Question

A block of mass 0.5 kg has its length 75 cm and breadth 25 cm. Then the moment of inertia about an axis perpendicular to the book and passing through the centre of gravity of the book is

• A. $\frac{10}{289}{kg-m}^2$
• B. $\frac{282}{10}{kg-m}^2$
• C. $\frac{10}{384}{kg-m}^2$
• D. $\frac{10}{483}{kg-m}^2$

Answer 1

The correct option is C $\frac{10}{384}{kg-m}^2$

$\begin{array}{c}\text{Given, length and breadth of rectangular}\\ \text{plate are}75cm\text{and}25cm\text{respectively.}\\ \text{Here,}{I}_{A{A}^{\prime }}=\frac{m{l}^2}{12}\text{and}{I}_{B{B}^{\prime }}=\frac{m{b}^2}{12}\end{array}$

$\text{Mass of plate}\left(m\right)=0.5kg$

$\begin{array}{c}\text{Thus, by Perpendicular ax's theorem:}\\ \begin{array}{cc}\text{Irequired}& =I_{A{A}^{\prime }}+I_{B{B}^{\prime }}\\ & =\frac{m}{12}(l^2+b^2)\\ & =\frac{0.5}{12}\times ({\left(\frac{3}{4}\right)}^2+{\left(\frac{1}{4}\right)}^2)kg-m^2\\ & =\frac{10}{384}kg-m^2\end{array}\end{array}$