AP EAMCET · PHYSICS · Thermal Properties of Matter
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
- A \(62.5^{\circ} \mathrm{C}\)
- B \(45^{\circ} \mathrm{C}\)
- C \(55^{\circ} \mathrm{C}\)
- D \(85^{\circ} \mathrm{C}\)
Answer & Solution
Correct Answer
(A) \(62.5^{\circ} \mathrm{C}\)
Step-by-step Solution
Detailed explanation
The temperature at the contact of the surface \[ \begin{aligned} & =\frac{K_1 d_2 \theta_1+K_2 d_1 \theta_2}{K_1 d_2+K_2 d_1} \\ & =\frac{2 K_2 d_2 \times 100+2 d_2 \times K_2 \times 25}{2 K_2 d_2+K_2 2 d_2} \\ & =\frac{200+50}{4}=62.5^{\circ} \mathrm{C} \end{aligned} \]
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