#### A metal wire of linear mass density of  9.8 g/m is stretched with a tension of 10 kg -wt  between two rigid supports 1 metre 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   The frequency of the alternating source is Option 1) Option 2) Option 3) Option 4)

As we learnt in

Standing wave in a string fixed at both ends -

- wherein

frequency of nth harmonic .

tension in string

mass / length

At resonance , frequency of vibration of wire become equal to frequency of  a.c.

For vibration of wire,$\upsilon = \frac{1}{2l}\sqrt{\frac{T}{\mu }}$

$\therefore \upsilon = \frac{1}{2\times 1}\sqrt{\frac{10\times 9.8}{9.8\times 10^{-3}}}= \frac{100}{2}= 50Hz$

Correct option is 1.

Option 1)

This is the correct option.

Option 2)

This is an incorrect option.

Option 3)

This is an incorrect option.

Option 4)

This is an incorrect option.

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