JEE Mains · Physics · STD 11 - System of particles and rotational motion
- JEE Mains 2022
- Medium
- Numerical
The radius of gyration of a cylindrical rod about an axis of rotation perpendicular to its length and passing through the center will be \(....m\) Given, the length of the rod is \(10 \sqrt{3} m\).
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(5\)
Step-by-step Solution
Detailed explanation
\(I =\frac{ m \ell^{2}}{12}= mk ^{2} \Rightarrow k ^{2}=\frac{\ell^{2}}{12} \Rightarrow k =\frac{\ell}{\sqrt{12}}=\frac{\ell}{2 \sqrt{3}}=\frac{10 \sqrt{3}}{2 \sqrt{3}}=5\)


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