JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
A solid sphere of mass \(500\,g\) and radius \(5\,cm\) is rotated about one of its diameter with angular speed of \(10\,rad \, s ^{-1}\). If the moment of inertia of the sphere about its tangent is \(x \times 10^{-2}\) times its angular momentum about the diameter. Then the value of \(x\) will be ..............
- A \(34\)
- B \(35\)
- C \(36\)
- D \(38\)
Answer & Solution
Correct Answer
(B) \(35\)
Step-by-step Solution
Detailed explanation
\(I _{ t }= x \times 10^{-2}\,L\) \(\frac{7}{5} mR ^2= x \times 10^{-2} \frac{2}{5} mR ^2 \omega\) \(\frac{7}{2 \omega}= x \times 10^{-2}=\frac{7}{2 \times 10}\)
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