KCET · Physics · Motion In One Dimension

Moment of Inertia of a thin uniform rod rotating about the perpendicular axis passing through its centre is \( I \). If the same rod is bent into a ring and its moment of inertia about its diameter is \( I^{\prime} \) then the ratio \( \frac{I}{I^{\prime}} \) is

A \( \frac{3}{2} \Pi^{2} \)
B \( \frac{8}{3} \Pi^{2} \)
C \( \frac{2}{3} \Pi^{2} \)
D \( \frac{5}{3} \Pi^{2} \)

Verified Solution

Answer & Solution

Correct Answer

(C) \( \frac{2}{3} \Pi^{2} \)

Step-by-step Solution

Detailed explanation

Moment of inertia of a thin uniform rod is given as
\(I=\frac{M L^{2}}{12}\)
Moment of inertia of a ring is given as
\(I^{\prime}=\frac{M R^{2}}{2}\)
Therefore \( \frac{I}{I}=\frac{\left(\frac{M L^{2}}{12}\right)}{\left(\frac{M R^{2}}{2}\right)}=\frac{L^{2}}{6 R^{2}} \)
Now, \( L=2 \Pi R \)
Thus \( \frac{I}{I}=\frac{(2 \Pi)^{2} R^{2}}{6 R^{2}}=\frac{4 \Pi^{2}}{6}=\frac{2 \Pi^{2}}{3} \)

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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