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MHT CET · Physics · Waves and Sound

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. A \(\frac{\mathrm{v}^{2} \varrho}{\mathrm{T}}\)
  2. B \(\frac{\mathrm{T}}{\mathrm{v}^{2} \varrho}\)
  3. C \(\mathrm{T}^{2} \varrho \mathrm{v}\)
  4. D \(\mathrm{Tv}^{2} \varrho\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{\mathrm{T}}{\mathrm{v}^{2} \varrho}\)

Step-by-step Solution

Detailed explanation

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