Question

A hoop and a solid cylinder of same mass and radius are made of a permanent magnetic material with their magnetic moments parallel to their respective axes. But the magnetic moment of hoop is twice that of solid cylinder. They are then placed in a uniform magnetic field in such a manner that their magnetic moments make a small angle with the field. If the oscillation periods of hoop and cylinder are $T_{h}and{T}_c$ respectively, then

• A. $T_h=T_c$
• B. $T_h=\frac{3}{2}{T}_c$
• C. $T_h=2T_c$
• D. $T_h=\frac{1}{2}{T}_c$

Answer 1

The correct option is A $T_h=T_c$

Time period, $T=2\pi \sqrt{\frac{I}{MB}}$

For hoop , $I_h=m{R}^2$ and $M_h=2M_c$

$\therefore T_h=2\pi \sqrt{\frac{I_h}{M_{h}B}}.......\left(1\right)$

For cylinder , $I_c=\frac{m{R}^2}{2}$

$\therefore T_c=2\pi \sqrt{\frac{I_c}{M_{c}B}}.....\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$

$\frac{T_h}{T_c}=\sqrt{\frac{I_h}{I_c}\times \frac{M_c}{M_h}}$

= $\sqrt{\frac{m{R}^2}{\left(\frac{m{R}^2}{2}\right)}\times \frac{M_c}{2M_c}}=1$

Thus, $T_h=T_c$

Hence, option (a) is the correct answer.