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

$\sqrt{2} K$

3 K

$\frac{4}{3} K$

$\frac{2}{3} K$

Solution

$\frac{4}{3} K$

Thermal conductivity equatio is given by

H = KA $\frac{T_{1} - T_{2}}{L}$

Where, ( T₁ - T₂ ) is the temperature difference two points, A is the area of cross-section and L is the length. As in question different materials having equal thickness and at constant temperature A, L and temperature difference remains constant.

Equivalent thermal conductivity of the compound slab,

K_eq = $\frac{l_{1} + l_{2}}{\frac{l_{1}}{K_{1}} + \frac{l_{2}}{K_{2}}}$

K_eq = $\frac{l + l}{\frac{l}{K} + \frac{l}{2 k}}$

= $\frac{2 l}{\frac{3 l}{2 K}}$

K_eq = $\frac{4}{3} K$

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Thermal Properties Of Matter

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

$\sqrt{2} K$

3 K

$\frac{4}{3} K$

$\frac{2}{3} K$

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For Daily Free Study Material Join wiredfaculty

Thermal Properties Of Matter

Consider a compound slab consisting of two different materials having equal lengths, thicknesses and thermal conductivities K and 2K respectively. The equivalent thermal conductivity of the slab is 2 K 3 K 43 K 23K

Some More Questions From Thermal Properties of Matter Chapter

A radiation of energy E falls normally on a perfectly reflecting surface. The momentum transferred to the surface is

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

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

$\sqrt{2} K$

3 K

$\frac{4}{3} K$

$\frac{2}{3} K$