A) \[4T\]
B) \[T/4\]
C) \[T/2\]
D) \[T\]
Correct Answer: C
Solution :
When magnet of length l is cut into four equal parts. then \[{m}'=\frac{m}{2}\] and \[{l}'=\frac{l}{2};\ \ \therefore \ {M}'=\frac{m}{2}\times \frac{l}{2}=\frac{ml}{4}=\frac{M}{4}\] New moment of inertia \[{I}'=\frac{\text{w}{{l}^{2}}}{12}=\frac{\frac{\text{w}}{4}.{{\left( \frac{1}{2} \right)}^{2}}}{12}=\frac{1}{16}.\frac{\text{w}{{l}^{2}}}{12}\] Here w is the mass of magnet. \[\therefore \,{I}'=\frac{1}{16}I\]; Time period of each part \[{T}'=2\pi \sqrt{\frac{{{I}'}}{{M}'{{B}_{H}}}}\] \[=2\pi \sqrt{\frac{I/16}{(M/4){{B}_{H}}}}=2\pi \sqrt{\frac{I}{4M{{B}_{H}}}}=\frac{T}{2}\]
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