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Standard XII

Physics

String Open at One End

A metal wire ...

Question

A metal wire of linear mass density of $9.8 g m$ is stretched with a tension of $10 k g - w t$ betweentwo rigid supports $1$ metre apart. The wire passes at its middle point between the poles of aper 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 D $50 H z$
Tension in the string $= 10 k g - w t$
Frequency of resonance= $\frac{1}{2 L} \sqrt{\frac{T}{μ}}$
where $μ$ is mass per unit length of wire.
Thus $n = \frac{1}{2} \sqrt{\frac{10 \times 9.8}{9.8 \times 10^{- 3}}} = 50 H z$

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The correct option is D 50HzTension in the string=10kg−wtFrequency of resonance=12L√Tμwhere μ is mass per unit length of wire.Thus n=12√10×9.89.8×10−3=50Hz

The correct option is D $50 H z$
Tension in the string $= 10 k g - w t$
Frequency of resonance= $\frac{1}{2 L} \sqrt{\frac{T}{μ}}$
where $μ$ is mass per unit length of wire.
Thus $n = \frac{1}{2} \sqrt{\frac{10 \times 9.8}{9.8 \times 10^{- 3}}} = 50 H z$