Two materials having coefficients of thermal conductivity ‘3K’ and ‘K’ and thickness ‘d’ and ‘3d’, respectively, are joined to form a slab as shown in the figure. The temperatures of the outer surfaces are ‘θ2’ and ‘θ1’ respectively, (θ2 θ1). The temperature of the interface is:

Question

Two materials having coefficients of thermal conductivity ‘3K’ and ‘K’ and thickness ‘d’ and ‘3d’, respectively, are joined to form a slab as shown in the figure. The temperatures of the outer surfaces are ‘θ2’ and ‘θ1’ respectively, (θ2 > θ1). The temperature of the interface is:

TuteeHUB · Accepted Answer

$\frac{{{\theta _2} + {\theta _1}}}{2}$

TuteeHUB · Answer

$\frac{{{\theta _1}}}{{10}} + \frac{{9{\theta _2}}}{{10}}$

TuteeHUB · Answer

$\frac{{{\theta _1}}}{6} + \frac{{5{\theta _2}}}{6}$

TuteeHUB · Answer

$\frac{{{\theta _1}}}{3} + \frac{{2{\theta _2}}}{3}$

1.	Two materials having coefficients of thermal conductivity ‘3K’ and ‘K’ and thickness ‘d’ and ‘3d’, respectively, are joined to form a slab as shown in the figure. The temperatures of the outer surfaces are ‘θ2’ and ‘θ1’ respectively, (θ2 > θ1). The temperature of the interface is:
A.	\(\frac{{{\theta _1}}}{{10}} + \frac{{9{\theta _2}}}{{10}}\)
B.	\(\frac{{{\theta _2} + {\theta _1}}}{2}\)
C.	\(\frac{{{\theta _1}}}{6} + \frac{{5{\theta _2}}}{6}\)
D.	\(\frac{{{\theta _1}}}{3} + \frac{{2{\theta _2}}}{3}\)
Answer» B. \(\frac{{{\theta _2} + {\theta _1}}}{2}\)

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