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Q.

Two slabs A and B of different materials but of the same thickness are joined end- to-end to form a composite slab. The thermal conductivities of A and B are K1 and K2, respectively. A steady temperature difference of 120C is maintained across the  composite slab. If K1=K22  the temperature difference across slab A is

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a

40C

b

60C

c

80C

d

100C

answer is C.

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Detailed Solution

The given situation can be shown as below. Rate of flow of heat will be equal in both the slabs,(12−x)K1=K2(x−0)12−x=2x⇒ x=40C    ∵K1=K22The temperature difference across slab,A=(12−x)=(12−4)=8∘C
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