A) 1 s
B) 2 s
C) 3 s
D) 4 s.
Correct Answer: D
Solution :
[d]: The time period T of oscillation of a magnet is given by\[T=2\pi \sqrt{\frac{I}{MB}}\] As I and M remain the same, \[\therefore \]\[T\propto \frac{1}{\sqrt{B}}\]or\[\frac{{{T}_{2}}}{{{T}_{1}}}=\sqrt{\frac{{{B}_{1}}}{{{B}_{2}}}}\] According to given problem, \[{{B}_{1}}=24\mu T,{{B}_{2}}=24\mu T-18\mu T=6\mu T,{{T}_{1}}=2s\] \[\therefore \]\[{{T}_{2}}=(2s)\sqrt{\frac{24\mu T}{6\mu T}}=4s\]You need to login to perform this action.
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