Question

A thin magnetic iron rod of length $30cm$ is suspended in a uniform magnetic field. Its time period of oscillation is $4s$. 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

• A. $\frac{4}{3}$
• B. $\sqrt{3}$
• C. $\frac{1}{\sqrt{3}}$
• D. $\frac{2}{\sqrt{3}}$

Answer 1

The correct option is A $\frac{4}{3}$

when a magnet is cut lengthwise, the pole strength remains same
$m=$ pole strength of original magnet
$u=mL=$magnetic moment of original magnet

for each cut part :
Pole strength $=m$
Magnetic moment of each part

$=m\frac{L}{3}=\frac{u}{3}$

(magnet cut in 3 equal parts)

let $M=$ moment of inertia of original magnet about perpendicular bisector axis

$I=\frac{\left(M\right){\left(L\right)}^2}{12}$

Let $I^{\prime }$ be the moment of inertia of each cut part

$I^{\prime }=\frac{\left(\frac{M}{3}\right){\left(\frac{L}{3}\right)}^2}{12}=\frac{1}{3^3}.\frac{M{L}^2}{12}=\frac{I}{3^3}$

for original magnet

\(T=2\pi \sqrt{\dfrac{I}}{\left ( \dfrac{\mu } ) \=\dfrac{1}{3}\times T={4}s\)

for one cut part, New time period

$T=2\pi \sqrt{\frac{\frac{I}{27}}{\left(\frac{\mu }{3}\right)\left(B_H\right)}}=\frac{1}{3}\times T=\frac{4}{3}s$


Hence option (d) is correct.