JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
Consider a uniform wire of mass \(M\) and length \(L\). It is bent into a semicircle. Its moment of inertia about a line perpendicular to the plane of the wire passing through the centre is :
- A \(\frac{1}{4} \frac{ ML ^{2}}{\pi^{2}}\)
- B \(\frac{2}{5} \frac{ ML ^{2}}{\pi^{2}}\)
- C \(\frac{ ML ^{2}}{\pi^{2}}\)
- D \(\frac{1}{2} \frac{ ML ^{2}}{\pi^{2}}\)
Answer & Solution
Correct Answer
(C) \(\frac{ ML ^{2}}{\pi^{2}}\)
Step-by-step Solution
Detailed explanation
\(\pi L \Rightarrow r =\frac{ L }{\pi}\) \(I = Mr ^{2}=\frac{ ML ^{2}}{\pi^{2}}\)
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