Q127 / 200 · NEET 2003 · Physics

A bar magnet is oscillating in the Earth's magnetic field with a period $T$. What happens to its period and motion if its mass is quadrupled?

  • A
    motion remains simple harmonic with time period $=\frac{T}{2}$
  • Correct answer
    B
    motion remains S.H.M with time period $=2 T$
  • C
    motion remains S.H.M. with time period $=4 T$
  • D
    motion remains S.H.M and period nearly constant

Solution

Time period of bar magnet $T$ $$ =2 \pi \sqrt{\frac{I}{14 H}} $$ Where $\mathrm{I}=$ M.I. $=\frac{m \mathrm{~L}^2}{12}$ $m=$ mass of magnet $\therefore T=2 \pi \sqrt{\frac{m L^2}{12 \mathrm{MH}}}$ $\Rightarrow T \propto \sqrt{m}$ $\therefore$ If mass is quadruplet then $T$ becomes twice.

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

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