A) 1 : 2
B) 72 : 1
C) 2 : 1
D) 1 : 72
Correct Answer: D
Solution :
As said,\[{{(KE)}_{rot}}\]remains same. i.e., \[\frac{1}{2}{{I}_{1}}\omega _{1}^{2}=\frac{1}{2}{{I}_{2}}\omega _{2}^{2}\]? \[\Rightarrow \]\[\frac{1}{2{{I}_{1}}}{{({{I}_{1}}{{\omega }_{1}})}^{2}}=\frac{1}{2{{I}_{2}}}{{({{I}_{2}}{{\omega }_{2}})}^{2}}\] \[\Rightarrow \]\[\frac{L_{1}^{2}}{{{I}_{1}}}=\frac{L_{2}^{2}}{{{I}_{2}}}\] \[\Rightarrow \]\[\frac{{{L}_{1}}}{{{L}_{2}}}=\sqrt{\frac{{{I}_{1}}}{{{I}_{2}}}}\] but \[{{I}_{1}}=I,\,{{I}_{2}}=2I\] \[\therefore \]\[\frac{{{L}_{1}}}{{{L}_{2}}}=\sqrt{\frac{I}{2I}}=\frac{1}{2}\] or \[{{L}_{1}}:{{L}_{2}}=1:\sqrt{2}\]
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