JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
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 :
- A \(\frac{\lambda \mathrm{L}^3}{16 \pi^2}\)
- B \(\frac{\lambda \mathrm{L}^3}{12}\)
- C \(\frac{\lambda \mathrm{L}^3}{4 \pi^2}\)
- D \(\frac{\lambda L^3}{8 \pi^2}\)
Answer & Solution
Correct Answer
(D) \(\frac{\lambda L^3}{8 \pi^2}\)
Step-by-step Solution
Detailed explanation
\(L=2 \pi R\) \(\mathrm{I}=\frac{\mathrm{MR}^2}{2}=\frac{\lambda \times \mathrm{L}}{2} \times\left(\frac{\mathrm{L}}{2 \pi}\right)^2=\frac{\lambda \mathrm{L}^3}{8 \pi^2}\)
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