A magnet is suspended horizontally in the earth's field. The period of oscillation in the place is T. If a piece of wood of the same moment of inertia as the magnet is attached to it, the new period of oscillation would be:

\frac{T}{\sqrt{2}}

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\frac{T}{2}

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\frac{T}{3}

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\sqrt{2} T

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Solution

The correct option is B $\sqrt{2} T$
$T = 2 π \sqrt{\frac{I}{M B}}$
So the frequency of oscillation is directly proportional to the square root of magnetic field
Moment of inertia after adding wooden piece is
$I_{2} = 2 I$
$\frac{T_{2}}{T_{1}} = \sqrt{\frac{I_{2}}{I_{1}}}$
$T_{2} = \sqrt{2} T$

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A magnet is suspended horizontally in the earth's field. The period of oscillation in the place is T. If a piece of wood of the same moment of inertia as the magnet is attached to it, the new period of oscillation would be:

The correct option is B √2TT=2π√IMBSo the frequency of oscillation is directly proportional to the square root of magnetic fieldMoment of inertia after adding wooden piece isI2=2IT2T1=√I2I1T2=√2T

The correct option is B $\sqrt{2} T$
$T = 2 π \sqrt{\frac{I}{M B}}$
So the frequency of oscillation is directly proportional to the square root of magnetic field
Moment of inertia after adding wooden piece is
$I_{2} = 2 I$
$\frac{T_{2}}{T_{1}} = \sqrt{\frac{I_{2}}{I_{1}}}$
$T_{2} = \sqrt{2} T$