Q128 / 180 · KCET 2015 · Physics
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
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\( \frac{8}{3} \Pi^{2} \)
- Correct answerC\( \frac{2}{3} \Pi^{2} \)
- D\( \frac{5}{3} \Pi^{2} \)
Solution
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 \( \frac{I}{I}=\frac{(2 \Pi)^{2} R^{2}}{6 R^{2}}=\frac{4 \Pi^{2}}{6}=\frac{2 \Pi^{2}}{3} \)
\[ I=\frac{M L^{2}}{12} \]
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 \( \frac{I}{I}=\frac{(2 \Pi)^{2} R^{2}}{6 R^{2}}=\frac{4 \Pi^{2}}{6}=\frac{2 \Pi^{2}}{3} \)
Hence, the correct answer is option: 'C'.