1. Time period of a magnet in a vibration magnetometer

   The time period is $T = 2\pi\sqrt{\frac{I}{MB_H}}$ where:

   • $I$ = moment of inertia of the magnet about the suspension axis,
   • $M$ = magnetic dipole moment,
   • $B_H$ = horizontal component of earth's magnetic field.

   Since $B_H$ is unchanged, $T \propto \sqrt{\frac{I}{M}}$

2. Original magnet

   Let:

   • mass of original magnet = $m$
   • length of original magnet = $L$
   • magnetic moment = $M$

   For a long bar magnet oscillating about a vertical axis through its center, $I = \frac{1}{12}mL^2$

   Given original time period: $T_1 = 2\,s$

3. Magnet cut along its length into 3 equal parts

   Cutting along length means:

   • length remains $L$
   • cross-sectional area becomes $\frac{1}{3}$ of original
   • hence mass of each piece becomes $\frac{m}{3}$
   • pole strength of each piece becomes $\frac{1}{3}$ of original
   • magnetic moment of each piece becomes $M' = \frac{M}{3}$

4. Three parts placed on each other with like poles together

   When stacked together with like poles together:

   • effective magnetic moment adds: $M_{\text{new}} = 3\times \frac{M}{3} = M$

   Also, the combination behaves like a single magnet of:

   • same length $L$
   • total mass $m$

   Therefore its moment of inertia is $I_{\text{new}} = \frac{1}{12}mL^2 = I$

5. New time period

   $T_2 \propto \sqrt{\frac{I_{\text{new}}}{M_{\text{new}}}} = \sqrt{\frac{I}{M}}$

   Hence, $T_2 = T_1 = 2\,s$

6. Option check

   • A: $2\sqrt3\,s$ ❌
   • B: $\frac{2}{3}\,s$ ❌
   • C: $2\,s$ ✅
   • D: $\frac{2}{\sqrt3}\,s$ ❌

Therefore, the correct answer is: $\boxed{2\,s}$ which corresponds to Option C.