Question

The moment of inertia of a thin uniform circular disc about one of its diameter is I. its moment of inertia about an axis tangent to it nd perpendicular to its plane is

• A. $\frac{2I}{3}$
• B. $2I$
• C. $\frac{I}{2}$
• D. $6I$

Answer 1

The correct option is D $6I$

$\begin{array}{c}\text{Given-* Moment of Inertia about one of its diameter is (I)}\\ \text{we Rnow that Moment of Inertia of a disc of mass (M)}\\ \text{and Radius (R) about an axis passing through its centre}\\ \text{and perpendicular to the plane. is -}\\ I_1=\frac{M{R}^2}{2}\end{array}$

$\begin{array}{c}\text{By perpendicular axis theorem, moment of inertia of}a\\ \text{disc about its diameter paratel to the plane is-}\\ I_{11}+I_{11}=I_L=\frac{M{R}^2}{2}\\ \Rightarrow I_{11}=\frac{M{R}^2}{4}=I\text{(given)}\end{array}$

$\begin{array}{c}\text{Now by parallel axis theorem, moment of inertia}\\ \text{about an axis tangent to the disc and parallel to the}\\ \text{plane will be-}\\ I^{\prime }=\frac{M{R}^2}{2}+M{R}^2=\frac{3}{2}M{R}^2=6I\end{array}$