A) \[{{2}^{1/4}}\]
B) \[{{2}^{1/2}}\]
C) 2
D) \[{{2}^{3/4}}\]
Correct Answer: C
Solution :
[c] Initially magnetic moment of system \[{{M}_{1}}=\sqrt{{{M}^{2}}+{{M}^{2}}}=\sqrt{2M}\] and moment of inertia \[{{I}_{1}}=I+I=2I\]. Finally when one of the magnet is removed then \[{{M}_{2}}=M\] and \[{{I}_{2}}=I\] So, \[T=2\pi \sqrt{\frac{I}{M\,\,\,{{B}_{H}}}}\] \[\frac{{{T}_{1}}}{{{T}_{2}}}=\sqrt{\frac{{{I}_{1}}}{{{I}_{2}}}\times \frac{{{M}_{2}}}{{{M}_{1}}}}=\sqrt{\frac{2I}{I}\times \frac{M}{\sqrt{2}M}}\] \[\Rightarrow \,\,\,{{T}_{2}}=\frac{{{2}^{5/4}}}{{{2}^{1/4}}}=2\,\sec \]
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