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(D) The time period $T$ of a bar magnet oscillating in a magnetic field is given by the formula: $T = 2\pi \sqrt{\frac{I}{MB}}$.Here,$M = 5 	imes 10^

Vedclass · Accepted Answer

$7.16 	imes 10^{-7} \, 	ext{kg} \cdot 	ext{m}^2$

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(D) The time period $T$ of a bar magnet oscillating in a magnetic field is given by the formula: $T = 2\pi \sqrt{\frac{I}{MB}}$.Here,$M = 5 	imes 10^{-5} \, 	ext{Wb} \cdot 	ext{m}$,$B = 8\pi 	imes 10^{-4} \, 	ext{T}$,and $T = 15 \, 	ext{s}$.Rearranging the formula to solve for the moment of inertia $I$:$I = \frac{T^2 MB}{4\pi^2}$.Substituting the given values:$I = \frac{(15)^2 	imes (5 	imes 10^{-5}) 	imes (8\pi 	imes 10^{-4})}{4\pi^2}$.$I = \frac{225 	imes 40\pi 	imes 10^{-9}}{4\pi^2} = \frac{225 	imes 10 	imes 10^{-9}}{\pi} = \frac{2250 	imes 10^{-9}}{3.14159}$.$I \approx 7.16 	imes 10^{-7} \, 	ext{kg} \cdot 	ext{m}^2$.

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