Two walls of thickness d1 and d2 and thermal conductivities K1 and K2 are in contact. In the steady state, if the temperatures at the outer surfaces are T1 and T2 the temperatures at the common wall is
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a
K1T1d2+K2T2d1K1d2+K2d1
b
K1T1+K2T2d1+d2
c
K1d1+K2d2T1+T2T1T2
d
K1T1d1+K2T2d2K1d1+K2d2
answer is A.
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Detailed Solution
Let the temperature of common wall be T. ThenK1AT1−Ttd1=K1AT−T1td2K1T1−Td1=K2T−T2d2d2K1T1−T=d1K2T−T2K1T1d2−K1d2T=d1K2T−d1K2T2K1T1d2+K2T2d1=TK1d2+d1K2T=K1T1d2+K2T2d1K1d2+K2d1