A uniform metal wire of length L,mass M,and d | vedclass

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(B) The speed of a transverse wave in a stretched wire is given by $V = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass de

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$\frac{T}{V^{2}\rho}$

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(B) The speed of a transverse wave in a stretched wire is given by $V = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.Linear mass density $\mu = \frac{M}{L}$.Also,mass $M = 	ext{Volume} 	imes 	ext{Density} = (A 	imes L) 	imes ho$,where $A$ is the area of cross-section.Therefore,$\mu = \frac{A 	imes L 	imes ho}{L} = Aho$.Substituting this into the wave speed formula: $V = \sqrt{\frac{T}{Aho}}$.Squaring both sides: $V^{2} = \frac{T}{Aho}$.Rearranging for the area of cross-section $A$: $A = \frac{T}{V^{2}ho}$.

$A$ uniform metal wire of length $L$,mass $M$,and density $\rho$ is under a tension $T$. If the speed of a transverse wave along the wire is $V$,then the area of cross-section of the wire is:

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