A wire of density 9 × 10–3 kg cm–3 is stretched between two clamps 1… | JEE Main 2020 (2 Sep, Shift 2)

Q85JEE Main 2020 · 2 SEP · SHIFT 2Medium

Stationary Waves & Vibrating Strings

A wire of density 9 $\times$ 10–3 kg cm–3 is stretched between two clamps 1 m apart. The resulting strain in the wire is 4.9 $\times$ 10–4. The lowest frequency of the transverse vibrations in the wire is: (Young’s modulus of wire Y = 9 $\times$ 1010 Nm–2), (to the nearest integer), _________

Numerical answer type. Work it out, then open the answer.

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Answer: 35

$\rho = 9\times10^{-3}\text{ kg/cm}^3 = 9\times10^{3}\text{ kg/m}^3$
Stress $= Y \times \text{strain} = 9\times10^{10}\times4.9\times10^{-4} = 4.41\times10^{7}\text{ N/m}^2$
$v = \sqrt{\dfrac{\text{stress}}{\rho}} = \sqrt{\dfrac{4.41\times10^7}{9\times10^3}} = 70\text{ m/s}$
$n = \dfrac{v}{2L} = \dfrac{70}{2} = 35\text{ Hz}$

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