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Standard XII

Physics

String Fixed at Both Ends

A metal wire ...

Question

A metal wire of linear mass density of $9.8 g / m$ is stretched with a tension of $10 k g - w t$ between two rigid supports $1 m$ apart. The wire passes at its middle point between the poles of a permanent magnet and it vibrates in resonance when carrying an alternating current of frequency $n$ . The frequency $n$ of the alternating source is

50 H z

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100 H z

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200 H z

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25 H z

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Solution

The correct option is C $50 H z$
When in resonance, the frequency of alternating current must be equal to the fundamental frequency of vibration of the metal wire.
Thus $f = \frac{1}{2 l} \sqrt{\frac{T}{μ}}$
$= \frac{1}{2} \sqrt{\frac{10 \times 9.8}{9.8 \times 10^{- 3}}} = 50 H z$

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The correct option is C 50HzWhen in resonance, the frequency of alternating current must be equal to the fundamental frequency of vibration of the metal wire.Thus f=12l√Tμ=12√10×9.89.8×10−3=50Hz

The correct option is C $50 H z$
When in resonance, the frequency of alternating current must be equal to the fundamental frequency of vibration of the metal wire.
Thus $f = \frac{1}{2 l} \sqrt{\frac{T}{μ}}$
$= \frac{1}{2} \sqrt{\frac{10 \times 9.8}{9.8 \times 10^{- 3}}} = 50 H z$