Question

Consider a uniform wire of mass $M$ and length $L$. It is bent into a semicircle. Its moment of inertia about a line perpendicular to the plane of the wire passing through the centre is

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

Answer 1

The correct option is D

$\frac{M{L}^2}{{\mathrm{\pi }}^2}$

Step 1. Given Data,

Mass of uniform wire $M$,

Length of uniform wire $L$,

We have to calculate the moment of inertia about a line perpendicular to the plane of the wire passing through the centre.

Step 2. Calculate the moment of inertia about a line perpendicular to the plane of the wire passing through the centre,

Since, wire is bent into semicircles,

We know that,

Length of circle $L=\mathrm{\pi R}$

$\Rightarrow R=\frac{L}{\mathrm{\pi }}$

Moment of inertia $I=M{R}^2$

$$\begin{gathered} \Rightarrow I=M{\left(\frac{L}{\mathrm{\pi }}\right)}^2 \\ \Rightarrow I={\frac{ML}{{\mathrm{\pi }}^2}}^2 \end{gathered}$$

Hence, the correct option is $\text{'}D\text{'}$.