ExamBro
ExamBro
JEE Mains · Physics · STD 11 - System of particles and rotational motion
  • JEE Mains 2022
  • Medium
  • Numerical

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 ...............

Pick an option to check your answer.
Verified Solution

Answer & Solution

Correct Answer

(A) \(8\)

Step-by-step Solution

Detailed explanation

\(\frac{ I _{1}}{ I _{2}}=\frac{2}{3}\left(\frac{\ell}{ r }\right)^{2}\)

\(=\frac{2}{3} \times 4 \pi^{2}=\frac{8 \pi^{2}}{3}\)

\(x =8\)

Same subject
Explore more questions on app