A) 8 s
B) 2 s
C) 1 s
D) 4 s
Correct Answer: C
Solution :
Time period of magnet in vibration magnetometer \[T=2\pi \sqrt{\frac{I}{M{{B}_{H}}}}\] where, \[I=\]moment of inertia of magnet \[\text{M =}\]magnetic moment \[{{\text{B}}_{\text{H}}}\text{=}\]horizontal component of the earths magnetic field. \[\therefore \] \[\frac{{{T}_{1}}}{{{T}_{2}}}=\sqrt{\frac{{{I}_{1}}}{{{I}_{2}}}\times \frac{{{M}_{2}}}{{{M}_{1}}}}\] or \[\frac{8}{{{T}_{2}}}=\sqrt{\frac{\frac{m{{l}^{2}}}{12}\times M/8}{m/8{{(l/8)}^{2}}/12\times M}}=\sqrt{\frac{8\times {{8}^{2}}}{8}}=8\] or \[{{T}_{2}}=\frac{8}{8}=1s\]You need to login to perform this action.
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