A rod of linear mass density ' $\lambda$ ' and length ' $L$ ' is bent to form a ring of radius 'R'. Moment…

A rod of linear mass density ' $\lambda$ ' and length ' $L$ ' is bent to form a ring of radius 'R'. Moment of inertia of ring about any of its diameter is :
  1. $\frac{\lambda \mathrm{L}^3}{16 \pi^2}$
  2. $\frac{\lambda \mathrm{L}^3}{12}$
  3. $\frac{\lambda \mathrm{L}^3}{4 \pi^2}$
  4. $\frac{\lambda L^3}{8 \pi^2}$

Solution

$L=2 \pi R$
$\mathrm{I}=\frac{\mathrm{MR}^2}{2}=\frac{\lambda \times \mathrm{L}}{2} \times\left(\frac{\mathrm{L}}{2 \pi}\right)^2=\frac{\lambda \mathrm{L}^3}{8 \pi^2}$

Asked in: JEE Main 2025 (08 Apr Shift 2)

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