Question

The moment of inertia of a disc of radius R and mass M about a diameter as axis is :

• A. $M{R}^2$
• B. $\frac{1}{2}M{R}^2$
• C. $\frac{1}{4}M{R}^2$
• D. $\frac{5}{4}M{R}^2$

Answer 1

Answer: (C)

$\frac{1}{4}M{R}^2$

$\begin{array}{c}\text{we know that moment of Inertia(I) about axis}\\ \text{perpendicular to the plane is -}\\ I_1=\frac{M{R}^2}{2}\end{array}$

$\begin{array}{c}\text{Now by perpendicular axis theorem-}\\ I_1=I_{AB}+I_{CD}\\ \text{and}{I}_{AB}=I_{CD}\text{(bothare diameters of Same disc)}\\ \Rightarrow I_1=2I_{AB}\\ \Rightarrow I_{AB}=\frac{I_1}{2}=\frac{M{R}^2}{4}\end{array}$