The time period of a small magnet in a horizontal plane is T. Another magnet B oscillates at the same place in a similar manner. The size of two magnets is the same but the magnetic moment of B is four times that of A. The time period of B will be-
Answer & explanation
Correct answer: option 2
$ \frac{T_A}{T_B} = \sqrt{\frac{M_B}{M_A}} $
$ T_B = T\sqrt {\frac{M_A}{4M_A}} = \frac{T}{2}$