A rod of linear mass density ' $\lambda$ ' and length ' $L$ ' is bent to form a ring of radius 'R'. Moment…
- $\frac{\lambda \mathrm{L}^3}{16 \pi^2}$
- $\frac{\lambda \mathrm{L}^3}{12}$
- $\frac{\lambda \mathrm{L}^3}{4 \pi^2}$
- $\frac{\lambda L^3}{8 \pi^2}$
Solution
$\mathrm{I}=\frac{\mathrm{MR}^2}{2}=\frac{\lambda \times \mathrm{L}}{2} \times\left(\frac{\mathrm{L}}{2 \pi}\right)^2=\frac{\lambda \mathrm{L}^3}{8 \pi^2}$
Asked in: JEE Main 2025 (08 Apr Shift 2)