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# Two rods of diffrent material with diffrent area of cross section and length of second rod is equal to twice of length of first rod, k2=4k1 and diameter of d2=d1/2 where d is the diameter  if two conductors . Find the Temperature at the junction

Eshan
2 years ago
Dear student,

For rods connected end to end, the rate of heat transfer through them must be same.

Hence$\dpi{80} \dfrac{K_1A_1}{l_1}(T_A-T)=\dfrac{K_2A_2}{l_2}(T-T_B)$
$\dpi{80} \implies \dfrac{K_1\pi d_1^2}{4l_1}(T_A-T)=\dfrac{K_2\pi d_2^2}{4l_2}(T-T_B)$
$\dpi{80} \implies 2(T_A-T)=(T-T_B)$
$\dpi{80} \implies T=\dfrac{2T_A+T_B}{3}$