A thin wire of length '$L$' and linear mass density '$m$' is bent into a circular ring (in $x$-$y$ plane) with centre '$C$' as shown in figure. The moment of inertia of the ring about an axis $yy'$ will be:

![](https://res.cloudinary.com/ddr1kuyb3/image/upload/v1786187159/questions/uoqfiqdlxvgt2ioi81yi.png)

Question

A thin wire of length '$L$' and linear mass density '$m$' is bent into a circular ring (in $x$-$y$ plane) with centre '$C$' as shown in figure. The moment of inertia of the ring about an axis $yy'$ will be:

![](https://res.cloudinary.com/ddr1kuyb3/image/upload/v1786187159/questions/uoqfiqdlxvgt2ioi81yi.png)

Accepted Answer

B. 1. First find the **mass of the wire** using the definition of linear mass density (mass per unit length).

$$
M = mL
$$

2. Since the wire of length $L$ is bent into a full circle, use the circumference formula to get the radius of the ring.

$$
L = 2\pi r \implies r = \frac{L}{2\pi}
$$

3. For a ring, the moment of inertia about a **diameter** (an axis in its own plane through the centre) is half that about the axis perpendicular to its plane.

$$
I_{cm} = \frac{1}{2}Mr^2
$$

4. The axis $yy'$ shown is **parallel to this diameter but shifted to the edge** of the ring, so apply the **parallel axis theorem** to shift from the centre to the rim.

$$
I_{yy'} = I_{cm} + Mr^2 = \frac{1}{2}Mr^2 + Mr^2 = \frac{3}{2}Mr^2
$$

5. Substituting $M = mL$ and $r = \frac{L}{2\pi}$ gives the final answer in terms of $m$ and $L$.

$$
I_{yy'} = \frac{3}{2}(mL)\left(\frac{L}{2\pi}\right)^2 = \frac{3mL^3}{8\pi^2}
$$

Hence, the answer is **option B**.

A thin wire of length 'LLL' and linear mass density 'mmm' is bent into a circular ring (in xxx-yyy plane) with centre 'CCC' as shown in figure. The moment of inertia of the ring about an axis yy′yy'yy′ will be:

A thin wire of length ' $L$ ' and linear mass density ' $m$ ' is bent into a circular ring (in $x$ - $y$ plane) with centre ' $C$ ' as shown in figure. The moment of inertia of the ring about an axis $yy'$ will be: