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 $10^{0} C$ . Now the end P is maintained at $10^{0} C$ , while the end S is heated and maintained at $400^{0} C$ . 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

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0.90 mm

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0.78 mm

2.34 mm

answer is A.

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Detailed Solution

$\begin{array}{l} S i n c e, b o t h a r e c o n n e c t e d i n s e r i e s, r a t e o f h e a t f l o w i s s a m e . \\ \Rightarrow \frac{2 k A (T - 10)}{L} = \frac{k A (400 - T)}{L} \\ \Rightarrow T = 140^{0} C \\ T e m p e r a t u r e G r a d i e n t o f P Q, \frac{△ T}{△ x} = \frac{140 - 10}{1} = 130 \\ S o, t e m p e r a t u r e a t d i s \tan c e x, T = 10 + 130 x \\ C o n s i d e r a n e l e m e n t o f l e n g t h d x a t d i s \tan c e x f r o m P . \\ I n c r e a s e i n i t s l e n g t h = d L = (d x) α (T - 10) \\ \Rightarrow \int d L = α 130 \int_{0}^{1} x d x \\ \Rightarrow △ L = 1.2 \times 10^{- 5} \times 130 \times \frac{1}{2} = 78 \times 10^{- 5} m = 0.78 m m \end{array}$