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(D) The time period of a magnet oscillating in a horizontal magnetic field is given by $T = 2\pi \sqrt{\frac{I}{MB_H}}$,where $I$ is the moment of ine

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$\sqrt{2}T$

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(D) The time period of a magnet oscillating in a horizontal magnetic field is given by $T = 2\pi \sqrt{\frac{I}{MB_H}}$,where $I$ is the moment of inertia,$M$ is the magnetic moment,and $B_H$ is the horizontal component of the Earth's magnetic field.From the formula,we see that $T \propto \sqrt{I}$.Initially,the moment of inertia is $I$. When a wooden block of the same moment of inertia $I$ is attached to the magnet,the total moment of inertia of the system becomes $I' = I + I = 2I$.The new time period $T'$ is given by $T' = 2\pi \sqrt{\frac{I'}{MB_H}} = 2\pi \sqrt{\frac{2I}{MB_H}}$.Therefore,$T' = \sqrt{2} 	imes (2\pi \sqrt{\frac{I}{MB_H}}) = \sqrt{2}T$.

$A$ magnet is suspended horizontally in the Earth's magnetic field. When it is made to oscillate in a horizontal plane,its time period is $T$. If a wooden block of the same moment of inertia as the magnet is attached to it,what will be the new time period of the system?

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