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The temperature of the two outer surfaces of a composite slab, consisting of two materials having coefficients of thermal conductivity $K \: and\: 2K$ and thickness $x \: and\: 4x,$ respectively are $T_{2} \: and\: T_{1},\left ( T_{2}> T_{1}\right )$ .The rate of heat transfer through the slab, in a steady state is $\left ( \frac{A\left ( T_{2} -T_{1}\right )K}{x} \right )f,$ with $f$ equal to
Option: 1
$1$

Option: 2
$1/2$

Option: 3
$2/3$

Option: 4
$1/3$

Answers (1)

best_answer

Here the two slabs are in series.

$\therefore\ \; R=R_{1}+R_{2}$

R = Thermal resistance

$R=\frac{l}{KA}$

$R_{1}=\frac{x}{K.A}$ & $R_{2}=\frac{4x}{2K.A}=\frac{2x}{K.A}$

$\therefore\ \; R=\frac{3x}{K.A}$

Rate of Heat Transfer $=\frac{\Delta T}{R}$

$=\frac{T_{1}-T_{2}}{3 \left(\frac{x}{KA} \right ) }= \left [ \frac{KA(T_{2}-T_{1})}{x} \right ].\frac{1}{3}$

$\therefore\ \;f=\frac{1}{3}$

Correct option is 4.

Posted by

Divya Prakash Singh

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