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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
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and ' $v^{\prime}$ is the speed of transverse wave along the wire. The area of cross-section of
the wire is
Solution:
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Verified Answer
The correct answer is:
$\frac{\mathrm{T}}{\mathrm{v}^{2} \varrho}$
$(\mathrm{C})$
$\begin{aligned} \mathrm{V} &=\sqrt{\frac{\mathrm{T}}{\mathrm{m}}} \end{aligned}\left[\mathrm{m}=\frac{\mathrm{M}}{\mathrm{L}}=\frac{\mathrm{AL} \rho}{\mathrm{L}}=\mathrm{A\rho}\right]$
$\therefore \mathrm{V}=\sqrt{\frac{\mathrm{T}}{\mathrm{A} \rho}}$
$\therefore \mathrm{V}^{2}=\frac{\mathrm{T}}{\mathrm{A} \rho}$
$\mathrm{A}=\frac{\mathrm{T}}{\mathrm{V}^{2} \rho}$
$\begin{aligned} \mathrm{V} &=\sqrt{\frac{\mathrm{T}}{\mathrm{m}}} \end{aligned}\left[\mathrm{m}=\frac{\mathrm{M}}{\mathrm{L}}=\frac{\mathrm{AL} \rho}{\mathrm{L}}=\mathrm{A\rho}\right]$
$\therefore \mathrm{V}=\sqrt{\frac{\mathrm{T}}{\mathrm{A} \rho}}$
$\therefore \mathrm{V}^{2}=\frac{\mathrm{T}}{\mathrm{A} \rho}$
$\mathrm{A}=\frac{\mathrm{T}}{\mathrm{V}^{2} \rho}$
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