TS EAMCET · Physics · Magnetic Properties of Matter
A short bar magnet having magnetic moment \(4 \mathrm{Am}^2\), placed in a vibrating magnetometer, vibrates with a time period of \(8 \mathrm{~s}\). Another short bar magnet having a magnetic moment \(8 \mathrm{Am}^2\) vibrates with a time period of \(6 \mathrm{~s}\). If the moment of inertia of the second magnet is \(9 \times 10^{-2} \mathrm{~kg}-\mathrm{m}^2\), the moment of inertia of the first magnet is (assume that both magnets are kept in the same uniform magnetic induction field.)
- A \(9 \times 10^{-2} \mathrm{~kg}-\mathrm{m}^2\)
- B \(8 \times 10^{-2} \mathrm{~kg}-\mathrm{m}^2\)
- C \(5.33 \times 10^{-2} \mathrm{~kg}-\mathrm{m}^2\)
- D \(12.2 \times 10^{-2} \mathrm{~kg}-\mathrm{m}^2\)
Answer & Solution
Correct Answer
(B) \(8 \times 10^{-2} \mathrm{~kg}-\mathrm{m}^2\)
Step-by-step Solution
Detailed explanation
We know that the time period of a vibrating bar magnet \(T=2 \pi \sqrt{\frac{I}{M B_H}}\) Given, \(T_1=8 \mathrm{~s}, l_1=I, M=4 \mathrm{Am}^2\) \(8=2 \pi \sqrt{\frac{1}{4 \times B_H}}\) Given,…
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