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(B) The period of oscillation of a magnet in a magnetic field is given by $T = 2\pi \sqrt{\frac{I}{MB}},$ where $I$ is the moment of inertia,$M$ is th

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$1/2$

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(B) The period of oscillation of a magnet in a magnetic field is given by $T = 2\pi \sqrt{\frac{I}{MB}},$ where $I$ is the moment of inertia,$M$ is the magnetic dipole moment,and $B$ is the magnetic field.For the original magnet of mass $m$ and length $\ell,$ $I = \frac{m\ell^2}{12}$ and magnetic moment is $M.$When the magnet is cut into two equal halves along its length,the new length is $\ell' = \ell/2$ and the new mass is $m' = m/2.$The new magnetic moment is $M' = \frac{M \cdot \ell'}{\ell} = M/2.$The new moment of inertia is $I' = \frac{m'(\ell')^2}{12} = \frac{(m/2)(\ell/2)^2}{12} = \frac{m\ell^2}{12 	imes 8} = \frac{I}{8}.$Now,the new period $T'$ is given by $T' = 2\pi \sqrt{\frac{I'}{M'B}} = 2\pi \sqrt{\frac{I/8}{(M/2)B}} = 2\pi \sqrt{\frac{I}{4MB}} = \frac{1}{2} 	imes 2\pi \sqrt{\frac{I}{MB}} = \frac{T}{2}.$Therefore,the ratio $T'/T = 1/2.$

$A$ thin rectangular magnet suspended freely has a period of oscillation equal to $T.$ Now it is broken into two equal halves (each having half of the original length) and one piece is made to oscillate freely in the same field. If its period of oscillation is $T',$ the ratio $T'/T$ is :-

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