A thin magnetic iron rod of length $30 \mathrm{~cm}$ is suspended in a uniform magnetic field. Its time…
A thin magnetic iron rod of length $30 \mathrm{~cm}$ is suspended in a uniform magnetic field. Its time period of oscillation is $4 \mathrm{~s}$. It is broken into three equal parts. The time period in seconds of oscillation of one part when suspended in the same magnetic field is
$\frac{1}{\sqrt{3}}$
$\frac{2}{\sqrt{3}}$
$\sqrt{3}$
$\frac{4}{\sqrt{3}}$
Solution
Time period of magnet
$
T=2 \pi \sqrt{\frac{I}{M H}}
$
$
M=\text { magnetic moment }=m \times l
$
$
I=\text { moment of inertia }=\frac{m l^2}{12}
$
When magnet is broken into three equal parts, Then, $M^{\prime}=m \times \frac{l}{3}=\frac{M}{3}$
$
I^{\prime}=\frac{m(l / 3)^2}{12}=\frac{m l^2}{9 \times 12}=\frac{I}{9}
$
$\therefore \quad$ Now, time period $T^{\prime}=2 \pi \sqrt{\frac{I^{\prime}}{M^{\prime} H}}$
$\begin{aligned} & =2 \pi \frac{\sqrt{I / 9}}{\frac{M}{3} H} \\ & =2 \pi \frac{\sqrt{I}}{M H \times 3} \\ & =\frac{T}{\sqrt{3}}=\frac{4}{\sqrt{3}} \mathrm{sec} .\end{aligned}$