Question

A copper wire is fixed between two rigid supports. It is stretched with negligible tension at ${30}^{0}C$ . The speed of transverse waves in the wire at 10 ${}^{0}C$ will be - ( density $d=90\times {10}^{3}kg/m^3$, Yong's = $1.3\times {10}^{11}$ N/$m^2$ temperature coefficient of expansion $\alpha =1.7\times {10}^{-5}{/}^{0}C$:-

• A. 210 m/s
• B. 110 m/s
• C. 90 m/s
• D. 70 m/s

Answer 1

The correct option is A 210 m/s

Given that,

Coefficient of expansion $\alpha =1.7\times {10}^{-5}{/}^{0}C$

Density $d=90\times {10}^{3}kg/m^3$

Young’s $Y=1.3\times {10}^{11}N/m^2$

Temperature $\Delta \theta ={30}^{0}C-{10}^{0}C={20}^{0}C$

We know that,

The coefficient of linear expansion is

$\alpha =\frac{dl}{l\times \Delta \theta }$

$\frac{dl}{l}=\alpha \Delta \theta$

Now, stress on the string

$\frac{F}{A}=Y\times \frac{dl}{l}$

$\frac{F}{A}=Y\alpha \Delta \theta$

$F=YA\alpha \Delta \theta$

Now, the speed of the transvers wave on the string

$v=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{F}{Ad}}$

$v=\sqrt{\frac{YA\alpha \Delta \theta }{Ad}}$

$v=\sqrt{\frac{Y\alpha \Delta \theta }{d}}$

$v=\sqrt{\frac{1.3\times {10}^{11}\times 1.7\times {10}^{-5}\times \left(20\right)}{90\times {10}^3}}$

$v=\sqrt{4911}$

$v=70.07$

$v=70.1m/s$

Hence, the speed is $70.1m/s$