Q. Two metal cubes $A$ and $B$ of same size are arranged as shown in the figure. The extreme ends of the combination are maintained at the indicated temperatures. The arrangement is thermally insulated. The coefficients of thermal conductivity of $A$ and $B$ are $300W/m℃$ and $200W/m℃$, respectively. After steady state is reached, the temperature of the interface will be
A) $45℃$
B) $90℃$
C) $30℃$
D) $60℃$
Explanation
(d) Temperature of interface $T = \frac{K_{1}\theta_{1} + K_{2}\theta_{2}}{K_{1} + K_{2}}$
$= \frac{300 \times 100 + 200 \times 0}{300 + 200} = 60℃$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.