A) 4s
B) 2s
C) \[\sqrt{2}\,s\]
D) 1 s
Correct Answer: D
Solution :
Time period of a freely suspended magnet \[T=2\pi \sqrt{\frac{I}{MH}}\] where I = moment of inertia M = magnetic dipole moment H= horizontal component of earth's magnetic field Given' \[T=2\pi \sqrt{\frac{I}{MH}}=2s\] ?.(i) When magnet is broken into two parts, then \[I'=\frac{\frac{m}{2}{{\left( \frac{l}{2} \right)}^{2}}}{12}=\frac{1}{8}.\frac{m{{l}^{2}}}{12}=\frac{I}{8}\] \[\therefore \] \[T'=2\pi \sqrt{\frac{(I/8)}{(M/2)H}}=2\pi \sqrt{\frac{I}{4MH}}\] \[=\frac{T}{2}=\frac{2}{2}=1s\]
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