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Q.

Two walls of thicknesses d1 and d2 and thermal conductivities K1 and K2 are in contact. In the steady state, if the temperatures at the outer surface are T1 and T2, the temperature at the common wall is:

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a

K1T1+K2T2d1+d2

b

K1T1d2+K2T2d1K1d2+K2d1

c

K1d1+K2d2T1+T2T1T2

d

K1T1d1+K2T2d2K1d1+K2d2

answer is B.

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Detailed Solution

Let temperature be T then K1A(T1-T)d1=K2A(T-T2)d2⇒T=K1T1d2+K2T2d1K1d2+K2d1
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