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Physics-

General

Easy

Question

The temperature of the two outer surfaces of a composite slab, consisting of two materials having coefficients of thermal conductivity $K$ and 2 $K$ and thickness $x$ and 4 $x$ , respectively are $T subscript 2 end subscript$ and $T subscript 1 end subscript left parenthesis T subscript 2 end subscript greater than T subscript 1 end subscript right parenthesis$ . The rate of heat transfer through the slab, in a steady state is $open parentheses fraction numerator A open parentheses T subscript 2 end subscript minus T subscript 1 end subscript close parentheses K over denominator x end fraction close parentheses f$ , with $f$ equals to

$1$
1/3
2/3
1/3

The correct answer is: 1/3

Let the temperature of common interface be $T ℃$ . Rate of heat flow
$H equals fraction numerator Q over denominator t end fraction equals blank fraction numerator K A increment T over denominator l end fraction$
$therefore H subscript 1 end subscript equals open parentheses fraction numerator Q over denominator t end fraction close parentheses subscript 1 end subscript$ = $fraction numerator 2 K A left parenthesis T minus T subscript 1 right parenthesis end subscript over denominator 4 x end fraction$
And $H subscript 2 end subscript equals open parentheses fraction numerator Q over denominator t end fraction close parentheses subscript 2 end subscript$ = $fraction numerator K A left parenthesis T subscript 2 end subscript minus T right parenthesis over denominator x end fraction$
In steady state, the rate of heat flow should be same in whole system $i e comma$
$H subscript 1 end subscript equals blank H subscript 2 end subscript$
$rightwards double arrow fraction numerator 2 K A left parenthesis T minus T subscript 1 end subscript right parenthesis over denominator 4 x end fraction$ = $fraction numerator K A left parenthesis T subscript 2 end subscript minus T right parenthesis over denominator x end fraction$
$rightwards double arrow fraction numerator T minus T subscript 1 end subscript over denominator 2 end fraction equals blank T subscript 2 end subscript minus T$
$rightwards double arrow T minus T subscript 1 end subscript equals blank 2 T subscript 2 end subscript minus 2 T$
$rightwards double arrow T equals fraction numerator 2 T subscript 2 end subscript plus T subscript 1 end subscript over denominator 3 end fraction$ (i)
Hence, heat flow from composite slab is
$H equals blank fraction numerator K A left parenthesis T subscript 2 end subscript minus T right parenthesis over denominator x end fraction$
= $fraction numerator K A over denominator x end fraction open parentheses T subscript 2 end subscript minus fraction numerator 2 T subscript 2 end subscript plus T subscript 1 end subscript over denominator 3 end fraction close parentheses equals blank fraction numerator K A over denominator 3 x end fraction blank left parenthesis T subscript 2 end subscript minus T subscript 1 end subscript right parenthesis$ (ii)
Accordingly, $H equals open square brackets fraction numerator A open parentheses T subscript 2 end subscript minus T subscript 1 end subscript close parentheses K over denominator x end fraction close square brackets f$ (iii)
By comparing Eqs. (ii) and (iii), we get
$rightwards double arrow f equals blank fraction numerator 1 over denominator 3 end fraction$

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