A thin bar magnet oscillating in a uniform magnetic field has a...

Let the initial mass, length, moment of inertia, and magnetic moment of the bar magnet be $m$, $L$, $I$, and $M$ respectively.
The initial time period is $T = 2\pi\sqrt{\dfrac{I}{MB}} = 4\text{ s}$

When the magnet is cut into two equal halves perpendicular to its length, the mass of each half becomes $m' = \dfrac{m}{2}$ and the length becomes $L' = \dfrac{L}{2}$.

The new moment of inertia for one half is:
$I' = \dfrac{m'(L')^2}{12} = \dfrac{\left(\dfrac{m}{2}\right)\left(\dfrac{L}{2}\right)^2}{12} = \dfrac{1}{8}\left(\dfrac{mL^2}{12}\right) = \dfrac{I}{8}$

The pole strength remains unchanged, but the magnetic length is halved. Thus, the new magnetic moment is:
$M' = M \times \dfrac{1}{2} = \dfrac{M}{2}$

The new time period $T'$ is:
$T' = 2\pi\sqrt{\dfrac{I'}{M'B}} = 2\pi\sqrt{\dfrac{\dfrac{I}{8}}{\dfrac{M}{2}B}} = 2\pi\sqrt{\dfrac{I}{4MB}} = \dfrac{1}{2} \left(2\pi\sqrt{\dfrac{I}{MB}}\right)$

$T' = \dfrac{T}{2} = \dfrac{4}{2} = 2\text{ s}$

Answer: $2\text{ s}$