JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
A solid sphere of radius 10 cm is rotating about an axis which is at a distance 15 cm from its centre. The radius of gyration about this axis is \(\sqrt{n}\) cm. The value of n is
- A 265
- B 100
- C 325
- D 125
Answer & Solution
Correct Answer
(A) 265
Step-by-step Solution
Detailed explanation
Let radius of gyration is k \( \Rightarrow mk^{2} = \frac{2}{5} mR^{2} + md^{2}\) \(k^{2} = \frac{2}{3} \times 10^{2} + 15^{2} = 265\) \((\sqrt{n})^{2} = 265 \Rightarrow n = 265\)
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