Question

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 1 m at 10 ⁰C. Now the end P is maintained at 10⁰0C, while the end S is heated and
maintained at 400 ⁰C. 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 x 10⁻⁵
K⁻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

Answer:

Let the temperature of the junction be T.

∴ rate of heat transfer =

dt

dQ

=

L

2KA(T−10)

=

L

KA(400−T)

⇒ 2(T−10)=400−T

or, T=140

0

C

Now, for the wire PQ, let us imagine a small length Δx at a distance x from the junction.

∴

Δx

ΔT

=

1

140−10

=130

So, temperature at distance x:

T=10+130x

or, T−10=130x

Increase in length of the small element Δx is expressed as:

dx

dy

=αΔT=α(T−10)

or,

dx

dy

=α×130x

Integrating both sides, we get:

∫

0

ΔL

dy=130α∫

0

L

xdx

or, ΔL=

2

130αx

2

=

0.78

130×1.2×10

−5

×1

m=0.78 mm