Question

A hoop and a solid cylinder of same mass and radius are made of a permanent magnetic material with their respective axes. But the magnetic moment of hoop is twice of solid cylinder. They are 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. $T_{h}=2T_{c}$
• B. $T_{h}=T_{c}$
• C. $T_{h}=0.5T_{c}$
• D. $T_{h}=1.5T_{c}$

Answer 1

Answer: (B)

Time period of any magnetic body place in an external magnetic field is:-
$T=2\pi \sqrt{\frac{I}{M B}}$
$T_{h o o p}=2\pi \sqrt{\frac{m R^{2}}{M B}}$ ...... $\left(\right.1\left.\right)$ $I_{h o o p}=mR^{2}$
$T_{c}=2\pi \sqrt{\frac{\frac{m R^{2}}{2}}{\frac{M}{2} \left(B\right)} \, }$ ...... $\left(\right.2\left.\right)$ $I_{\text{Solid cylinder}}=\frac{m R^{2}}{2}$
From $\left(\right.1\left.\right)$ and $\left(\right.2\left.\right)$
$T_{h}=T_{c}$