The length $l$ of a magnet is large compared to its width and breadth. The time period of its oscillation in a vibration magnetometer is $2 \mathrm{s}$. The magnet is cut into three equal parts of length $\frac{l}{3}$ each. If these parts are placed on each other with their like poles together, then the time period of this combination is

Given: Time period of oscillation of magnet in a vibration magnetometer $T=2 \mathrm{s}$

Time period of oscillation of the magnet is given as,$T=2\pi \sqrt{\frac{I}{MB}} ...\left(1\right)$

Where I is the moment of inertia of magnet $I=\frac{m{l}^2}{12}$

where l is the length of the magnet

M is the magnetic moment $M=m_p\times l$

here $m_p$ is the magnetic pole strength.

Putting the above value in equation (i), we get

$T=2\pi \sqrt{\frac{m{l}^2}{12\times m_p\times l\times B}}$

$T \alpha \sqrt{\frac{l}{m_p}}$

In the second case, the cut parts are placed on each other with their like poles together.

In the second situation, the length is decreased by $3$ times, and pole strength is increased by $3$ times, so the time period will decrease by $3$ times by using the above relation.

Thus, the new time period will be $=\frac{T}{3}=\frac{2}{3} \mathrm{s}$