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 co-efficient 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 :
0.78 $$mm$$
Solution :
The heat flowrate is same
$$\eqalign{ & \therefore \,\,\frac{{KA\left( {400 - T} \right)}}{\ell } = \frac{{2\,KA\left( {T - 10} \right)}}{\ell } \cr & \therefore \,\,T = {140^ \circ }C \cr} $$
The temperature gradient access $$Pd$$ is
$$\eqalign{ & \frac{{dT}}{{dx}} = \frac{{140 - 10}}{1} \cr & \therefore \,\,dt = 130\,dx \cr} $$
Therefore change temperature at a cross - section $$M$$ distant $$'x'$$ from $$P$$ is
$$\eqalign{ & dl = dx\alpha \,\Delta T \cr & = dx\,\alpha \left( {130x} \right) \cr & \therefore \int {dl = 130\alpha \int\limits_0^1 {xdx} } \cr & \therefore \,\,\Delta l = 130 \times 1.2 \times {10^{ - 5}} \times \frac{1}{2} \cr & = 78 \times {10^{ - 5}}m \cr} $$
The heat flowrate is same
$$\eqalign{ & \therefore \,\,\frac{{KA\left( {400 - T} \right)}}{\ell } = \frac{{2\,KA\left( {T - 10} \right)}}{\ell } \cr & \therefore \,\,T = {140^ \circ }C \cr} $$
The temperature gradient access $$Pd$$ is
$$\eqalign{ & \frac{{dT}}{{dx}} = \frac{{140 - 10}}{1} \cr & \therefore \,\,dt = 130\,dx \cr} $$
Therefore change temperature at a cross - section $$M$$ distant $$'x'$$ from $$P$$ is
$$\eqalign{ & dl = dx\alpha \,\Delta T \cr & = dx\,\alpha \left( {130x} \right) \cr & \therefore \int {dl = 130\alpha \int\limits_0^1 {xdx} } \cr & \therefore \,\,\Delta l = 130 \times 1.2 \times {10^{ - 5}} \times \frac{1}{2} \cr & = 78 \times {10^{ - 5}}m \cr} $$
