JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
The figure shows two solid discs with radius \({R}\) and \({r}\) respectively. If mass per unit area is same for both, what is the ratio of \({MI}\) of bigger disc around axis \({AB}\) (Which is \(\perp\) to the plane of the disc and passing through its centre) of MI of smaller disc around one of its diameters lying on its plane? Given \('M'\) is the mass of the larger disc. (\(MI\) stands for moment of inertia)
- A \(2 r^{4}: R^{4}\)
- B \(2 {R}^{2}: {r}^{2}\)
- C \(2 {R}^{4}: {r}^{4}\)
- D \({R}^{2}: r^{2}\)
Answer & Solution
Correct Answer
(C) \(2 {R}^{4}: {r}^{4}\)
Step-by-step Solution
Detailed explanation
Ratio of moment of inertia \(=\frac{\frac{1}{2} MR ^{2}}{\frac{1}{4} mr ^{2}}\) \(=\frac{2 \sigma \pi R^{2} R^{2}}{\sigma \pi r^{2} r^{2}}=\frac{2 R^{4}}{r^{4}}\)
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