MHT CET · Physics · Rotational Motion
The ratio of radii of gyration of a circular ring and circular disc of the same mass and radius, about an axis passing through their centres and perpendicular to their planes is
- A \(1: \sqrt{2}\)
- B \(2: 1\)
- C \(\sqrt{2}: 1\)
- D \(3: 2\)
Answer & Solution
Correct Answer
(C) \(\sqrt{2}: 1\)
Step-by-step Solution
Detailed explanation
\(\mathrm{I}=\mathrm{MK}^2\)
For a ring, \(\mathrm{MR}^2=\mathrm{MK}_{\mathrm{r}}{ }^2 \quad \therefore \mathrm{K}_{\mathrm{r}}=\mathrm{R}\)
For a disc, \(\frac{\mathrm{MR}^2}{2}=\mathrm{MK}_{\mathrm{d}}^2 \quad \therefore \mathrm{K}_{\mathrm{d}}=\frac{\mathrm{R}}{\sqrt{2}}\)
\(\frac{\mathrm{K}_{\mathrm{r}}}{\mathrm{K}_{\mathrm{d}}}=\sqrt{2}\)
For a ring, \(\mathrm{MR}^2=\mathrm{MK}_{\mathrm{r}}{ }^2 \quad \therefore \mathrm{K}_{\mathrm{r}}=\mathrm{R}\)
For a disc, \(\frac{\mathrm{MR}^2}{2}=\mathrm{MK}_{\mathrm{d}}^2 \quad \therefore \mathrm{K}_{\mathrm{d}}=\frac{\mathrm{R}}{\sqrt{2}}\)
\(\frac{\mathrm{K}_{\mathrm{r}}}{\mathrm{K}_{\mathrm{d}}}=\sqrt{2}\)
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