Explanation

(b) When magnet is divided into two equal parts, the magnetic dipole moment
$\ \ \ \ \ \ \ \ \ \ M^{'} =$ pole strength$\times \frac{l}{2} = \frac{M}{2}$
(pole strength remains same)
Also, the mass of magnet becomes half, ie,
$\ \ \ \ \ \ \ \ \ \ m^{'} = \frac{m}{2}$
Moment of inertia of magnet
$\ \ \ \ \ \ \ \ \ \ \ \ \ I = \frac{ml^{2}}{12}$
New moment of ineria
$\ \ \ \ \ \ \ \ \ \ \ \ I^{'} = \frac{1}{12}\left( \frac{m}{2} \right)\left( \frac{l}{2} \right)^{2} = \frac{ml^{2}}{12 \times 8}$
$\therefore\ \ \ \ \ \ \ \ I^{'} = \frac{I}{8}$
Now, $T = 2\pi\sqrt{\left( \frac{I}{MB_{H}} \right)}$
$\ \ \ \ \ \ \ \ T^{'} = 2\pi\sqrt{\left( \frac{I^{'}}{M^{'}B_{H}} \right)}\ = 2\pi\sqrt{\left( \frac{I/8}{MB_{H}/2\ } \right)\ }\ $
$\therefore\ \ \ T^{'} = \frac{T}{2} \Longrightarrow \frac{T^{'}}{T} = \frac{1}{2}$