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