Question

A square plate of edge $a/2$ is cut out from a uniform square plate of edge $a$ as shown in the figure. The mass of the remaining portion is $M$ . The moment of inertia of the shaded portion about an axis passing through $O$ (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. $\frac{1}{6}M{a}^2$

Answer 1

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

Let $m_1=$ mass of the square plate of side ${}^{\prime }a/2^{\prime }$
and $m_2=$ mass of the square of side ${}^{\prime }{a}^{\prime }$
Then $m_1=\sigma {\left(\frac{a}{2}\right)}^2$ and $m_2=\sigma \left(a{)}^2\right(\sigma \text{being the areal density) and}$ $m_2-m_1=M$

$\Rightarrow I=\frac{m_2{a}^2}{6}-\{\frac{m_1(a/2{)}^2}{6}+m_1{\left(\frac{a}{4}\right)}^2\}$
$=\frac{\sigma a^4}{6}-\{\frac{\sigma (a/2{)}^4}{6}+\sigma {\left(\frac{a}{2}\right)}^2\cdot {\left(\frac{a}{4}\right)}^2\}$

$\Rightarrow I=\sigma a^4\left\{\frac{27}{12\times 16}\right\}$

Also $M=\sigma (1-\frac{1}{4})a^2\Rightarrow \sigma =\frac{4}{3}\frac{M}{a^2}$

$\Rightarrow I=\left(\frac{4M}{3a^2}\right)\cdot a^4\left\{\frac{27}{12\times 16}\right\}\Rightarrow I=\frac{3M{a}^2}{16}$