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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?
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The correct answer is:
motion remains S.H.M with time period $=2 T$
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.
$$
=2 \pi \sqrt{\frac{I}{14 H}}
$$
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.
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