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 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 &#

Step-by-step solution

Heat flow from P to Q d Q d t = 2 K A T - 10 1 Heat flow from Q to S d Q d t = K A ( 400 - T ) 1 At steady state heat flow is same in whole combination 2 K A T - 10 1 = K A 400 - T 2 T - 20 = 400 - T 3 T = 420 T = 140 o Temp of junction is 140 o C Temp at a distance x from end P Is T x = 130 x + 10 o Change in length dx is dy d y = α d x T x - 10 ∫ 0 ∆ y d y = ∫ 0 1 α d x 130 x + 10 - 10 ∆ y = α x 2 2 × 130 0 1 ∆ y = 1.2 × 10 - 5 × 65 ∆ y = 78.0 × 10 - 5 m = 0.78 m m