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(D) The time period $T$ of oscillation of a magnet is given by $T = 2\pi \sqrt{\frac{I}{MB}}$,where $I$ is the moment of inertia,$M$ is the magnetic m

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$4$

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(D) The time period $T$ of oscillation of a magnet is given by $T = 2\pi \sqrt{\frac{I}{MB}}$,where $I$ is the moment of inertia,$M$ is the magnetic moment,and $B$ is the horizontal magnetic field.Since $I$ and $M$ are constant,we have $T \propto \frac{1}{\sqrt{B}}$,which implies $\frac{T_2}{T_1} = \sqrt{\frac{B_1}{B_2}}$.Given $B_1 = 24 \, \mu T$ and $T_1 = 2 \, s$.The new magnetic field is $B_2 = B_1 - 18 \, \mu T = 24 \, \mu T - 18 \, \mu T = 6 \, \mu T$.Substituting the values: $\frac{T_2}{2} = \sqrt{\frac{24}{6}} = \sqrt{4} = 2$.Therefore,$T_2 = 2 	imes 2 = 4 \, s$.

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