A combination of two identical bar magnets are placed one over the other such that they are perpendicular and bisect each other. The time period of oscillation in a horizontal magnetic field is 2^5/4 seconds. One of the magnets is removed and if the other magnet oscillates in the same field, then the time period in seconds is

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2^5/4

2^1/2

2^1/4

answer is C.

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Detailed Solution

$\begin{aligned} T = 2 π \sqrt{\frac{I}{MB}}; \frac{T_{1}}{T_{2}} = \sqrt{\frac{M_{2}}{M_{1}} \cdot \frac{I_{1}}{I_{2}}} = \sqrt{\frac{M}{\sqrt{2} M} \frac{2 I}{I}} \\ T_{2} = \frac{2^{5 / 4}}{2^{1 / 4}} = 2 \sec \end{aligned}$

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