Question

The ends Q and R of two thin wires, PQ and RS, are soldered (joined) together. Initially each of the of wire has a length of 1m at `10^@C.` Now the end P is maintained at `10^@C,` while the ends S is heated and maintained at `400^@C.` The system is thermally insultated 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 `1.2xx10^-5K^-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

Correct Answer - A
Let A be the arrea of cross-section of each wire and `T .^(@)C` will be the temperatre of junction of two wires

Rate of heat flow from Q to P is,
`(dQ)/(dt) = (2KA(T-10))/(1)`
Rate of heat flow from S to R is
`(dQ)/(dt) =(KA(400-T))/(1)`
At steady state, teh rate of flow of heat is same in two wires. So
`(2KA(T-10))/(1) = (KA(400-T))/(1)`
or `2T-20 =400-T` or `3T =420`
or `T=140^(@)C`
Thus temperature of junction `=140^(@)C`
Temperature difference Q and P `=140-10 =130^(@)C`
Let x be the temperature difference per unit length of wire PQ.
Temperature of element dx at a distance x from end P is `T_(x) =(130 x+10) .^(@)C`
If dy is the change in length of element dx due to thermal expansiion, then
`dy =alpha dx(T_(x) - 10) =alpha dx (130 x + 10 - 10)`
`:. dy =130 alpha x dx`
Total change in length of PQ is
`Delta y =int_(0)^(1) 130 alpha x dx = 130 alpha((x^(2))/(2))_(0)^(1) =65 alpha`
`=65xx1.2xx10^(-5) = 78.0xx10^(-5) m=0.78 mm`