A) \[{{2}^{1/4}}\]
B) \[{{2}^{1/2}}\]
C) 2
D) \[{{2}^{3/4}}\]
Correct Answer: C
Solution :
Initially magnetic moment of system \[{{M}_{1}}=\sqrt{{{M}^{2}}+{{M}^{2}}}=2M\] 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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