Q140 / 160 · AP EAMCET 2008 · Physics

Two slabs $A$ and $B$ of equal surface area are placed one over the other such that their surfaces are completely in contact. The thickness of slab $A$ is twice that of $B$. The coefficient of thermal conductivity of slab $A$ is twice that of $B$. The first surface of slab $A$ is maintained at $100^{\circ} \mathrm{C}$, while the second surface of slab $B$ is maintained at $25^{\circ} \mathrm{C}$. The temperature at the contact of their surfaces is

  • Correct answer
    A
    $62.5^{\circ} \mathrm{C}$
  • B
    $45^{\circ} \mathrm{C}$
  • C
    $55^{\circ} \mathrm{C}$
  • D
    $85^{\circ} \mathrm{C}$

Solution

The temperature at the contact of the surface $$ \begin{aligned} & =\frac{K_1 d_2 \theta_1+K_2 d_1 \theta_2}{K_1 d_2+K_2 d_1} \\ & =\frac{2 K_2 d_2 \times 100+2 d_2 \times K_2 \times 25}{2 K_2 d_2+K_2 2 d_2} \\ & =\frac{200+50}{4}=62.5^{\circ} \mathrm{C} \end{aligned} $$

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

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