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JEE Mains · Physics · STD 11 - System of particles and rotational motion
  • JEE Mains 2023
  • Medium
  • Numerical

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 \(................\).

Pick an option to check your answer.
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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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