JEE Mains · Physics · STD 11 - System of particles and rotational motion
- JEE Mains 2025
- Hard
- MCQ
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 :
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(\frac{\lambda L^3}{8 \pi^2}\)
Step-by-step Solution
Detailed explanation
\(L=2 \pi R\)
\(\mathrm{I}=\frac{\mathrm{MR}^2}{2}=\frac{\lambda \times \mathrm{L}}{2} \times\left(\frac{\mathrm{L}}{2 \pi}\right)^2=\frac{\lambda \mathrm{L}^3}{8 \pi^2}\)
\(\mathrm{I}=\frac{\mathrm{MR}^2}{2}=\frac{\lambda \times \mathrm{L}}{2} \times\left(\frac{\mathrm{L}}{2 \pi}\right)^2=\frac{\lambda \mathrm{L}^3}{8 \pi^2}\)
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