MHT CET · Physics · Rotational Motion
The moment of inertia of a thin uniform rod about a perpendicular axis passing through one of its ends is 'I'. Now, the rod is bent in a ring and its moment of inertia about diameter is ' \(\mathrm{I}_{1}^{\prime}\). Then \(\frac{\mathrm{I}}{\mathrm{I}_{1}}\) is
- A \(\frac{8 \pi^{2}}{3}\)
- B \(\frac{11 \pi^{2}}{3}\)
- C \(\frac{4 \pi^{2}}{3}\)
- D \(\frac{\pi^{2}}{3}\)
Answer & Solution
Correct Answer
(A) \(\frac{8 \pi^{2}}{3}\)
Step-by-step Solution
Detailed explanation
If \(M\) is the mass of the rod and \(L\) is its length, then \(I=\frac{M L^{2}}{3}\) Radius of the circular ring is given by
\(\begin{aligned}
2 \pi r=L &=\frac{L}{2 \pi} \\
\therefore I_{1} &=\frac{M r^{2}}{2}=\frac{M}{2} \cdot \frac{L^{2}}{4 \pi^{2}} \\
\therefore \frac{1}{I_{1}} &=\frac{8 \pi^{2}}{3}
\end{aligned}\)
\(\begin{aligned}
2 \pi r=L &=\frac{L}{2 \pi} \\
\therefore I_{1} &=\frac{M r^{2}}{2}=\frac{M}{2} \cdot \frac{L^{2}}{4 \pi^{2}} \\
\therefore \frac{1}{I_{1}} &=\frac{8 \pi^{2}}{3}
\end{aligned}\)
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