# NEET Questions Solved

If two metallic plates of equal thicknesses and thermal conductivities ${\mathrm{K}}_{1}$ and ${\mathrm{K}}_{2}$ are put together face to face and a common plate is constructed, then the equivalent thermal conductivity of this plate will be

(a) $\frac{{\mathrm{K}}_{1}{\mathrm{K}}_{2}}{{\mathrm{K}}_{1}+{\mathrm{K}}_{2}}$                     (b) $\frac{2{\mathrm{K}}_{1}{\mathrm{K}}_{2}}{{\mathrm{K}}_{1}+{\mathrm{K}}_{2}}$
(c) $\frac{{\left({\mathrm{K}}_{1}^{2}+{\mathrm{K}}_{2}^{2}\right)}^{3/2}}{{\mathrm{K}}_{1}{\mathrm{K}}_{2}}$              (d) $\frac{{\left({\mathrm{K}}_{1}^{2}+{\mathrm{K}}_{2}^{2}\right)}^{3/2}}{2{\mathrm{K}}_{1}{\mathrm{K}}_{2}}$

(b) In series

${\mathrm{R}}_{\mathrm{eq}}={\mathrm{R}}_{1}+{\mathrm{R}}_{2}⇒\frac{2\mathrm{l}}{{\mathrm{K}}_{\mathrm{eq}}\mathrm{A}}=\frac{\mathrm{l}}{{\mathrm{K}}_{1}\mathrm{A}}+\frac{\mathrm{l}}{{\mathrm{K}}_{2}\mathrm{A}}\phantom{\rule{0ex}{0ex}}⇒\frac{2}{{\mathrm{K}}_{\mathrm{eq}}}=\frac{1}{{\mathrm{K}}_{1}}+\frac{1}{{\mathrm{K}}_{2}}⇒{\mathrm{K}}_{\mathrm{eq}}=\frac{2{\mathrm{K}}_{1}{\mathrm{K}}_{2}}{{\mathrm{K}}_{1}+{\mathrm{K}}_{2}}$

Difficulty Level:

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