Two plates $A$ and $B$ of equal surface area are placed one on top of the other to form a composite plate of the same surface area. The thickness of $A$ and $B$ are $4.0$ cm and $6.0$ cm respectively. The temperature of the exposed surface of plate $A$ is $- 10$ $^{o}$ C and that of the exposed surface of plate B is $10$ $^{0}$ C. Neglect heat loss from the edges of the composite plate, the temperature of the contact surface is $T_{1}$ if the plates A and B are made of the same material and $T_{2}$ if their thermal conductivities are in the ratio $2 : 3$ then

T_{1} = - 4^{0} C

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T_{1} = - 2^{0} C

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T_{2} = - 3^{0} C

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T_{2} = 0^{0} C

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Solution

The correct options are
B $T_{1} = - 2^{0} C$
D $T_{2} = 0^{0} C$
We use the resistance approach for this problem.
The heat resistance is given by,
$R = \frac{L}{K A}$
If the two plates are made of the same plate,
$\frac{K_{a} A (T_{1} + 10)}{4} = - \frac{K_{b} A (T_{1} - 10)}{6}$
Solving this for $T_{1}$ gives, it equal to -2
Similarly, if the two plates are made of materials with thermal conductivities in the ratio, 2:3,
$\frac{K_{a} A (T_{1} + 10)}{4} = - \frac{K_{b} A (T_{1} - 10)}{6}$
With $\frac{K_{a}}{K_{b}} = \frac{2}{3}$
Solving this, gives $T_{2} = 0$

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