Q. The time period of oscillation of a freely suspended bar magnet with usual notations is given by
A) $T = 2\pi\sqrt{\frac{I}{MB_{H}}}$
B) $T = 2\pi\sqrt{\frac{MB_{H}}{I}}$
C) $T = \sqrt{\frac{I}{MB_{H}}}$
D) $T = 2\pi\sqrt{\frac{B_{H}}{MI}}$
Explanation
The correct answer is Option A: $T = 2\pi\sqrt{\frac{I}{MB_{H}}}$.
A detailed explanation for this question is being prepared.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.