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(B) The time period $T$ of a vibrating bar magnet in a uniform magnetic field $B$ is given by $T = 2\pi \sqrt{\frac{I}{MB}}$,where $I$ is the moment o

Vedclass · Accepted Answer

$8 	imes 10^{-2} 	ext{ kg-m}^2$

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(B) The time period $T$ of a vibrating bar magnet in a uniform magnetic field $B$ is given by $T = 2\pi \sqrt{\frac{I}{MB}}$,where $I$ is the moment of inertia and $M$ is the magnetic moment.For the first magnet: $T_1 = 8 	ext{ s}$,$M_1 = 4 	ext{ Am}^2$,$I_1 = I$.$8 = 2\pi \sqrt{\frac{I}{4B}} \implies 64 = 4\pi^2 \frac{I}{4B} = \frac{\pi^2 I}{B} \quad (i)$For the second magnet: $T_2 = 6 	ext{ s}$,$M_2 = 8 	ext{ Am}^2$,$I_2 = 9 	imes 10^{-2} 	ext{ kg-m}^2$.$6 = 2\pi \sqrt{\frac{9 	imes 10^{-2}}{8B}} \implies 36 = 4\pi^2 \frac{9 	imes 10^{-2}}{8B} = \frac{\pi^2 (9 	imes 10^{-2})}{2B} \quad (ii)$Dividing $(i)$ by $(ii)$:$\frac{64}{36} = \frac{\pi^2 I / B}{\pi^2 (9 	imes 10^{-2}) / 2B} = \frac{I 	imes 2}{9 	imes 10^{-2}}$$\frac{16}{9} = \frac{2I}{9 	imes 10^{-2}}$$16 = \frac{2I}{10^{-2}} \implies 16 	imes 10^{-2} = 2I$$I = 8 	imes 10^{-2} 	ext{ kg-m}^2$.

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