Q. Two rods of the same length and diameter having thermal conductivities $K_{1}$ and $K_{2}$ are joined in parallel. The equivalent thermal conductivity of the combination is
A) $\frac{K_{1}K_{2}}{K_{1\ }{+ \ K}_{2}}$
B) $K_{1\ }{+ \ K}_{2}$
C) $\frac{K_{1}K_{2}}{2}$
D) $\sqrt{K_{1}}K_{2}$
Explanation
(c) Let $n$ slabs each of length $l$, areas
$A_{1},\ A_{2},\ A_{3},\ldots\ldots.,A_{n}$ and thermal conductivities
$K_{1},\ K_{2},\ K_{3},\ldots\ldots.,K_{n}$are connected in parallel, then,
$K_{eq} = \frac{K_{1} + K_{2} + K_{3} + \ldots + K_{n}}{n}$
For two slabs of equal area $K_{eq} = \frac{K_{1} + K_{2}}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.