As we know that The time period of oscillation is,
$$
\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{I}}{\mathrm{MB}}}
$$
where,
$\mathrm{M}=$ magnetic moment of the magnet
$\mathrm{B}=$ uniform magnetic field in which magnet is oscillating. if a magnet of magnetic moment $\mathrm{M}$ is cut into $n$ equal parts then magnetic moment $\mathrm{M}$ of all equal parts is $\mathrm{nM}^{\prime}=\mathrm{M}$ Magnetic dipole moment for two parts of magnet $\left(M^{\prime}=M / 2\right)$ So, magnetic moment of each parts of magnet is $\mathrm{M}^{\prime}=\mathrm{M} / 2$ Here, moment of inertia $\mathrm{I}=\frac{\mathrm{m} l^2}{12}$.
When magnet is cut into two equal pieces, perpendicular to length, then moment of inertia of each piece of magnet about an axis perpendicular to length passing through its centre is
$$
\mathrm{I}^{\prime}=\frac{m^{\prime} d^{\prime 2}}{12}
$$
$$
\begin{aligned}
&\mathrm{I}^{\prime}=\frac{\mathrm{m}}{2} \frac{(l / 2)^2}{12} \quad \quad\left(\because l^{\prime}=\frac{l}{2}\right) \\
&\mathrm{I}^{\prime}=\frac{\mathrm{m} l^2}{12} \times \frac{1}{8}=\frac{\mathrm{I}}{8} \quad\left[\because \mathrm{I}=\frac{m l^2}{12}\right]
\end{aligned}
$$
So, $\quad I^{\prime}=\frac{\mathrm{I}}{8}$
Magnetic dipole moment
$$
M^{\prime}=\mathrm{M} / 2
$$
Its time period of oscillation is
$$
\begin{gathered}
\mathrm{T}^{\prime}=2 \pi \sqrt{\frac{\mathrm{I}^{\prime}}{\mathrm{M}^{\prime} \mathrm{B}}}=2 \pi \sqrt{\frac{\mathrm{I} / 8}{\mathrm{M} / 2) \mathrm{B}}}=\frac{2 \pi}{2} \sqrt{\frac{\mathrm{I}}{\mathrm{MB}}} \\
\mathrm{T}^{\prime}=\frac{\mathrm{T}}{2}
\end{gathered}
$$