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**Hint:**The question has multiple correct options. We can solve this problem using the concept of rate of heat flow. The plates are placed one on top of another, this implies that the plates are connected in series. In series, the rate of heat flow through each plate will be constant. Assume the temperature at the contact is of some value and solve.

**Complete step by step solution:**

We are given with two plates which are placed one on the top on another, they have equal surface areas. As discussed in the hint, the plates will be in series combination and the rate of heat flow through plates will be equal.

The rate of heat flow is given as

$\dfrac{Q}{t} = \dfrac{{KA({T_{final}} - {T_{initial}})}}{t}$

Here, $\dfrac{Q}{t}$ is the rate of heat flow

K is the conductivity of the material

A is the area of the surface

t is the thickness of the plate.

First case, when the plates A and B are made of the same material.

${K_A} = {K_B} = K$

Let the temperature at the contact be ${T_1}$ , the rate of heat flow through the upper plate (A) will be

${\left( {\dfrac{Q}{t}} \right)_A} = \dfrac{{KA({T_1} + 10)}}{4}$

As the temperature at the contact be ${T_1}$ , the rate of heat flow through the lower plate (B) will be

${\left( {\dfrac{Q}{t}} \right)_B} = \dfrac{{KA(10 - {T_1})}}{6}$

As there is no heat loss to the surrounding, therefore the amount of heat transferred must be equal and the rate of heat flow must also be equal.

{\left( {\dfrac{Q}{t}} \right)_A} = {\left( {\dfrac{Q}{t}} \right)_B}

$ \Rightarrow \dfrac{{KA({T_1} + 10)}}{4} = \dfrac{{KA(10 - {T_1})}}{6}$

Solving this we get.

$6({T_1} + 10) = 4(10 - {T_1})$

$ \Rightarrow 6{T_1} + 4{T_1} = 40 - 60$

$ \Rightarrow 10{T_1} = - 20$

$ \Rightarrow {T_1} = - 2$

Therefore, the temperature of the contact surface when the plates A and B are made of the same material is ${T_1} = - {2^0}C$

Thus, option B is the correct option.

Second case, when the plates A and B are made of the different material.

\dfrac{{{K_A}}}{{{K_B}}} = \dfrac{2}{3}

Let the temperature at the contact be ${T_2}$ , the rate of heat flow through the upper plate (A) will be

${\left( {\dfrac{Q}{t}} \right)_A} = \dfrac{{K{A_A}({T_2} + 10)}}{4}$

As the temperature at the contact be ${T_1}$ , the rate of heat flow through the lower plate (B) will be

${\left( {\dfrac{Q}{t}} \right)_B} = \dfrac{{{K_B}A(10 - {T_2})}}{6}$

As there is no heat loss to the surrounding, therefore the amount of heat transferred must be equal and the rate of heat flow must also be equal.

{\left( {\dfrac{Q}{t}} \right)_A} = {\left( {\dfrac{Q}{t}} \right)_B}

$ \Rightarrow \dfrac{{{K_A}A({T_2} + 10)}}{4} = \dfrac{{{K_B}A(10 - {T_2})}}{6}$

$ \Rightarrow \left( {\dfrac{{{K_A}}}{{{K_B}}}} \right)\dfrac{{({T_2} + 10)}}{4} = \dfrac{{(10 - {T_2})}}{6}$

Substituting the value of \dfrac{{{K_A}}}{{{K_B}}} = \dfrac{2}{3} , we get

$ \Rightarrow \left( {\dfrac{2}{3}} \right)\dfrac{{({T_2} + 10)}}{4} = \dfrac{{(10 - {T_2})}}{6}$

Solving this we get.

$ \Rightarrow \left( {\dfrac{2}{3}} \right)\dfrac{{({T_2} + 10)}}{4} = \dfrac{{(10 - {T_2})}}{6}$

$ \Rightarrow ({T_2} + 10) = (10 - {T_2})$

$ \Rightarrow {T_2} = {0^0}C$

Therefore, the temperature of the contact surface when the plates A and B are made of the same material is ${T_2} = {0^0}C$

Thus, option D is also the correct option.

**Option B and D are the correct options.**

**Note:**In series combination the rate of heat flow through plates will be equal. Also, there is no heat loss to the surrounding. Similar to rate of heat flow, we have thermal resistance is the ratio of the temperature difference between the two faces of a material to the rate of heat flow per unit area.

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