A) 1 s
B) 2 s
C) 3 s
D) 4 s
Correct Answer: D
Solution :
The time period T of oscillation of a magnet is given by \[T=2\pi \,\sqrt{\frac{l}{MB}}\] where, l = moment of inertia of the magnet about axis of rotation \[\operatorname{M} = magnetic moment of the magnet\] \[\operatorname{B}= uniform magnetic field\] As the l, B remain the same \[\therefore \,\,\,\,\,\,\,\,T\propto \frac{1}{\sqrt{B}}\,\,or\,\,\frac{{{T}_{2}}}{{{T}_{1}}}=\sqrt{\frac{{{B}_{1}}}{{{B}_{2}}}}\] Given that \[{{\operatorname{B}}_{1}}=24\mu T,\,\,{{B}_{2}}=\,\,24\,\mu T-\,\,18\mu T\,\,=\,\,6\mu T\] \[{{T}_{1}}=\,\,2\,s\] \[\therefore \,\,\,\,\,\,{{T}_{2}}=(2s)\sqrt{\frac{24\,\mu T}{6\,\mu T}}=4\,s\]
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