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

Question

Consider a compound slab consisting of two different materials having equal thicknesses and thermal conductivities K and 2 K, respectively. The equivalent thermal conductivity of the slab is (A) (2/3) K (B) √ 2 K (C) 3K (D) ( 4/3) K

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/464ae1a11a7d0a03a09a970499eec6a6-.png" /> (2 ℓ/ K eq ( A ))=(ℓ/2 KA )+(ℓ/ KA ) (. series connection .R = R 1+ R 2) ⇒ K eq = (4/3) K

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

$\frac{2}{3} K$

$2 K$

3K

$\frac{4}{3} K$

$\frac{2 ℓ}{K _{e q} ( A )} = \frac{ℓ}{2 K A} + \frac{ℓ}{K A}$
$($ series connection $R = R_{1} + R_{2})$
$\Rightarrow K_{e q} = \frac{4}{3} K$

Q. 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

32​K

2​K

3K

34​K

Keq​(A)2ℓ​=2KAℓ​+KAℓ​ ( series connection R=R1​+R2​) ⇒Keq​=34​K

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

$\frac{2}{3} K$

$2 K$

$\frac{4}{3} K$

$\frac{2 ℓ}{K _{e q} ( A )} = \frac{ℓ}{2 K A} + \frac{ℓ}{K A}$
$($ series connection $R = R_{1} + R_{2})$
$\Rightarrow K_{e q} = \frac{4}{3} K$