Question

A composite slab is prepared by pasting two plates of thicknesses $\mathrm{L}_1$ and $\mathrm{L_2}$ and thermal conductivities $\mathrm{K}_1$ and $\mathrm{K}_{{2}}$. The slabs have equal Cross-sectional area. Find the equivalent conductivity of the slab.

Accepted Answer

$\mathrm{K}_{\mathrm{eq}}=\frac{\mathrm{L}_1+\mathrm{L}_2}{\frac{\mathrm{L}_1}{\mathrm{~K}_1}+\frac{\mathrm{L}_2}{\mathrm{~K}_2}}$

Answer

$\mathrm{K}_{\mathrm{eq}}=\frac{\mathrm{L}_1+\mathrm{L}_2}{\mathrm{K}_1 \mathrm{K}_2}$

Answer

$\mathrm{K}_{\mathrm{eq}}=\frac{4}{\mathrm{K}_1+\mathrm{K}_2}+\frac{\mathrm{L}_2}{\mathrm{K}_1+\mathrm{K}_2}$

Answer

$\mathrm{K}_{	ext {eq }}=\frac{\mathrm{L}_1 \mathrm{L}_2}{\mathrm{L}_1+\mathrm{L}_2}$

JEE PHYSICS MCQ: THERMAL PROPERTIES OF MATTER

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