A) 1 : 2
B) \[\sqrt{2}\,:1\]
C) 2 : 1
D) \[1:\sqrt{2}\]
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 \[{{I}_{1}}=I,\,\,{{I}_{2}}=2I\] \[\therefore \frac{{{L}_{1}}}{{{L}_{2}}}=\sqrt{\frac{I}{2I}}=\frac{1}{\sqrt{2}}\] or \[{{L}_{1}}:{{L}_{2}}=1:\sqrt{2}\]
You need to login to perform this action.
You will be redirected in
3 sec
