A) \[2s\]
B) \[2/3\text{ }s\]
C) \[2\sqrt{3}\text{ }s\]
D) \[2/\sqrt{3}\text{ }s\]
Correct Answer: B
Solution :
The time period of oscillations of magnet \[T=2\pi \sqrt{\left( \frac{I}{MH} \right)}\] where,\[l=\]moment of inertia of magnet\[=\frac{m{{L}^{2}}}{12}\] (\[\because \]magnet can be consider as a thin rod of length L and m, being the mass of magnet) \[M=pole\text{ }strength\times L\] and H = horizontal component of the earth's magnetic field When the three equal parts of magnet are placed on one another with their like poles together, then the new moment of inertia, \[l'=\frac{1}{12}\left( \frac{m}{3} \right)\times {{\left( \frac{L}{3} \right)}^{2}}\times 3=\frac{1}{12}\frac{m{{L}^{{}}}}{9}=\frac{l}{9}\] and the new magnetic moment, \[M'=\]pole strength \[\times \frac{L}{3}\times 3=\]pole strength \[t\times L\] \[=M\] Hence, \[T'=2\pi \sqrt{\left( \frac{l/9}{MH} \right)}\] \[\Rightarrow \] \[T'=\frac{1}{3}\times T\] \[T'=\frac{2}{3}s\]
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