Question

A thin rectangular magnet suspended freely has a period of oscillation equal to T. Now it is broken into two equal halves (each having half of the original length) and one piece is made to oscillate freetly in the same field. If its period of oscillation is T', then ratio $\frac{T^{\prime }}{T}$ is

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

Answer 1

The correct option is D $\frac{1}{2}$

When a magnet is cut along perpendicular bisector, the magnetic length of each part is half of original length but the pole strength remains same.

m = pole strength of original magnet
Magnetic moment of original magnet,
$\mu =m\left(2L\right)$

For each half cut part:

Pole strength = m

Magnetic moment of each part

${\mu }^{\prime }=mL$

$\Rightarrow {\mu }^{\prime }=\frac{m\left(2L\right)}{2}=\frac{\mu }{2}$

let, M = mass of original magnet, I=moment of inertia of original magnet about perpendicular bisector axis

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

Let I' be the new moment of inertia of each cut half part

$I^{\prime }=\frac{\left(\frac{M}{2}\right)(L{)}^2}{12}=\frac{1}{8}.\frac{m{L}^2}{3}=\frac{1}{8}$

Time period of oscillation of original magnet,

$T=2\pi \sqrt{\frac{I}{\mu B_H}}$

$(I$ = moment of inertia of oscillating magnet, $\mu$= magnetic moment, $B_H$ = magnetic field)

New time period for each half part

$T^{\prime }=2\pi \sqrt{\frac{\frac{I}{8}}{\left(\frac{\mu }{2}\right)\left(B_H\right)}}$

$\Rightarrow T^{\prime }=\frac{1}{2}.2\pi \sqrt{\frac{I}{\mu B_H}}$

$\Rightarrow T^{\prime }=\frac{T}{2}$

$\Rightarrow T^{\prime }=\frac{1}{2}$

Hence, the correct option is (C)