Q.A rectangular magnet suspended freely has a period of oscillation equal to T. Now it is broken into two equal halves (each having half of the original length) and one piece is made to oscillate freely. Its period of oscillation is $T_c$. The ratio $T_c/T$ is (A) $\dfrac{1}{2\sqrt2}$ (B) $\dfrac12$ (C) $\sqrt2$ (D) $\dfrac14$

[!TLDR] Halving the length and mass makes $I'=I/8$ and $m'=m/2$, so $T'=T\sqrt{(I'/m')/(I/m)}=T …

For a uniform rectangular bar magnet of mass M and length L, oscillating about a transverse axis through its centre, the moment of inertia is $I=\dfrac{ML^2}{12}$, and the magnetic moment is $m=q_mL$ (pole strength times length). When the magnet is broken into two equal halves along its length, each half has mass $M/2$ and length $L/2$: its moment of inertia (about its OWN centre, the new axis of oscillation) is $I'=\dfrac{(M/2)(L/2)^2}{12}=\dfrac{ML^2}{96}=\dfrac{I}{8}$, and its magnetic moment is $m'=q_m(L/2)=\dfrac{m}{2}$ (the pole strength $q_m$ itself is unchanged, since it depends on the material's magnetisation and cross-se …

Work out how the moment of inertia ($I\propto ML^2$) and the magnetic moment ($m\propto L$) each individually scale when the magnet is halved in both mass and leng …

Assuming the period is unchanged, or assuming it scales the same way as the length alone, instead of carefully tracking how BOTH I (which drops by a facto …