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

Rate of the flow from Q to P `(dQ)/(dt)=(2KA(T-10))/(1)`
Rate of heat flow S to R `(dQ)/(dt)=(KA(400-T))/(1)`
At steady state rate of heat flow is same :

`therefore(2KA(T-10))/(1)=(KA(400-T)or2T-20=400-T`
or `3T=420thereforeT=140^(@)C`
Temperature of junction is 140°C
Temperature at a distance x from end P is `T_(X)=(130X+10^(@))`
`[Use(KA)/(1)(140-10)=(KA)/(X)(T_(X)-10)]`
Change in length dx is suppose dy

Then `dY=alphadx(T_(X)-10)`
`int_(0)^(DeltaY)dY=int_(0)^(1)alphadx(130X+10-10)`
`DeltaY=[(alphax^(2))/(2)xx130]_(0)^(1),DeltaY=1.2xx10^(-5)xx65,DeltaY=78.0xx10^(-5)m=0.78mm`