Question

A square plate of edge $\frac{a}{2}$ is cut out from a uniform square plate of edge ${}^{\prime }{a}^{\prime }$ as shown in figure. The mass of the remaining portion is $M$. The moment of inertia of the shaded portion about an axis passing through ${}^{\prime }{O}^{\prime }$ (centre of the square of side a) and perpendicular to plane of the plate is:

• A. $\frac{9}{64}M{a}^2$
• B. $\frac{3}{16}M{a}^2$
• C. $\frac{5}{12}M{a}^2$
• D. $\frac{M{a}^2}{6}$

Answer 1

The correct option is B $\frac{3}{16}M{a}^2$

$\begin{array}{c}\text{First, let's consider the square plate of side}a.\\ \text{Then, its moment of inertia about the axis}\\ \text{perpendicularly passing through}O\text{,}\end{array}$

$\begin{array}{cc}& \text{will be}\left(\frac{m^{\prime }{a}^2}{6}\right)=I^{\prime }[\begin{array}{c}\text{By Perpendicular axis}\\ \text{theorem]}\end{array}\\ & \text{[Where}{m}^{\prime }\text{is mass of this Plate].}\end{array}$

$\begin{array}{c}\text{Let,}{m}^{\prime \prime }\text{be the mass of smaller removed portion (side}=\frac{a}{2}\text{)}\\ \text{Then, its moment of inpritia about same axis}=\frac{m^{\prime \prime }{\left(\frac{a}{2}\right)}^2}{6}+m^{\prime \prime }{\left(\frac{a}{4}\right)}^2\\ \text{[By Parallel axis theorem].}\end{array}$

$\text{So, required moment of inertia}\begin{array}{cc}& =I^{\prime }-I^{\prime \prime }\\ & =\frac{m^{\prime }{a}^2}{6}\frac{m^{\prime \prime }{a}^2}{24}-\frac{m^{\prime \prime }{a}^2}{16}\\ & =\frac{m^{\prime }{a}^2}{6}-\frac{5}{48}{m}^{\prime \prime }{a}^2\end{array}$

$\text{Here,}{m}^{\prime }-m^{\prime \prime }=M,--\left(i\right)\text{and Also,}{m}^{\prime \prime }=\frac{{\left(\frac{a}{2}\right)}^2}{(a{)}^2}\cdot m^{\prime }=\frac{m\prime }{4}$

$\text{Thus,}{m}^{\prime }=\frac{4}{3}m\text{and}{m}^{\prime \prime }=\frac{m}{3}$

$\text{Thus, moment of imertia of remaining Portion}=\left(\frac{4}{3}M\right)\frac{a^2}{6}-\frac{5}{48}\cdot \left(\frac{m}{3}\right)a^2$

$=\frac{3}{16}m{a}^2$