JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
Four identical solid spheres each of mass \('m'\) and radius \('a'\) are placed with their centres on the four corners of a square of side \('b'\). The moment of inertia of the system about one side of square where the axis of rotation is parallel to the plane of the square is
- A \(\frac{4}{5} ma ^{2}+2 mb ^{2}\)
- B \(\frac{8}{5} m a^{2}+m b^{2}\)
- C \(\frac{8}{5} m a^{2}+2 m b^{2}\)
- D \(\frac{4}{5} ma ^{2}\)
Answer & Solution
Correct Answer
(C) \(\frac{8}{5} m a^{2}+2 m b^{2}\)
Step-by-step Solution
Detailed explanation
\(I =2 \times\left(\frac{2}{5} ma ^{2}\right)+2 \times\left(\frac{2}{5} ma ^{2}+ mb ^{2}\right)\) \(I =\frac{8}{5} ma ^{2}+2 mb ^{2}\)
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