Time period of oscillation of a bar magnet in a uniform magnetic field is T_o. The magnet is cut into 3 equal parts transverse to its length. The new time period of oscillation of one part will be:

Let us understand how the time period of oscillation of a bar magnet changes when it is cut. The time period of oscillation $T_o$ of a bar magnet of magnetic moment $M$ and moment of inertia $I$ in a uniform magnetic field $B$ is given by the formula:

$T_o=2\pi \sqrt{\frac{I}{MB}}$

Let the original mass of the bar magnet be $m$ and its original length be $L$. The moment of inertia of a uniform bar magnet about an axis passing through its center and perpendicular to its length is:

$I=\frac{m{L}^2}{12}$

When the magnet is cut into 3 equal parts transverse to its length (perpendicular to its length):

1. The mass of each new part becomes:
$m^{\prime }=\frac{m}{3}$

2. The length of each new part becomes:
$L^{\prime }=\frac{L}{3}$

3. The new moment of inertia $I^{\prime }$ of one of these parts is:
$I^{\prime }=\frac{m^{\prime }{\left(L^{\prime }\right)}^2}{12}=\frac{\left(\frac{m}{3}\right){\left(\frac{L}{3}\right)}^2}{12}=\frac{1}{27}\left(\frac{m{L}^2}{12}\right)=\frac{I}{27}$

4. Since the magnetic pole strength remains the same when cut transversely, the new magnetic moment $M^{\prime }$ of one part is proportional to its length:
$M^{\prime }=\frac{M}{3}$

Now, let us calculate the new time period of oscillation $T^{\prime }$ for one of the cut parts in the same magnetic field $B$:

$T^{\prime }=2\pi \sqrt{\frac{I^{\prime }}{M^{\prime }B}}$

Substituting the values of $I^{\prime }$ and $M^{\prime }$:

$T^{\prime }=2\pi \sqrt{\frac{\frac{I}{27}}{\frac{M}{3}B}}=2\pi \sqrt{\frac{I}{MB}·\frac{3}{27}}$

$T^{\prime }=2\pi \sqrt{\frac{I}{MB}·\frac{1}{9}}=\frac{1}{3}\left(2\pi \sqrt{\frac{I}{MB}}\right)=\frac{T_o}{3}$

Thus, the new time period of oscillation of one part will be $T_o/3$.