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:

$\frac{T_o}{3}$

The correct answer is Option (3) → $\frac{T_o}{3}$

$T_o$, Initial time period of Magnet = $2π\sqrt{\frac{I}{MB}}$

where,

I = Moment of inertia

M = Magnetic moment of Bar Magnet

B = Magnetic field

When the Magnet is cut into 3 pieces,

$M'=\frac{M}{3}$ and $I'=\frac{I}{3^3}=\frac{I}{27}$

$T_f=2π\sqrt{\frac{\frac{I}{27}}{\frac{M}{3}×B}}=2π\sqrt{\frac{3}{27}\frac{I}{MB}}$

$=\frac{2πI}{MB}×\frac{1}{3}$

$=\frac{T_o}{3}$