MHT CET · Physics · Rotational Motion
Two rings of same mass 'M' and radius 'R' are so placed that their centre is common and their planes are perpendicular to each other. The moment of inertia of the system about an axis passing through the centre and perpendicular to any one ring is
- A \(\frac{3 \mathrm{MR}^{2}}{2}\)
- B \(\frac{\mathrm{MR}^{2}}{2}\)
- C \(\frac{2 \mathrm{MR}^{2}}{3}\)
- D \(\mathrm{MR}^{2}\)
Answer & Solution
Correct Answer
(A) \(\frac{3 \mathrm{MR}^{2}}{2}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{MR}^{2}+\frac{\mathrm{MR}^{2}}{2}=\frac{3 \mathrm{MR}^{2}}{2}\)
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