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T_o is time period of oscillation of the needle in uniform horizontal magnetic field (B_H) of the earth. If another magnetic field is applied perpendicular to B_Hsuch that the needle deflects by $θ$ for its equilibrium. Now period of oscillation of the needle is T. The relation between T and T_o is

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$T^{2} = T_{0}^{2} cosθ$

$T = T_{0}^{} cosθ$

$T = \frac{T_{0}^{}}{cosθ}$

$T^{2} = \frac{T_{0}^{2}}{cosθ}$

answer is A.

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Detailed Solution

$T_{0} α \frac{1}{\sqrt{B_{H}}}; T_{0} α \frac{1}{\sqrt{B^{2} + B_{H}^{2}}} \Rightarrow \frac{T^{2}}{T_{0}^{2}} = \frac{1}{\sqrt{1 + {Tan}^{2} θ}} = cosθ \Rightarrow T^{2} = T_{0}^{2} cosθ$

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