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(C) The time period of an oscillating magnet in a horizontal plane is given by $T = 2\pi \sqrt{\frac{I}{MB_H}}$,where $B_H = B \cos \phi$ is the horiz

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$\frac{9}{4\sqrt{3}}$

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(C) The time period of an oscillating magnet in a horizontal plane is given by $T = 2\pi \sqrt{\frac{I}{MB_H}}$,where $B_H = B \cos \phi$ is the horizontal component of the Earth's magnetic field.Thus,$T \propto \frac{1}{\sqrt{B \cos \phi}}$,which implies $B \propto \frac{1}{T^2 \cos \phi}$.Given $T_1 = 2 \ s$,$\phi_1 = 30^o$ and $T_2 = 3 \ s$,$\phi_2 = 60^o$.The ratio of the resultant magnetic fields is $\frac{B_1}{B_2} = \frac{T_2^2 \cos \phi_2}{T_1^2 \cos \phi_1}$.Substituting the values: $\frac{B_1}{B_2} = \left( \frac{3}{2} ight)^2 	imes \frac{\cos 60^o}{\cos 30^o} = \frac{9}{4} 	imes \frac{1/2}{\sqrt{3}/2} = \frac{9}{4\sqrt{3}}$.

$A$ magnet oscillating in a horizontal plane has a time period of $2 \ s$ at a place where the angle of dip is $30^o$ and $3 \ s$ at another place where the angle of dip is $60^o$. The ratio of resultant magnetic fields at the two places is

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