Question

A hoop and a solid cylinder of same mass and radius are made of a permanent magnetic material with their magnetic
moment parallel to 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 $T_h\text{and}{T}_c$ respectively then

• A. $T_h=T_c$
• B. $T_h=2T_c$
• C. $T_h=1.5T_c$
• D. $T_h=0.5T_c$

Answer 1

The correct option is A $T_h=T_c$

Using, time period of oscillation,
$T=2\pi \sqrt{\frac{I}{MB}}$
Where,
$M=$ magnetic moment,
$I=$ moment of inertia and
$B=$ magnetic field

$T_h=2\pi \sqrt{\frac{m{R}^2}{\left(2MB\right)}}$

$T_c=2\pi \sqrt{\frac{\frac{1}{2}m{R}^2}{MB}}$

Clearly,$T_h=T_c$