Practice portal › Thermal Properties of Matter › Conduction and Thermal Resistance

Consider a compound slab consisting of two different materials having equal thicknesses and thermal conductivities K and 2K respectively. The equivalent thermal conductivity of the slab is

Asked in CBSE AIPMT 2003 · Series and parallel resistance

• (1) 2/3K
• (2) √2 K
• (3) 3K
• (4) 4/3K

Step-by-step solution

Why the other options are wrong

• (1) comes from forgetting that the combined slab is twice as thick as either layer: writing L/(K_eq)=L/K+L/(2K) instead of (2L)/(K_eq) gives (2K)/3, exactly half the right answer
• (2) comes from the geometric mean of K and 2K, since √K×2K=√2 K; thermal resistances in series add, they do not combine as a geometric mean
• (3) comes from adding the two conductivities, K+2K, which is the rule for two slabs side by side carrying separate heat currents, not for one behind the other

Reading this solution needs no sign-in. You only sign in to practise against the clock and keep your progress.