A uniform metal wire has length 'L', mass 'M' and density ' $\mathrm{Q}$ '. It is under tension 'T" and '…

A uniform metal wire has length 'L', mass 'M' and density ' $\mathrm{Q}$ '. It is under tension 'T" and ' $v^{\prime}$ is the speed of transverse wave along the wire. The area of cross-section of the wire is
  1. $\frac{\mathrm{v}^{2} \varrho}{\mathrm{T}}$
  2. $\frac{\mathrm{T}}{\mathrm{v}^{2} \varrho}$
  3. $\mathrm{T}^{2} \varrho \mathrm{v}$
  4. $\mathrm{Tv}^{2} \varrho$

Solution

$(\mathrm{C})$ $\begin{aligned} \mathrm{V} &=\sqrt{\frac{\mathrm{T}}{\mathrm{m}}} \end{aligned}\left[\mathrm{m}=\frac{\mathrm{M}}{\mathrm{L}}=\frac{\mathrm{AL} \rho}{\mathrm{L}}=\mathrm{A\rho}\right]$ $\therefore \mathrm{V}=\sqrt{\frac{\mathrm{T}}{\mathrm{A} \rho}}$ $\therefore \mathrm{V}^{2}=\frac{\mathrm{T}}{\mathrm{A} \rho}$ $\mathrm{A}=\frac{\mathrm{T}}{\mathrm{V}^{2} \rho}$

Asked in: MHT CET 2020 (19 Oct Shift 2)

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