JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
Solid sphere \(A\) is rotating about an axis \(PQ\). If the radius of the sphere is \(5\,cm\) then its radius of gyration about \(P Q\) will be \(\sqrt{x} \;cm\). The value of \(x\) is \(................\).
- A \(110\)
- B \(55\)
- C \(10.48\)
- D \(100\)
Answer & Solution
Correct Answer
(A) \(110\)
Step-by-step Solution
Detailed explanation
\(I _{ cm }=\frac{2}{5} MR ^2\) \(I _{ PQ }= I _{ cm }+ md ^2\) \(I _{ PQ }=\frac{2}{5} mR ^2+ m (10\,cm )^2\) For radius of gyration \(I _{ PQ }= mk ^2\) \(k ^2=\frac{2}{5} R ^2+(10\,cm )^2\) \(=\frac{2}{5}(5)^2+100\) \(=10+100=110\) \(k =\sqrt{110}\,cm\) \(x =110\)
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