TS EAMCET · Physics · Rotational Motion

A thin uniform wire of mass ' \(m\) ' and linear mass density ' \(\rho\) ' is bent in the form of a circular loop. The moment of inertia of the loop about its diameter is

A \(\frac{m^2}{4 \pi^2 \rho^2}\)
B \(\frac{m^3}{4 \rho^2}\)
C \(\frac{m^3}{8 \pi^2 \rho^2}\)
D \(\frac{m^3}{8 \rho^2}\)

Verified Solution

Answer & Solution

Correct Answer

(C) \(\frac{m^3}{8 \pi^2 \rho^2}\)

Step-by-step Solution

Detailed explanation

About the diameter moment of inertia of the loop \(=\) moment of inertia of the ring \(\mathrm{I}=\frac{1}{2} \mathrm{mR}^2\) where \(R=\) radius of the loop Here, \(\rho=\frac{m}{1}\) or, \(\rho=\frac{m}{2 \pi R} \Rightarrow R=\frac{m}{2 \pi \rho}\)…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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