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 $10$ ${}^{o}C$. Now the end P is maintained at $10$ ${}^{o}C$, while the end S is heated and maintained at $400$ ${}^{o}C$. 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

The correct option is A $0.78$ mm

Let the temp be $T$.

Rate of transfer of heat in the rod is given as,.

$\frac{\Delta Q}{\Delta t}=kA\frac{\Delta T}{\Delta x}$

${\left(\frac{\Delta Q}{\Delta t}\right)}_{PQ}={\left(\frac{\Delta Q}{\Delta t}\right)}_{RS}$

$\to K_{PQ}A\frac{(T-10)}{L}=K_{RS}A\frac{(400-T)}{L}$

$\to 2K_{RS}A\frac{(T-10)}{L}=K_{RS}A\frac{(400-T)}{L}$

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

$\to 3T=420$

$T={140}^{o}C$

for wire the temp gradient

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

Temp at the point on rod $PQ$ at the end $P$ is

$T_p=10+130x$

$(T_p-10)=130x$......(1)

Let the length of a small element be $dx$.

$\therefore \frac{dy}{dx}=\alpha (T_p-10)=\alpha 130x$.

on Integrating we get

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

$\to \Delta L=\frac{130\alpha L^2}{2}$

$\to \Delta L=\frac{130\times 1.2\times {10}^{-5}\times 1}{2}\Rightarrow 78\times {10}^{-5}\Rightarrow 0.78mm$