Explanation

(b) From $T = 2\pi\sqrt{\frac{I}{MB}},4 = 2\pi\sqrt{\frac{I}{MB}}$.
When it is cut into two equal parts in length, mass of each part becomes $1/2$, $\ I = mass\ \frac{(length)^{2}}{12}$ becomes $\frac{1}{8}th$ and $M\ becomes\frac{1}{\ 2}.$
$T^{'} = 2\pi\sqrt{\frac{I/8}{(M/2)B}}$
$\therefore\ \ = \frac{1}{2}\left( 2\pi\sqrt{\frac{I}{MB}} \right)$
$= \frac{1}{2}T = 2$s