A) 1 : 2
B) \[\sqrt{5}\,:\,\sqrt{6}\]
C) 2 : 3
D) \[\frac{\sqrt{3}}{2}\]
Correct Answer: D
Solution :
Radius of gyration\[K=\sqrt{\frac{I}{m}}\] (where J is moment of inertia) \[{{K}_{disc}}=\sqrt{\frac{\frac{1}{2}m{{R}^{2}}+m{{R}^{2}}}{m}}=\sqrt{\frac{3}{2}}R\] \[{{K}_{ring}}=\sqrt{\frac{m{{R}^{2}}+m{{R}^{2}}}{m}}=\sqrt{2}R\] \[\therefore \] \[\frac{{{K}_{disc}}}{{{K}_{ring}}}=\frac{\sqrt{\frac{3}{2}}}{\sqrt{2}}=\frac{\sqrt{3}}{2}\]
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