From a uniform square plate of side a and mass m, a square portion DEFG of side $\frac{a}{2}$ is removed. Then, the moment of inertia of  remaining portion about the axis AB is


(a) $\frac{7m{a}^2}{16}$                    (b) $\frac{3m{a}^2}{16}$                    (c) $\frac{3m{a}^2}{4}$                       (d) $\frac{9m{a}^2}{16}$
 

Dear Student

Consider the following diagram. 





Moment of Inertia (M.I) of the full square plate about HFG= MI about EFI= $\frac{m{a}^2}{12}$

By Parallel axis theorem,

MI of full square about AHB=


​$$\begin{gathered} I_0+m{d}^{2 }; where I_0= MI about EFI \\ d= dis\tan ce between axes i.e., dis\tan ce between EFI and AHB \\ i.e.,\frac{a}{2} \\ \Rightarrow I_{full square, AHB}=\frac{m{a}^2}{12}+m(\frac{a}{2}{)}^2 =\frac{m{a}^2}{3} \end{gathered}$$

Moment of inertia of small square(to be removed) about IK

= $I{ }_{small square, IK}=\frac{\left({\displaystyle \frac{m}{4}}\right)({\displaystyle \frac{a}{2}}{)}^2}{12}=\frac{m{a}^2}{192}$
By parallel axis theorem again,

Moment of inertia of small square about AHB

​$$\begin{gathered} \Rightarrow I_{small square, AB}=I_{small square}+\frac{m}{4}(dis\tan ce between IK and AHB) \\ \Rightarrow =\frac{m{a}^2}{192}+(\frac{m}{4}\times (\frac{a}{2}+\frac{a}{4}{)}^2) \\ = \frac{28m{a}^2}{192} \end{gathered}$$

Now when the small square is removed,

$$\begin{gathered} I_{remaining,AB}=I_{full square, AB}-I_{small square,AB} \\ \Rightarrow I_{required}=\frac{m{a}^2}{3}-\frac{28m{a}^2}{192}=\frac{3m{a}^2}{16} \end{gathered}$$

hence option B is correct.


note, whenever something is removed or added from/to a body and we are dealing with Moment of Inertia, then we can apply principle of superposition. But we must take care that we are writing M.I about same axes for all elements. ​ removed mass can be treated as negative mass, hence MI is subtracted, as in this question.