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Physics-

General

Easy

Question

A cylinder of radius $r$ and thermal conductivity $K subscript 1 end subscript$ is surrounded by a cylindrical shell of linear radius $r$ and outer radius $2 r$ , whose thermal conductivity is $K subscript 2 end subscript.$ There is no loss of heat across cylindrical surfaces, when the ends of the combined system are maintained at temperatures $T subscript 1 end subscript a n d T subscript 2 end subscript$ . The effective thermal conductivity of the system, in the steady state is

$\frac{K_{1} K_{2}}{K_{1} + K_{2}}$
$K_{1} + K_{2}$
$\frac{K_{1} + 3 K_{2}}{4}$
$\frac{3 K_{1} + K_{2}}{4}$

The correct answer is: $fraction numerator K subscript 1 end subscript plus 3 K subscript 2 end subscript over denominator 4 end fraction$

$open parentheses fraction numerator increment Q over denominator increment t end fraction close parentheses subscript i n n e r end subscript plus open parentheses fraction numerator increment Q over denominator increment t end fraction close parentheses subscript o u t e r end subscript equals open parentheses fraction numerator increment Q over denominator increment t end fraction close parentheses subscript t o t a l end subscript$
$fraction numerator K subscript 1 end subscript pi r to the power of 2 end exponent open parentheses T subscript 2 end subscript minus T subscript 1 end subscript close parentheses over denominator l end fraction plus blank fraction numerator K subscript 2 end subscript pi open square brackets open parentheses 2 r close parentheses to the power of 2 end exponent minus r to the power of 2 end exponent close square brackets open parentheses T subscript 2 end subscript minus T subscript 1 end subscript close parentheses over denominator l end fraction$
$equals fraction numerator K pi open parentheses 2 r close parentheses to the power of 2 end exponent open parentheses T subscript 2 end subscript minus T subscript 1 end subscript close parentheses over denominator l end fraction$
or $open parentheses K subscript 1 end subscript plus 3 K subscript 2 end subscript close parentheses fraction numerator pi r to the power of 2 end exponent open parentheses T subscript 2 end subscript minus T subscript 1 end subscript close parentheses over denominator l end fraction blank equals blank fraction numerator K pi blank 4 blank r to the power of 2 end exponent left parenthesis T subscript 2 end subscript minus T subscript 1 end subscript right parenthesis over denominator l end fraction$
$therefore blank K equals fraction numerator K subscript 1 end subscript plus 3 K subscript 2 end subscript over denominator 4 end fraction$

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