Question

Four point masses, each of mass m, are placed at the four corners of a square of edge ‘a’. The moment of inertia of the system about a line perpendicular to the plane of the square and passing through the centre of the square is

• A. $\frac{m{a}^2}{2}$
• B. $m{a}^2$
• C. $2m{a}^2$
• D. $4m{a}^2$

Answer 1

The correct option is C $2m{a}^2$

Perpendicular distance of every particle from the given line is $\frac{a}{\sqrt{2}}$.
The moment of inertia of one particle is therefore, $m{\left(\frac{a}{\sqrt{2}}\right)}^2=\frac{1}{2}m{a}^2$.
$\therefore$ The moment of inertia of the system is, $4\times \frac{1}{2}m{a}^2=2m{a}^2$.