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# A composite slab consists of two slabs  A and  B of different materials but of the same thickness placed in contact as shown in figure. The thermal conductivity of  A and B are  ${k}_{1}$ and ${k}_{2}$ respectively. A steady temperature difference of ${12}^{\circ }\text{C}$ is maintained across the composite slab. If  ${k}_{1}=\frac{{k}_{2}}{2}$, the temperature difference across slab  A will be:

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a
4∘C
b
8∘C
c
12∘C
d
16∘C
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detailed solution

Correct option is B

Let temperature of junction be  To∘CTemperature difference across slab  A=T1−To∘CTemperature difference across slab  B=To−T2∘CSince both slabs are connected in series, heat current through both will be equal k1AT1−T0l2=k2ATo−T2l2 As k1=k22⇒k22T1−To=To−T2k2⇒T1−To=2To−2T2⇒T1+2T2=3T0….(1) As T1−T2=12∘C⇒T2=T1−12…..(2) From (1) and (2), we get ⇒T1+2T1−12=3To⇒3T1−24=3T0⇒T1−T0=8∘C

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