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(C) The time period of oscillation of a magnetic dipole in a uniform magnetic field is given by $T = 2\pi \sqrt{\frac{I}{MB}}$,where $I$ is the moment

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$\frac{1}{2}$

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(C) The time period of oscillation of a magnetic dipole in a uniform magnetic field is given by $T = 2\pi \sqrt{\frac{I}{MB}}$,where $I$ is the moment of inertia,$M$ is the magnetic moment,and $B$ is the magnetic field.For the original magnet of mass $m$ and length $l$,$I = \frac{ml^2}{12}$ and $M = m_s l$ (where $m_s$ is pole strength).When the magnet is cut into two equal halves along its length,for each piece: mass $m' = \frac{m}{2}$,length $l' = \frac{l}{2}$,and pole strength $m_s' = m_s$.New moment of inertia $I' = \frac{m' (l')^2}{12} = \frac{(m/2) (l/2)^2}{12} = \frac{1}{8} \left( \frac{ml^2}{12} \right) = \frac{I}{8}$.New magnetic moment $M' = m_s l' = m_s (l/2) = \frac{M}{2}$.The new time period $T' = 2\pi \sqrt{\frac{I'}{M'B}} = 2\pi \sqrt{\frac{I/8}{(M/2)B}} = 2\pi \sqrt{\frac{I}{4MB}} = \frac{1}{2} \left( 2\pi \sqrt{\frac{I}{MB}} \right) = \frac{T}{2}$.Therefore,the ratio $\frac{T'}{T} = \frac{1}{2}$.

$A$ 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 $T'$,then the ratio $\frac{T'}{T}$ is

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