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(D) The time period of an oscillating magnet in a uniform magnetic field is given by $T = 2 \pi \sqrt{\frac{I}{MH}}$.Here,$M$ is the magnetic moment $

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$\frac{4}{\sqrt{3}}$

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(D) The time period of an oscillating magnet in a uniform magnetic field is given by $T = 2 \pi \sqrt{\frac{I}{MH}}$.Here,$M$ is the magnetic moment $(M = m 	imes l)$ and $I$ is the moment of inertia $(I = \frac{m l^2}{12})$,where $m$ is the pole strength and $l$ is the length.When the rod is broken into three equal parts,the new length is $l' = \frac{l}{3}$ and the new pole strength remains $m$ (as the cross-section is unchanged).Thus,the new magnetic moment is $M' = m 	imes \frac{l}{3} = \frac{M}{3}$.The new moment of inertia is $I' = \frac{m(l/3)^2}{12} = \frac{m l^2}{9 	imes 12} = \frac{I}{9}$.The new time period $T'$ is given by $T' = 2 \pi \sqrt{\frac{I'}{M' H}}$.Substituting the values: $T' = 2 \pi \sqrt{\frac{I/9}{(M/3)H}} = 2 \pi \sqrt{\frac{I}{3MH}} = \frac{T}{\sqrt{3}}$.Given $T = 4 \ s$,the new time period is $T' = \frac{4}{\sqrt{3}} \ s$.

$A$ thin magnetic iron rod of length $30 \ cm$ is suspended in a uniform magnetic field. Its time period of oscillation is $4 \ s$. It is broken into three equal parts. The time period in seconds of oscillation of one part when suspended in the same magnetic field is

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