Question

The period of osillation of a freely suspended bar magnet is 4 second. If it is cut into two equal parts in length, then the time period of each part will be

• A. $4sec$
• B. $2sec$
• C. $0.5sec$
• D. $0.25sec$

Answer 1

The correct option is B $2sec$

We know that, Torque on bar

$T=\overrightarrow{\mu }\times \overrightarrow{B}$

$I\alpha =-\mu B\sin \theta$

$I\alpha =-\mu B\theta$ $\sin \theta \approx$ (for very small $\theta$)

$I\frac{d^2\theta }{d{t}^2}+\mu B\theta =0$

$\frac{d^2\theta }{d{t}^2}+\left(\frac{\mu B}{I}\right)\theta =0$

from S.H.M

$w^2=\frac{\mu B}{I}\Rightarrow T=2\pi \sqrt{\frac{I}{\mu B}}$

$T\alpha \sqrt{\frac{I}{\mu }}$

$\frac{{\mu }_1}{{\mu }_2}=\frac{2m\left(l_1\right)}{2m\left(l_2\right)}=\frac{2m\left(l\right)}{2m\left(1/2\right)}=2$

$\frac{I_1}{I_2}=\frac{\rho l_1{b}_1(l_1^2+b_1^2)/12}{\rho l_2{b}_2(l_2^2+b_2^2)/12}$

$\frac{I_1}{I_2}=\left(\frac{l_1}{l_2}\right)\left(\frac{b_1}{b_2}\right)\left(\frac{l_1^2}{l_2^2}\right)\left(\frac{1+(b_1/l_1{)}^2}{1+(b_2/l_2{)}^2}\right)$

$\frac{l_1}{l_2}=2$ $b_1=b_2$ and $l_1>>b_1$

So, $\frac{l}{b}\approx 0$

$\therefore \frac{I_1}{I_2}=\left(2\right)\left(1\right)(2{)}^2=8$

$T\alpha \sqrt{\frac{I}{\mu }}$

$\frac{T_1}{T_2}=\sqrt{\frac{I_1}{I_2}}\sqrt{\frac{{\mu }_2}{{\mu }_1}}$

$\frac{T_1}{T_2}=\sqrt{8}\sqrt{\frac{1}{2}}$ $(\because \frac{{\mu }_1}{{\mu }_2}=2)$

$\frac{T_1}{T_2}=\sqrt{4}\Rightarrow T_2=\frac{T_1}{2}$

$T_2=\frac{4}{2}=2\sec$

$\therefore$ option B is correct