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 :

• A. $ml{l}^2$
• B. $2m{l}^2$
• C. $3m{l}^2$
• D. $\sqrt{3}m{l}^2$

Answer 1

Answer: (C)

Moment of inertia of point mass
$=$ mass $\times (\text{ Perpendicular distance from axis }{)}^2$


Moment of Inertia
$=m(0{)}^2+m(l\sqrt{2}{)}^2+m{\left(\frac{l}{\sqrt{2}}\right)}^2+m{\left(\frac{l}{\sqrt{2}}\right)}^2$
$=3m{l}^2$