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. 0.90mm
C. 1.56 mm
D. 2.34 mm

Answer

A

Explanation

First we need to find temperature of junction. Let temperature of junction be T. Both the wires are connected in series, hence same heat current flows through them.
`(dQ)/(dt) = (2KA (T - 10) )/(L) = (KA (400 - T) )/(L)`
` rArr 2 (T - 10) = 400 - T`
` 3T = 420`
`T = 140^@C`
for wire PQ

In case of steady state we can assume that temperature gradient remains constant along the length of rod, hence temperature gradient can be written as follows:
`(dT)/(dx) = (140 - 10)/(1) = 130` ....(i)
Let us select a segment at a distance x of width dx. Temperature at distance x can be written as follows:
T = 10 + 130x
Relation between length and temperature can be written as follows:
`L = L_0 (1+ alpha T)`
` rArr (L - L_0)/(L_0) = alpha Delta T`
` rArr (Delta L)/(L) = alpha Delta T`
Further let us assume that length of segment dx changes by dy, when temperature changes by dT, then from above equation we can write the following:
`(dy)/(dx) = alpha dT`
`(dy)/(dx) = alpha (130x)`
` int_(0)^(Delta L) dy = 130 alpha int_(0)^(L) x dx`
` Delta L = 130 alpha [x/2]_0^1`
`Delta L = (130 xx 1.2 xx 10^(-5) xx 1)/(2)`
` Delta L = 78 xx 10^(-5) m = 0.78 mm`