A wire of density 9 × 10 ^ – 3 kg cm ^ – 3 is stretched between two...

Fundamental frequency in the string

$f = \dfrac{1}{2ℓ}\sqrt{\dfrac{T}{\mu}} = \dfrac{1}{2ℓ}\sqrt{\dfrac{T}{ρA}} = \dfrac{1}{2ℓ}\dfrac{\sqrt{Y \Delta ℓ}}{ρℓ}$
$f = \dfrac{1}{2ℓ}\sqrt{\dfrac{Y \Delta ℓ}{ρℓ}}$
$= \dfrac{1}{2}\sqrt{\dfrac{9 \times {10}^{10} \times 4 . 9 \times {10}^{- 4}}{9 \times {10}^{- 3} \times {10}^{- 6}}} = 35$