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

• Correct answer
  A

  $\frac{l}{\sqrt{2}}$
• B

  $\frac{l}{2}$
• C

  $l$
• D

  $\sqrt{2} l$

Solution

The given situation is shown in figure.

Moment of inertia of the systems about an axis perpendicular to the square and passing through the centre is given as $$ \begin{aligned} I &=m\left(\frac{l}{\sqrt{2}}\right)^{2}+m\left(\frac{l}{\sqrt{2}}\right)^{2}+m\left(\frac{l}{\sqrt{2}}\right)^{2}+m\left(\frac{l}{\sqrt{2}}\right)^{2} \\ &=4 \frac{m l^{2}}{2}=2 m l^{2} \end{aligned} $$ If $k$ be the radius of gyration, then $$ I=4 m k^{2} $$ $\therefore$ From Eegs. (i) und (ii), we have $$ \begin{aligned} 2 m l^{2} &=4 m k^{2} \\ \Rightarrow \quad k &=\frac{1}{\sqrt{2}} \end{aligned} $$

Hence, the correct answer is option: 'A'.