JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
The moment of inertia of a uniform thin rod about a perpendicular axis passing through one end is \(I_{1}\). The same rod is bent into a ring and its moment of inertia about a diameter is \(I _{2}\). If \(\frac{ I _{1}}{ I _{2}}\) is \(\frac{ x \pi^{2}}{3}\), then the value of \(x\) will be ...............
- A \(8\)
- B \(7\)
- C \(2\)
- D \(1\)
Answer & Solution
Correct Answer
(A) \(8\)
Step-by-step Solution
Detailed explanation
\(\frac{ I _{1}}{ I _{2}}=\frac{2}{3}\left(\frac{\ell}{ r }\right)^{2}\) \(=\frac{2}{3} \times 4 \pi^{2}=\frac{8 \pi^{2}}{3}\) \(x =8\)
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