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

Answer 1

The correct option is A $0.78mm$

Let the temperature of the junction be $T$.

$\therefore$ rate of heat transfer $=\frac{dQ}{dt}=\frac{2KA(T-10)}{L}=\frac{KA(400-T)}{L}$

$\Rightarrow$ $2(T-10)=400-T$

or, $T={140}^{0}C$

Now, for the wire PQ, let us imagine a small length $\Delta x$ at a distance $x$ from the junction.

$\therefore$ $\frac{\Delta T}{\Delta x}=\frac{140-10}{1}=130$

So, temperature at distance $x$:

$T=10+130x$

or, $T-10=130x$

Increase in length of the small element $\Delta x$ is expressed as:

$\frac{dy}{dx}=\alpha \Delta T=\alpha (T-10)$

or, $\frac{dy}{dx}=\alpha \times 130x$

Integrating both sides, we get:

${\int }_0^{\Delta L}dy=130\alpha {\int }_0^{L}xdx$

$\because L=1m$ (Given)

or, $\Delta L=\frac{130\alpha x^2}{2}=\frac{130\times 1.2\times {10}^{-5}\times 1}{0.78}m=0.78mm$