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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 and $M$ is

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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 and $M$ is the magnetic dipole moment.For the original magnet of mass $m$ and length $\ell$,$I = \frac{m\ell^2}{12}$ and $M = q_m \ell$ (where $q_m$ is pole strength).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 pole strength $q_m$ remains the same.New magnetic moment $M' = q_m \ell' = q_m (\ell/2) = M/2$.New moment of inertia $I' = \frac{m'(\ell')^2}{12} = \frac{(m/2)(\ell/2)^2}{12} = \frac{m\ell^2}{12 	imes 8} = I/8$.The new period of oscillation is $T' = 2\pi \sqrt{\frac{I'}{M'B}} = 2\pi \sqrt{\frac{I/8}{(M/2)B}} = 2\pi \sqrt{\frac{1}{4} \cdot \frac{I}{MB}} = \frac{1}{2} T$.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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