Question

Select One Correct Option from the following : The end 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 $10^\circ c$.Now the end P is maintained at $10^\circ c$ while the end S is heated and maintained at$400^\circ c$. 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 is1.2 $10^{-5 } k^{-1}$ , 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:

correct answer is 0.78 mm

Explanation:

Let the temperature of the junction be T.

we know that

Rate of heat transfer = dQ / dt =2KA(T-10) / L = KA(400 -T) / L

⇒           2(T-10) = 400 -T

⇒           2T - 20 = 400 - T

⇒                  3T = 420

⇒                      T = 140°

Now

For the wire PQ

Let us imagine a small Δx at the distance X from the junction.

Thus

ΔT / Δx = ( 140 - 10 ) / 1 = 130°

So

temperature at distance X is given as

T = 10 +130x

⇒ T-10 =130x

For increasing in length of small element Δx  the expression is given as

dy/dx = αΔT = α(T-10)

By putting T-10 =130x we get

dy/dx =  α×130x

⇒   dy = (α×130x)dx

Integrating on both sides we get

ΔL = 130αx²/2 = ( 130×1.2×10^(-5) × 1 ) / 2 = 0.78 mm

So correct answer is 0.78 mm