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 $1\, m$ 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)

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

Change in length of element $= dx \alpha dT$
Where $dT =\frac{390 x }{3}+10-10$
$=\frac{390 x }{3}$
$d \ell=\frac{\alpha \times 390 x }{3} dx$
$\Delta \ell=\alpha \frac{390}{3}\left[\frac{ x ^{2}}{2}\right]_{0}^{1}$
$=\frac{1.2 \times 10^{-5} \times 390}{3} \times \frac{1}{2}$
$\Delta \ell=\frac{1.2 \times 390}{6} \times 10^{-5} \times 10^{3} mm$
$=0.78\, mm$