A uniform metal wire has length ' $L$ ', mass ' $M$ ' and cross-sectional area ' $A$ '. It is under tension…
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
$\frac{A T}{V^2}$
$\frac{T}{A^2 V}$
$\frac{T}{V^2 A}$
$\frac{V^2}{A^2 T}$
Solution
For transverse waves on a wire $V=\sqrt{\frac{T}{m}}$
The mass per unit length can be written as $m=\frac{M I}{L}=\frac{A L \rho}{L}=A \rho$
$\begin{aligned}
& \therefore V=\sqrt{\frac{T}{m}}=\sqrt{\frac{T}{A \rho}} \\
& \therefore V^2=\frac{T}{A \rho} \\
& \therefore A=\frac{T}{V^2 \rho} \text { SO, density }=T / V^{\wedge} 2 A
\end{aligned}$