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 Th and Tc respectively, then -
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 Th and Tc respectively, then -
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For small oscillations of a magnetic dipole, T = 2 pi sqrt(I/(mu B)); compare I/mu for the two bodies.
Updated On: Nov 5, 2026
- Th = 1.5 Tc
- Th = Tc
- Th = 2Tc
- Th = 0.5 Tc
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The Correct Option is B
Solution and Explanation
For small-angle oscillations, \(I\ddot\theta=-\mu B\theta\), so \(T=2\pi\sqrt{\frac{I}{\mu B}}\).
Moments of inertia in the intended solution: hoop \(I_h=MR^2\), solid cylinder \(I_c=\frac12MR^2\), so \(I_h=2I_c\).
Given \(\mu_h=2\mu_c\).
\(\frac{T_h}{T_c}=\sqrt{\frac{I_h}{I_c}\cdot\frac{\mu_c}{\mu_h}}=\sqrt{2\cdot\frac12}=1\).
So \(T_h=T_c\).
Answer: (2) Th = Tc
Moments of inertia in the intended solution: hoop \(I_h=MR^2\), solid cylinder \(I_c=\frac12MR^2\), so \(I_h=2I_c\).
Given \(\mu_h=2\mu_c\).
\(\frac{T_h}{T_c}=\sqrt{\frac{I_h}{I_c}\cdot\frac{\mu_c}{\mu_h}}=\sqrt{2\cdot\frac12}=1\).
So \(T_h=T_c\).
Answer: (2) Th = Tc
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