Question

Four particle each of mass 'm' are placed at the corners of a square of side length '1'. The radius of gyration of the system about an axis perpendicular to the plane of square and passing through its centre is

• A. $\frac{1}{\sqrt{2}}$
• B. $\frac{1}{2}$
• C. $1$
• D. $\sqrt{2}l$

Answer 1

The correct option is A $\frac{1}{\sqrt{2}}$

The distance of mass from the center of square is $r=\frac{1}{\sqrt{2}}$

So the total moment of inertia of system $I=4m{r}^2=2m{l}^2$

Let the radius of gyration be $k$

$I=2m{l}^2=4m{k}^2$

$\Rightarrow k=\frac{1}{\sqrt{2}}$