Four small objects each of mass $m$ are fixed at the corners of a rectangular wire-frame of negligible mass and of sides $a$ and $b(a>b)$ If the wire frame is now roated about an axig passing along the side of length $b,$ then the moment of inertia of the system for this axis of rotation is

• Correct answer
  A

  $2 \mathrm{ma}^{2}$
• B

  $4 \mathrm{ma}^{2}$
• C

  $2 m\left(a^{2}+b^{2}\right)$
• D

  $2 m\left(a^{2}-b^{2}\right)$

Solution

The given situation can be shown as Moment of inertia of the system about side of length $b$ say $C D$ is $=M .I .$ of mass at $A$ about $C D+M .I .$ of mass at $B$ mass ot $D$ about $C D$ $$ \begin{array}{l} =m(a)^{2}+m(a)^{2}+m(0)^{2}+m(0)^{2} \\ =2 m a^{2} \end{array} $$

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