MHT CET · Physics · Rotational Motion
The ratio of radii of gyration of a ring to a disc (both circular) of same radii and mass, about a tangential axis perpendicular to the plane is
- A \(\frac{2}{\sqrt{3}}\)
- B \(\frac{\sqrt{2}}{1}\)
- C \(\frac{\sqrt{3}}{\sqrt{2}}\)
- D \(\frac{2}{\sqrt{5}}\)
Answer & Solution
Correct Answer
(A) \(\frac{2}{\sqrt{3}}\)
Step-by-step Solution
Detailed explanation
\(I_{\text {Ring }}=M R^{2}+M h^{2}=M R^{2}+M R^{2}=2 M R^{2}\)
\(I_{\text {Disc }}=\frac{1}{2} M R^{2}+M h^{2}=\frac{1}{2} M R^{2}+M R^{2}=\frac{3}{2} M R^{2}\)
\(\therefore k_{\text {ring }}=\sqrt{2} R\)
\(\quad k_{\text {disc }}=\sqrt{\frac{3}{2}} R\)
\(\therefore \frac{k_{\text {ring }}}{k_{\text {disc }}}=\frac{\sqrt{2} R}{\sqrt{3}}=\frac{2}{\sqrt{3}} R\)
\(I_{\text {Disc }}=\frac{1}{2} M R^{2}+M h^{2}=\frac{1}{2} M R^{2}+M R^{2}=\frac{3}{2} M R^{2}\)
\(\therefore k_{\text {ring }}=\sqrt{2} R\)
\(\quad k_{\text {disc }}=\sqrt{\frac{3}{2}} R\)
\(\therefore \frac{k_{\text {ring }}}{k_{\text {disc }}}=\frac{\sqrt{2} R}{\sqrt{3}}=\frac{2}{\sqrt{3}} R\)
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