Question:
Four equal masses, $\mathrm{m}$ each are placed at the corners of a square of length
$(l)$ as shown in the figure. The moment of inertia of the system about an axis passing through $\mathrm{A}$ and parallel to $\mathrm{DB}$ would be :
Correct Option: , 3
Solution:
(3)
Moment of inertia of point mass
$=$ mass $\times(\text { Perpendicular distance from axis })^{2}$
Moment of Inertia
$=\mathrm{m}(0)^{2}+\mathrm{m}(l \sqrt{2})^{2}+\mathrm{m}\left(\frac{l}{\sqrt{2}}\right)^{2}+\mathrm{m}\left(\frac{l}{\sqrt{2}}\right)^{2}$
$=3 \mathrm{~m} /{ }^{2}$
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