JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
The radius of gyration of a uniform rod of length \(l,\) about an axis passing through a point \(\frac{l}{4}\) away from the centre of the rod, and perpendicular to it, is
- A \(\frac{1}{8} l\)
- B \(\sqrt{\frac{7}{48}} l\)
- C \(\sqrt{\frac{3}{8}} l\)
- D \(\frac{1}{4} l\)
Answer & Solution
Correct Answer
(B) \(\sqrt{\frac{7}{48}} l\)
Step-by-step Solution
Detailed explanation
\(\mathrm{m} \frac{l^{2}}{12}+\mathrm{m} \frac{l^{2}}{16}=\mathrm{mk}^{2}\) \(\frac{7 l^{2}}{48}=k^{2}\)
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