^2 + m(d_B)^2 + m(d_D)^2 + m(d_C)^2$$I = m(0)^2 + m\left(\frac{l}{\sqrt{2}}\right)^2 + m\left(\frac{l}{\sqrt{2}}\right)^2 + m(\sqrt{2}l)^2$$I = 0 + m\left(\frac{l^2}{2}\right) + m\left(\frac{l^2}{2}\right) + 2ml^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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