A stretched wire emits a fundamental note of $256 H z$ . Keeping the stretching force constant and reducing the length of wire by $10 c m$ , the frequency becomes $320 H z$ , the original length of the wire is:

100 c m

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50 c m

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400 c m

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200 c m

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Solution

The correct option is B $50 c m$
frequency of fundamental note, $υ = \frac{1}{2 L} \sqrt{\frac{T}{m}}$
In first case, $256 = \frac{1}{2 L} \sqrt{\frac{T}{m}}$ ......(i)
In second case, $320 = \frac{1}{2 (L - 10)} \sqrt{\frac{T}{m}}$ ......(ii)
dividing (ii) and (i) we get,
$\frac{320}{256} = \frac{2 L}{2 (L - 10)}$ or $\frac{L}{L - 10} = \frac{5}{4}$

$∴ L = 50 c m$

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A stretched wire emits a fundamental note of 256Hz. Keeping the stretching force constant and reducing the length of wire by 10cm, the frequency becomes 320Hz, the original length of the wire is:

The correct option is B 50cmfrequency of fundamental note, υ=12L√TmIn first case, 256=12L√Tm ......(i)In second case, 320=12(L−10)√Tm ......(ii)dividing (ii) and (i) we get,320256=2L2(L−10) or LL−10=54∴L=50cm

A stretched wire emits a fundamental note of $256 H z$ . Keeping the stretching force constant and reducing the length of wire by $10 c m$ , the frequency becomes $320 H z$ , the original length of the wire is: