Q122 / 200 · NEET 2002 · Physics

Two bar magnets having geometry with magnetic moments $M$ and $2 M$ are first placed in such a way that their similar poles are same side then its time period of oscillation is $T_1$. Now the polarity of one of the magnet is reversed then the time period of oscillation is $T_2$, so:

  • Correct answer
    A
    $T_1 < T_2$
  • B
    $T_1=T_2$
  • C
    $T_1>T_2$
  • D
    $T_2=\infty$

Solution

We can write $T=2 \pi \sqrt{\frac{1}{M B}} \Rightarrow T \propto \frac{1}{\sqrt{M}}$ Case I : $M_1=2 M+M$ Case II : $M_2=2 M-M$ $\begin{aligned} & \therefore \frac{T_1}{T_2}=\sqrt{\frac{M}{3 M}}=\frac{1}{\sqrt{3}} \\ & \Rightarrow T_2=\sqrt{3} T_1 \end{aligned}$

Hence, the correct answer is option: 'A'.

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