JEE Advanced · Physics · 12. Thermal Properties

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 1m at $10^{o} C .$ Now the end P is maintained at $10^{o} C,$ while the end S is heated and maintained at $400^{o} C .$ 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 is $1.2 \times 10^{- 5} K^{- 1},$ 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

Verified Solution

Answer & Solution

Correct Answer

(A) 0.78 mm

Step-by-step Solution

Detailed explanation

Heat flow from P to Q

\frac{d Q}{d t} = \frac{2 K A (T - 10)}{1}

Heat flow from Q to S

\frac{d Q}{d t} = \frac{K A (400 - T)}{1}

At steady state heat flow is same in whole combination

\frac{2 K A (T - 10)}{1} = K A (400 - T)

2 T - 20 = 400 - T

3 T = 420

T = 140^{o}

Temp of junction is

140^{o} C

Temp at a distance x from end P
Is

T_{x} = (130 x + 10^{o})

Change in length dx is dy

d y = α d x (T_{x} - 10)

\int_{0}^{∆ y} d y = \int_{0}^{1} α d x (130 x + 10 - 10)

∆ y = {[\frac{α x^{2}}{2} \times 130]}_{0}^{1}

∆ y = 1.2 \times 10^{- 5} \times 65

∆ y = 78.0 \times 10^{- 5} m = 0.78 m m

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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	List-I		List-II
(i)	$\frac{1}{\sqrt{2}} (\sin ω t \hat{j} + \cos ω t \hat{k})$	(p)	$0$
(ii)	$\frac{1}{\sqrt{2}} (\sin ω t \hat{i} + \cos ω t \hat{j})$	(q)	$- \frac{α}{4} \hat{i}$
(iii)	$\frac{1}{\sqrt{2}} (\sin ω t \hat{i} + \cos ω t \hat{k})$	(r)	$\frac{3 α}{4} \hat{i}$
(iv)	$\frac{1}{\sqrt{2}} (\cos ω t \hat{j} + \sin ω t \hat{k})$	(s)	$\frac{α}{4} \hat{j}$
		(t)	$- \frac{3 α}{4} \hat{i}$