The ends Q and R of two thin wires, PQ and RS, are soldered (joined) together. Initially each of the wires…

1. 0.78 mm
2. 0.90 mm
3. 1.56 mm
4. 2.34 mm

Solution

Heat flow from P to Q
$\frac{dQ}{dt}=\frac{2KA\left(T-10\right)}{1}$
Heat flow from Q to S
$\frac{dQ}{dt}=\frac{KA (400-T)}{1}$
At steady state heat flow is same in whole combination
$\frac{2KA\left(T-10\right)}{1}=KA\left(400-T\right)$
$2T-20=400-T$
$3T=420$
$T={140}^o$

Temp of junction is ${140}^{o}C$
Temp at a distance x from end P
Is $T_x=\left(130x+{10}^o\right)$
Change in length dx is dy
$dy=\alpha dx\left(T_x-10\right)$
$\underset{0}{\overset{∆y}{\int }}dy=\underset{0}{\overset{1}{\int }}\alpha dx\left(130x+10-10\right)$
$∆y={\left[\frac{\alpha x^2}{2}\times 130\right]}_0^1$
$∆y=1.2\times {10}^{-5}\times 65$
$∆y=78.0\times {10}^{-5}m=0.78 mm$

Asked in: JEE Advanced 2016 (Paper 2)

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