Question

The time period of oscillation of a bar magnet in an internal magnetic field is $T$. If it is cut into two parts along its length & then along its breadth (Total four parts), its new time period is (Ignore earth's magnetic field).

Answer 1

We know that, $T=2\pi \sqrt{\frac{I}{MB}}$.

When it is cut into four parts along length & breadth
$M^{\prime }=\frac{M}{4}$

& new $I^{\prime }=\frac{1}{12}\left(\frac{m}{4}\right){\left(\frac{L}{2}\right)}^2=\frac{I}{16}$

Where $m$ is the mass of the bar magnet.

Hence, $T^{\prime }=2\pi \sqrt{\frac{I}{16\times \frac{M}{4}B}}=\frac{T}{2}$

So, $\left(C\right)$ is the correct answer.