TS EAMCET · Physics · Thermal Properties of Matter
Two slabs \(A\) and \(B\) of different materials but of the same thickness are joined end to end to form a composite slab. The thermal conductivities of \(A\) and \(B\) are \(k_1\) and \(k_2\) respectively. A steady temperature difference of \(12^{\circ} \mathrm{C}\) is maintained across the composite slab. If \(k_1=\frac{k_2}{2}\), the temperature difference across slab \(A\) is
- A \(4^{\circ} \mathrm{C}\)
- B \(6^{\circ} \mathrm{C}\)
- C \(8^{\circ} \mathrm{C}\)
- D \(10^{\circ} \mathrm{C}\)
Answer & Solution
Correct Answer
(C) \(8^{\circ} \mathrm{C}\)
Step-by-step Solution
Detailed explanation
The given situation can be shown as Rate of flow of heat will be equal in both the slabs \(\begin{aligned} & \therefore & (12-x) K_1 & =K_2(x-0) \\ & & 12-x & =2 x \quad\left(\because K_1=\frac{K_2}{2}\right) \\ & & x & =4^{\circ} \mathrm{C} \end{aligned}\) The temperature…
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