Question

The speed of a transverse wave, going on a wire having a length $50cm$ and mass $5.0g$, is $80m/s$. The area of cross-section of the wire is $1.0m{m}^2$ and its Young's modulus is $16\times {10}^{11}$ $N/m^2$. Find the extension of the wire over its natural length.

• A. $0.01mm$
• B. $0.02mm$
• C. $0.002mm$
• D. $0.05mm$

Answer 1

Answer: (B)

$0.02mm$

$\begin{array}{c}\text{Given,}\\ \text{wave speed on wire}=80m/s\\ \text{Length of wire (L)}=50cm\\ \text{mass of wire (M)}=5.0gram\\ \text{Area of cross section}=1.0{mm}^2\end{array}$

$\text{young's modulus}=1.6\times {10}^{11}N/m^2$

$\Delta L=?$

$\text{Now,Mass per unit lenght =}\frac{M}{L}$

$\begin{array}{cc}=& \frac{5\times {10}^{-3}Kg}{50\times {10}^{-2}m}\\ =& 0.01Kg/m\end{array}$

$\begin{array}{c}\text{we know,}\\ \text{wave velocity on wive}=\sqrt{\frac{T}{\mu }}..T=\text{Tension on wive}\\ T=v^2{\mu }_1=\dots \text{(i)}\end{array}$

$\begin{array}{c}\text{we also know,}\\ Y=\frac{\text{stress}}{\text{strain}}\\ \begin{array}{cc}Y& =\frac{T\cdot L}{A\cdot \Delta L}\\ \Delta L& =\frac{\text{T L}}{\text{AY}}\\ \text{from (i)}& \\ & =\frac{V^2\cdot \mu \cdot L}{AY}\\ & =\frac{80\times 80\cdot 0.01\times 50\times {10}^{-2}}{1\times {10}^{-6}\times 16\times {10}^{11}}=0.02mm\end{array}\end{array}$