Two bar magnets having same geometry with magnetic moments M and 2M are firstly placed in such a way that their similar poles are on the same side, and then its time period of oscillation is T₁. Now the polarity of one of the magnets is reversed and the time period of oscillation is T₂, then

Question

Accepted Answer

(1) T₁<T₂. Given: two magnets of the same geometry, moments M and 2M, tied together and oscillating in the horizontal field H. Idea: the moments of inertia always add; the magnetic moments add or subtract depending on the polarity. Total inertia in both cases: Iₜₒₜ=I+I=2I. Like poles together (sum position): net moment =2M+M=3M, so T₁=2π√(2I)/(3MH). One reversed (difference position): net moment =2M-M=M, so T₂=2π√(2I)/(MH). Ratio: (T₁)/(T₂)=√M/(3M)=1/(√3). Since 1/√3<1, T₁<T₂.

Two bar magnets having same geometry with magnetic moments M and 2M are firstly placed in such a way that their similar poles are on the same side, and then its time period of oscillation is T₁. Now the polarity of one of the magnets is reversed and the time period of oscillation is T₂, then

Step-by-step solution

Why the other options are wrong