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

Options

A. `0.78"mm"`
B. `0.90"mm"`
C. `1.56"mm"`
D. `2.34"mm"`

Answer

A

Explanation

Heat flow from `P "to" Q`
`(dQ)/(dt) = (2KA(T-10))/(1)`
Heat flow form `Q "to" S (dQ)/(dt) = (KA(400-T))/(1)`
At steady state heat flow is same in whole combination `(2KA(T-10))/(1) = KA(400-T)`
`2T - 20 = 400 -T`
`3T = 420`
`T= 140^(@)`

Temp of junction is `140^(@)C`
Temp at a distance x from end `P` is `T_(x) = (130x + 10^(@))`
Change in length dx is dy `dy = alphadx(T_(x)-10)`
`underset(0)overset(Deltay)intdy = underset(0)overset(1)intalphadx(130x+10-10)`
`Deltay = [(alphax^(2))/(2)xx 130]_(0)^(1)`
`Deltay= 1.2 xx 10^(-5) xx 65`
`Deltay = 78.0 xx 10^(-5)m = 0.78"mm"`