A) 2 s
B) 4 s
C) 8 s
D) None of these
Correct Answer: B
Solution :
As, \[T=2\pi \sqrt{\left( \frac{I}{M{{B}_{H}}} \right)}\]where, I = moment of inertia of magnet M = magnetic moment of magnet \[{{\operatorname{B}}_{H}}=earths horizontal component\] When it is broken \[I=\frac{I}{2}\] and mass will be \[\frac{m}{2}\] (If m is total mass) Now, after cutting \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,M'=\frac{M}{2}\] [m is tide strength] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,M'=\frac{M}{2}\] and \[I'=\frac{1}{12}M{{l}^{2}}[Thin\,\,bar]\] after breaking \[I'=\frac{1}{12}\left( \frac{M}{2} \right)\left( \frac{{{l}^{2}}}{4} \right)=\frac{l}{8}\] \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,T'=2\,\pi \frac{\sqrt{(l/8)}}{(M/2){{B}_{H}}}\] \[=\,\,\,\frac{T}{2}=\frac{8}{2}=4\,s\]
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