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A uniform metal wire has length ' $L$ ', mass ' $M$ ' and cross-sectional area ' $A$ '. It is under tension ' $T$ ' and ' $V$ ' is the speed of transverse wave along the wire. The density of the wire
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The correct answer is:
$\frac{T}{V^2 A}$
For transverse waves on a wire $V=\sqrt{\frac{T}{m}}$
The mass per unit length can be written as $m=\frac{M}{L}=\frac{A L \rho}{L}=A \rho$
$\begin{aligned} & \therefore V=\sqrt{\frac{T}{m}}=\sqrt{\frac{T}{A \rho}} \\ & \therefore V^2=\frac{T}{A \rho} \\ & \therefore A=\frac{T}{V^2 \rho}\end{aligned}$
The mass per unit length can be written as $m=\frac{M}{L}=\frac{A L \rho}{L}=A \rho$
$\begin{aligned} & \therefore V=\sqrt{\frac{T}{m}}=\sqrt{\frac{T}{A \rho}} \\ & \therefore V^2=\frac{T}{A \rho} \\ & \therefore A=\frac{T}{V^2 \rho}\end{aligned}$
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