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Physics-

General

Easy

Question

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

$\frac{K_{1} K_{2}}{K_{1} + K_{2}}$
$\frac{2 K_{1} K_{2}}{K_{1} + K_{2}}$
$\frac{{(K_{1}^{2} + K_{2}^{2})}^{3 / 2}}{K_{1} K_{2}}$
$\frac{{(K_{1}^{2} + K_{2}^{2})}^{3 / 2}}{2 K_{1} K_{2}}$

The correct answer is: $fraction numerator 2 K subscript 1 end subscript K subscript 2 end subscript over denominator K subscript 1 end subscript plus K subscript 2 end subscript end fraction$

In series, $R subscript e q end subscript equals R subscript 1 end subscript plus R subscript 2 end subscript rightwards double arrow fraction numerator 2 l over denominator K subscript e q end subscript A end fraction equals fraction numerator l over denominator K subscript 1 end subscript A end fraction plus fraction numerator l over denominator K subscript 2 end subscript A end fraction$
$rightwards double arrow fraction numerator 2 over denominator K subscript e q end subscript end fraction equals fraction numerator 1 over denominator K subscript 1 end subscript end fraction plus fraction numerator 1 over denominator K subscript 2 end subscript end fraction rightwards double arrow K subscript e q end subscript equals fraction numerator 2 K subscript 1 end subscript K subscript 2 end subscript over denominator K subscript 1 end subscript plus K subscript 2 end subscript end fraction$

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The graph, shown in the adjacent diagram, represents the variation of temperature $left parenthesis T right parenthesis$ of two bodies, $x$ and $y$ having same surface area, with time ( $t$ ) due to the emission of radiation. Find the correct relation between the emissivity and absorptivity power of the two bodies.

physics-General

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