Two materials having coefficients of thermal conductivity $3 K$ and $K$ and thickness $d$ and $3 d$ respectively, are joined to form a slab as shown in the figure. The temperatures of the outer surfaces are $θ_{2}$ and $θ_{1}$ respectively, $(θ_{2} > θ_{1}) .$ The temperature at the interface is

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Let the temperature of the junction T ° C . Rate of heat flow in Rod 1 = rate of heat flow in Rod 2 3 kA d θ 2 - T = kA 3 d T - θ 1 ⇒ 9 θ 2 - T = T - θ 1 ⇒ 10 T = 9 θ 2 + θ 1 ⇒ T = 9 θ 2 + θ 1 10 = θ 1 10 + 9 θ 2 10

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