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(B) The time period of oscillation of a magnet in a magnetic field is given by $T = 2\pi \sqrt{\frac{I}{M B_H}}$,where $I$ is the moment of inertia an

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$0.36 	imes 10^{-6} \,T$

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(B) The time period of oscillation of a magnet in a magnetic field is given by $T = 2\pi \sqrt{\frac{I}{M B_H}}$,where $I$ is the moment of inertia and $M$ is the magnetic moment.This implies $T \propto \frac{1}{\sqrt{B_H}}$,or $B_H \propto \frac{1}{T^2}$.At the first place,the frequency is $40 	ext{ oscillations/min}$,so the time period $T_1 = \frac{60}{40} = 1.5 \,s$. The magnetic field is $(B_H)_1 = 0.1 	imes 10^{-5} \,T = 10^{-6} \,T$.At the second place,the time period $T_2 = 2.5 \,s$.Using the ratio $\frac{(B_H)_2}{(B_H)_1} = \left( \frac{T_1}{T_2} ight)^2$,we get:$(B_H)_2 = (B_H)_1 	imes \left( \frac{1.5}{2.5} ight)^2$$(B_H)_2 = 10^{-6} 	imes \left( \frac{3}{5} ight)^2 = 10^{-6} 	imes \frac{9}{25} = 10^{-6} 	imes 0.36 = 0.36 	imes 10^{-6} \,T$.

$A$ magnet makes $40$ oscillations per minute at a place having magnetic field intensity of $0.1 \times 10^{-5} \,T$. At another place,it takes $2.5 \,s$ to complete one vibration. The value of earth's horizontal field at that place is:

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