The length of magnet is large compared to its width and breadth. The time period of its oscillation in a vibration magnetometer is 2 s. The magnet is cut along its length into three equal parts and three parts are then placed on each other with their like poles together. The time period of this combination will be

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⅔ s

$2 / \sqrt{3} s$

2 s

$2 \sqrt{3} s$

answer is B.

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Detailed Solution

We know that
$\begin{array}{l} T = 2 π \sqrt{(\frac{K}{MH})} \\ M^{'} = \frac{M}{3} + \frac{M}{3} + \frac{M}{3} = M \\ K^{'} = \frac{m}{12} {(\frac{1}{3})}^{2} = \frac{1}{9} (\frac{m}{12}) = \frac{1}{9} K \\ ∴ \frac{T^{'}}{T} = \sqrt{(\frac{K^{'}}{K})} = \frac{1}{3} \end{array} or T^{'} = T \times (\frac{1}{3}) = \frac{2}{3} s$