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 and T_c respectively, then:

पर्याय

• T_h = T_c 

• T_h = 2 T_c 

• T_h = 1.5 T_c 

• T_h = 0.5 T_c 

Answer 1

T_h = T_c 

Explanation:

Using, time /oscillation period,

T = 2π `sqrt("I"/"MB")`

Where, M = magnetic moment, I moment of inertia and B = magnetic field

T_h = 2π `sqrt("MR"^2/((2"MB")))`

T_c = 2π `sqrt((1//2"mR"^2)/"MB")` 

Clearly, T_h = T_c