Question

A square plate of edge $a/2$ 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 ${}^{\prime }{O}^{\prime }$ (center 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

Answer: (B)

$\frac{3}{16}M{a}^2$

Let $m_1=$ mass of square plate of side $a$

$\Rightarrow m_2=$ mass of square plate of side $\frac{a}{2}$

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

$=\frac{\sigma a^4}{6}-\{\frac{\sigma {\left(\frac{a}{2}\right)}^4}{6}+\sigma {\left(\frac{a}{2}\right)}^2{\left(\frac{a}{4}\right)}^2\}$

$=\sigma a^4\{\frac{1}{6}-\frac{1}{16\times 6}-\frac{1}{4\times 10}\}$

$=\sigma a^4\left\{\frac{(2\times 16)-2-3}{16\times 12}\right\}$

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

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

$\Rightarrow \sigma =\frac{4}{3}\frac{M}{a^2}$

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

$\Rightarrow I=\frac{3}{16}M{a}^2$

Hence, the answer is $\frac{3}{16}M{a}^2.$