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Question: Two bars of equal length and the same cross-sectional area but of different thermal conductivities, ...

Two bars of equal length and the same cross-sectional area but of different thermal conductivities, k1 and k2, are joined end to end as shown in figure. One end of the composite bar is maintained at temperature Th where as the opposite end is held at Tc. If there are no heat losses from the sides of the bars, the temperature Tj of the junction is given by –

A

k2k1\frac { \mathrm { k } _ { 2 } } { \mathrm { k } _ { 1 } } (Th+Tc)2\frac{(T_{h} + T_{c})}{2}

B

k2k1+k2\frac{k_{2}}{k_{1} + k_{2}} (Th + Tc)

C

(Th+Tc)2\frac{(T_{h} + T_{c})}{2}

D

1k1+k2\frac{1}{k_{1} + k_{2}} (k1Th + k2Tc)

Answer

1k1+k2\frac{1}{k_{1} + k_{2}} (k1Th + k2Tc)

Explanation

Solution

t µ Kr2l\frac{Kr^{2}}{\mathcal{l}}

t1 = t2

K1r12l1\frac{K_{1}r_{1}^{2}}{\mathcal{l}_{1}}= K2r22l2\frac{K_{2}r_{2}^{2}}{\mathcal{l}_{2}}