The time period of oscillation of magnet in a vibrating magnetometer is $1.5$ sec. The time period of oscillation of another magnet similar in size and mass but having one-fourth the magnetic moment than that of the first magnet oscillating at the same place will be

0.75

sec

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1.5

sec

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3.0

sec

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6.0

sec

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Solution

The correct option is C $3.0$ sec
$\begin{matrix} \frac{T_{1}}{T_{2}} = \sqrt{\frac{M_{2}}{M_{1}}} = \sqrt{\frac{M_{1}}{\frac{4}{M_{1}}}} = \frac{T_{1}}{T_{2}} = \frac{1}{2} \Rightarrow T_{2} = 3 \end{matrix}$

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The time period of oscillation of magnet in a vibrating magnetometer is 1.5 sec. The time period of oscillation of another magnet similar in size and mass but having one-fourth the magnetic moment than that of the first magnet oscillating at the same place will be

The correct option is C 3.0 secT1T2=√M2M1=  ⎷M14M1=T1T2=12⇒T2=3

The time period of oscillation of magnet in a vibrating magnetometer is $1.5$ sec. The time period of oscillation of another magnet similar in size and mass but having one-fourth the magnetic moment than that of the first magnet oscillating at the same place will be

The correct option is C $3.0$ sec
$\begin{matrix} \frac{T_{1}}{T_{2}} = \sqrt{\frac{M_{2}}{M_{1}}} = \sqrt{\frac{M_{1}}{\frac{4}{M_{1}}}} = \frac{T_{1}}{T_{2}} = \frac{1}{2} \Rightarrow T_{2} = 3 \end{matrix}$