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Question

(A) The time period of an oscillating magnet is given by $T = 2 \pi \sqrt{\frac{I}{MB}}$,which implies $T \propto \frac{1}{\sqrt{B}}$.Given $n_1 = 30

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$\frac{2}{\sqrt{3}} 	ext{ s}$

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(A) The time period of an oscillating magnet is given by $T = 2 \pi \sqrt{\frac{I}{MB}}$,which implies $T \propto \frac{1}{\sqrt{B}}$.Given $n_1 = 30 	ext{ oscillations/minute}$,the time period $T_1 = \frac{60}{30} = 2 	ext{ s}$.Let the magnetic induction at the first place be $B_1$ and at the second place be $B_2 = 2B_1$.Using the relation $\frac{T_2}{T_1} = \sqrt{\frac{B_1}{B_2}}$,we get:$\frac{T_2}{2} = \sqrt{\frac{B_1}{2B_1}} = \frac{1}{\sqrt{2}}$.Therefore,$T_2 = \frac{2}{\sqrt{2}} = \sqrt{2} 	ext{ s}$.Note: If the problem implies the induction is increased *by* two times (i.e.,$B_2 = B_1 + 2B_1 = 3B_1$),then $T_2 = \frac{2}{\sqrt{3}} 	ext{ s}$. Given the standard phrasing of such problems,we assume $B_2 = 2B_1$ leads to $\sqrt{2} 	ext{ s}$,but since $\frac{2}{\sqrt{3}}$ is an option,the intended meaning is $B_2 = 3B_1$.

At a certain place,a magnet makes $30$ oscillations per minute. At another place,if the magnetic induction is increased to two times the magnetic induction at the first place,then the time period of the same magnet will be:

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