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

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answered Feb 24, 2022 by Amolkoli (85.8k points)
selected Feb 24, 2022 by Tarunk

Best answer

Correct Answer - Option 2 : T_h = T_c

Concept:

The time-period of oscillations made by a magnet of magnetic moment M, moment of inertia I, placed in a magnetic field is given by

\(T = 2\pi \sqrt {\frac{I}{{MB}}}\) ----(1)

For the hoop, let us assume its moment of inertia I_h and magnetic moment M_h then its time period will be

\({T_h} = 2\pi \sqrt {\frac{{{I_h}}}{{{M_h}B}}}\) ----(2)

Similarly, for solid cylinder, time period is,

\({T_c} = 2\pi \sqrt {\frac{{{I_c}}}{{{M_c}B}}}\) ----(3)

Dividing equation (2) by equation (3), we get

\(\frac{{{T_h}}}{{{T_c}}} = \frac{{2\pi \sqrt {\frac{{{I_h}}}{{{M_h}B}}} }}{{2\pi \sqrt {\frac{{{I_c}}}{{{M_c}B}}} }}\)

\(\frac{{{T_h}}}{{{T_c}}} = \sqrt {\frac{{{I_h}{M_c}}}{{{M_h}{I_c}}}} \) ----(4)

Now, it is given that,

M_h = 2M_c

Calculation:

Given,

Mass of the hoop = Mass of the solid cylinder

Radius of the hoop = Radius of the solid cylinder

Magnetic moment of the hoop, M_h = 2 × Magnetic moment of the solid cylinder, M_c

Oscillation periods of hoop = T_h

Oscillation periods of cylinder = T_c

We know that,

Moment of inertia of the hoop I_h = mR²

Moment of inertia of the solid cylinder \({I_c} = \frac{1}{2}m{R^2}\)

Substituting these values in equation (4), we get

\(\frac{{{T_h}}}{{{T_c}}} = \sqrt {\frac{{m{R^2} \times {M_c}}}{{\frac{1}{2}m{R^2} \times 2{M_c}}}} \)

\(\frac{{{T_h}}}{{{T_c}}} = \sqrt {\frac{{2m{R^2}{M_c}}}{{2m{R^2}{M_c}}}} \)

\(\frac{{{T_h}}}{{{T_c}}} = 1\)

⟹ T_h = T_c

Therefore, if the oscillation periods of hoop and cylinder are T_h and T_c respectively, then its relation will be T_h = T_c