Q20 (2024)
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 :
{{IMG1}}
- $\frac{3\mathrm{mL}^3}{8\pi^2}$
- $\frac{3\mathrm{mL}^3}{8\pi}$
- $\frac{3\mathrm{mL}^2}{8\pi^2}$
- $\frac{3\mathrm{mL}^2}{8\pi}$
### Related Formula
$M = m L$
$$L = 2\pi r \Rightarrow r = \frac{L}{2\pi}$$
$$I_{yy'} = I_{\text{dia}} + M r^2 = \frac{1}{2} M r^2 + M r^2 = \frac{3}{2} M r^2$$
### Core Logic
$$I_{yy'} = \frac{3}{2} (mL) \left(\frac{L}{2\pi}\right)^2 = \frac{3mL^3}{8\pi^2}$$
### Pattern Recognition
Parallel axis theorem applied to tangential axis in the plane of the ring.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: System of Particles and Rotational Motion
