The period of oscillation of a magnet in a vibration magnetometer is 2 sec. The period of oscillation of magnet whose magnetic moment is four times that of the first magnet is

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0.5 sec

answer is A.

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Detailed Solution

$\begin{array}{l} T = 2 π \sqrt{(\frac{I}{{MB}_{H}})} \\ T^{'} = 2 π \sqrt{(\frac{I}{4 {MB}_{H}})} = \frac{1}{2} [2 π \sqrt{\frac{I}{({MB}_{H})}}] \\ = \frac{1}{2} \times 2 = 1 second \end{array}$

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