Q185 / 225 · JEE Main 2004 · Physics

The temperature of 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 $T_2$ 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 Figure

  • A
    1
  • B
    $1 / 2$
  • C
    $2 / 3$
  • Correct answer
    D
    $1 / 3$

Solution

$\Delta \mathrm{q}=\frac{\mathrm{kA}}{\mathrm{x}}\left[\mathrm{T}_2-\frac{2 \mathrm{~T}_2-\mathrm{T}_1}{3}\right]$ $=\frac{\mathrm{kA}}{3 \mathrm{x}}\left[\mathrm{T}_2-\mathrm{T}_1\right]$

Hence, the correct answer is option: 'D'.

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