A) \[\sqrt{4}\,\,:\,\,\sqrt{3}\]
B) \[\sqrt{3}\,\,:\,\,\sqrt{4}\]
C) \[\sqrt{3}\,\,:\,\,\sqrt{2}\]
D) \[\sqrt{2}\,\,:\,\,\sqrt{3}\]
Correct Answer: A
Solution :
\[{{K}_{disc}}=\frac{1}{2}m\nu _{d}^{2}\left( 1+\frac{{{k}^{2}}}{{{R}^{2}}} \right)\,\,=\,\,\frac{3}{4}m\nu _{d}^{2}\] \[\left[ As\,\,\frac{{{k}^{2}}}{{{R}^{2}}}=\frac{1}{2}\,\,\,\,\,\,\,\,\,for\,\,disc \right]\] \[{{K}_{ring}}=\frac{1}{2}m{{\nu }_{r}}\left( 1+\frac{{{k}^{2}}}{{{R}^{2}}} \right)=m\nu _{r}^{2}\] \[\left[ As\,\,\frac{{{k}^{2}}}{{{R}^{2}}}=1\,\,\,\,\,\,\,\,\,\,for\,\,ring \right]\] According to problem \[{{K}_{disc}}=\,\,{{K}_{ring}}\] \[\Rightarrow \,\,\,\,\,\,\frac{3}{4}m\nu _{d}^{2}=m\nu _{r}^{2}\,\,\,\,\,\,\,\,\Rightarrow \,\,\,\,\frac{{{v}_{d}}}{{{v}_{r}}}=\sqrt{\frac{4}{3}}\]
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