A) \[\frac{1}{2\sqrt{2}}\]
B) \[\frac{1}{2}\]
C) 2
D) \[\frac{1}{4}\]
Correct Answer: B
Solution :
When magnet is divided into two equal parts, the magnetic dipole moment M? = pole strength \[\times \frac{l}{2}=\frac{M}{2}\] (pole strength remains same) Also, the mass of magnet becomes half, then \[m'=\frac{m}{2}\] Moment of inertia of magnet \[l=\frac{m{{l}^{2}}}{12}\] (\[\because \] magnet is equivalent to thin rod of length \[l\]) New moment of inertia \[l'=\frac{1}{12}\left( \frac{m}{2} \right){{\left( \frac{l}{2} \right)}^{2}}=\frac{m{{l}^{2}}}{12\times 8}\] \[\left( \because \,m'=\frac{m}{2}\,and\,l'=\frac{l}{2} \right)\] \[\therefore \] \[l'=\frac{l}{8}\] Now, \[T=2\pi \sqrt{\left( \frac{l}{MB} \right)}\] \[T'=2\pi \sqrt{\left( \frac{l'}{M'B} \right)}=2\pi \sqrt{\left( \frac{l/8}{MB/2} \right)}\] \[\therefore \] \[T'=\frac{T}{2}\,\,\,\,\Rightarrow \,\,\,\frac{T'}{T}=\frac{1}{2}\]
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