A) \[\frac{1}{2\sqrt{2}}\]
B) \[\frac{1}{2}\]
C) 2
D) \[\frac{1}{4}\]
Correct Answer: B
Solution :
| The time period of a rectangular magnet oscillating in earth’s magnetic field is given by \[\operatorname{T}=2\pi \sqrt{\frac{I}{\mu {{B}_{H}}}}\] |
| where I = Moment of inertia of the rectangular magnet |
| \[\mu = Magnetic moment\] |
| \[{{\operatorname{B}}_{H}}\,\,=\] Horizontal component of the earth’s magnetic field |
| Case 1: \[\operatorname{T}=\,\,2\pi \sqrt{\frac{I}{\mu {{B}_{H}}}}\, where\,\,\,\,I=\frac{1}{12}\,M{{\ell }^{2}}\] |
| Case 2: Magnet is cut into two identical pieces such that each piece has half the original length. Then |
| \[T'=2\pi \sqrt{\frac{I'}{\mu '\,{{B}_{H}}}}\] |
| where \[I'=\frac{1}{12}\left( \frac{M}{2} \right){{\left( \frac{\ell }{2} \right)}^{2}}=\frac{I}{8}\,\,and\,\,\mu '=\frac{\mu }{2}\] |
| \[\therefore \,\,\,\frac{T'}{T}=\sqrt{\frac{I'}{\mu }\times \frac{\mu }{l}}\,\,\,=\,\,\,\sqrt{\frac{I/8}{\mu /2}\times \frac{\mu }{I}}=\sqrt{\frac{1}{4}}\,\,=\,\,\frac{1}{2}\] |
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