The speed of a transverse wave travelling in a wire of length $50 \mathrm{~cm}$, cross-sectional area $1…
The speed of a transverse wave travelling in a wire of length $50 \mathrm{~cm}$, cross-sectional area $1 \mathrm{~mm}^2$ and mass $5 \mathrm{~g}$ is $80 \mathrm{~ms}^{-1}$. The Young's modulus of the material of the wire is $4 \times 10^{11} \mathrm{Nm}^{-2}$. The extension in the length of the wire is
$8 \times 10^{-5} \mathrm{~m}$
$8 \times 10^{-4} \mathrm{~m}$
$16 \times 10^{-5} \mathrm{~m}$
$16 \times 10^{-4} \mathrm{~m}$
Solution
Given, length of the wire,
$
\begin{aligned}
l & =50 \mathrm{~cm} \\
& =5 \times 10^{-2} \mathrm{~m}
\end{aligned}
$
Cross-sectional area of wire,
$
\begin{aligned}
A & =1 \mathrm{~mm}^2 \\
& =1 \times 10^{-6} \mathrm{~m}^2
\end{aligned}
$
and mass of the wire, $m=5 \mathrm{~g}$
Speed of transverse wave, $v=80 \mathrm{~m} / \mathrm{s}$
Now, speed of transverse wave given as
$
\begin{array}{rlrl}
& \therefore & v & =\sqrt{\frac{T}{\mu}}=\sqrt{\frac{T}{m} \times l} \\
\Rightarrow & v & =\sqrt{\frac{T}{m} \times l}
\end{array}
$
$\Rightarrow \quad v^2=\frac{T}{m} \times l$
Now, Young's modulus, $Y=\frac{\frac{T}{A}}{\frac{\Delta l}{l}}$
From Eqs. (i) and (ii), we get
$
\therefore \quad \Delta l=\frac{v^2 m}{A Y}
$
Putting the given values, we get
$
\begin{gathered}
\Delta l=\frac{80^2 \times 5 \times 10^{-3}}{1 \times 10^{-6} \times 4 \times 10^{11}} \mathrm{~m} \\
\Delta l=8 \times 10^{-5} \mathrm{~m}
\end{gathered}
$
or
So, the extension in the length of the wire is $\Delta l=8 \times 10^{-5} \mathrm{~m}$
