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(A) Let the mass of the first magnet be $m_1 = m$ and the second be $m_2 = 2m$. Let the lengths be $l_1 = l$ and $l_2 = 2l$.The moment of inertia of a

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$\frac{1}{2 \sqrt{2}}$

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(A) Let the mass of the first magnet be $m_1 = m$ and the second be $m_2 = 2m$. Let the lengths be $l_1 = l$ and $l_2 = 2l$.The moment of inertia of a bar magnet about its center is $I = \frac{m l^2}{12}$.Thus, $I_1 = \frac{m l^2}{12}$ and $I_2 = \frac{(2m)(2l)^2}{12} = \frac{8 m l^2}{12} = 8 I_1$.The maximum torque is given by $	au_{max} = M B$. Since $	au_{max,1} = 	au_{max,2}$, we have $M_1 B = M_2 B$, so $M_1 = M_2 = M$.The time period of oscillation is $T = 2 \pi \sqrt{\frac{I}{M B}}$.The ratio of time periods is $\frac{T_1}{T_2} = \sqrt{\frac{I_1}{I_2} \cdot \frac{M_2}{M_1}}$.Substituting the values, $\frac{T_1}{T_2} = \sqrt{\frac{I_1}{8 I_1} \cdot \frac{M}{M}} = \sqrt{\frac{1}{8}} = \frac{1}{2 \sqrt{2}}$.

Two small magnets have their masses and lengths in the ratio $1:2$. The maximum torques experienced by them in a uniform magnetic field are the same. For small oscillations, the ratio of their time periods is

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