Question

The ends $Q$ and $R$ of two thin wires, $PQ$ and $RS$, are soldered (joined) together. Initially each of the wires has a length of $1m$ at $10^{\circ }C$. Now the end $P$ is maintained at $10^{\circ }C$, while the end $S$ is heated and maintained at $400^{\circ }C$. The system is thermally insulated from its surroundings. If the thermal conductivity of wire $PQ$ is twice that of the wire RS and the coefficient of linear thermal expansion of $PQ$ is $1.2\times 10^{-5}{K}^{-1}$, the change in length of the wire $PQ$ is

• A. 0.78 mm
• B. 0.90 mm
• C. 1.56 mm
• D. 2.34 mm

Answer 1

Answer: (A)

$K_{PQ}=2K_{RS}$
${\alpha }_{PQ}=1.2\times 10^{-5}{K}^{-1}$
$\frac{dH}{dt}=\frac{390}{3R}=KA\frac{dT}{dx}$
$\frac{dT}{dx}=\frac{390}{3RKA}$
$=\frac{390}{3\times \frac{1}{KA}\times KA}$
$\frac{dT}{dx}=\frac{390}{3}$
$T=\frac{390x}{3}+10$

Change in length of element $=dx\alpha dT$
Where $dT=\frac{390x}{3}+10-10$
$=\frac{390x}{3}$
$d\ell =\frac{\alpha \times 390x}{3}dx$
$\Delta \ell =\alpha \frac{390}{3}{\left[\frac{x^2}{2}\right]}_0^1$
$=\frac{1.2\times 10^{-5}\times 390}{3}\times \frac{1}{2}$
$\Delta \ell =\frac{1.2\times 390}{6}\times 10^{-5}\times 10^{3}mm$
$=0.78mm$