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(C) Let the magnetic moments of magnets $A$ and $B$ be $M_A = 2M$ and $M_B = M$ respectively. Let their moments of inertia be $I_A = I_B = I$ (since t

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$1: \sqrt{3}$

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(C) Let the magnetic moments of magnets $A$ and $B$ be $M_A = 2M$ and $M_B = M$ respectively. Let their moments of inertia be $I_A = I_B = I$ (since they are geometrically similar and have the same mass/dimensions).When like poles are kept together,the net magnetic moment is $M_{net1} = M_A + M_B = 2M + M = 3M$. The total moment of inertia is $I_{net} = I_A + I_B = 2I$.The time period is given by $T = 2\pi \sqrt{\frac{I_{net}}{M_{net}B_H}}$.Thus,$T_1 = 2\pi \sqrt{\frac{2I}{3MB_H}}$.When unlike poles are kept together,the net magnetic moment is $M_{net2} = M_A - M_B = 2M - M = M$.The total moment of inertia remains $I_{net} = 2I$.Thus,$T_2 = 2\pi \sqrt{\frac{2I}{MB_H}}$.Taking the ratio $T_1 / T_2 = \sqrt{\frac{2I}{3MB_H} / \frac{2I}{MB_H}} = \sqrt{\frac{1}{3}} = 1 : \sqrt{3}$.

Two bar magnets $A$ and $B$ are geometrically similar but the magnetic moment of $A$ is twice that of $B$. $T_1$ is the time period of oscillation when their like poles are kept together. When unlike poles are kept together,the time period of oscillation is $T_2$. The ratio $T_1: T_2$ will be

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