Two bar magnets oscillate in a horizontal plane in earth’s magnetic field with time periods of $3 s$ and $4 s$ respectively. If their moments of inertia are in the ratio of $3 : 2$ , then the ratio of their magnetic moments will be
$27 : 16$
$2 : 1$
$1 : 3$
$8 : 3$

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Solution

The correct option is D $8 : 3$
Time period of oscillation of bar magnet in earth's magnetic field is given by
$T = 2 π \sqrt{\frac{I}{M B_{H}}}$
$\Rightarrow T \propto \sqrt{\frac{I}{M}}$
$\Rightarrow \frac{T_{1}}{T_{2}} = \sqrt{\frac{I_{1}}{I_{2}}} \sqrt{\frac{M_{2}}{M_{1}}}$

$\Rightarrow \frac{3}{4} = \sqrt{\frac{3}{2}} \sqrt{\frac{M_{2}}{M_{1}}}$

$\Rightarrow \frac{M_{1}}{M_{2}} = \frac{3}{2} \times \frac{16}{9} = \frac{8}{3}$

Hence, option $(B)$ is correct.

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