KCET · Physics · Thermal Properties of Matter
The coefficient of thermal conductivity of copper is 9 times that of steel. In the composite cylindrical bar shown in the figure, what will be the temperature at the junction of copper and steel?
- A \(75^{\circ} \mathrm{C}\)
- B \(67^{\circ} \mathrm{C}\)
- C \(25^{\circ} \mathrm{C}\)
- D \(33^{\circ} \mathrm{C}\)
Answer & Solution
Correct Answer
(A) \(75^{\circ} \mathrm{C}\)
Step-by-step Solution
Detailed explanation
Temperature of interface
\(\theta=\frac{\mathrm{K}_{1} \theta_{1} \mathrm{l}_{2}+\mathrm{K}_{2} \theta_{2} \mathrm{l}_{1}}{\mathrm{~K}_{1} \mathrm{l}_{2}+\mathrm{K}_{2} \mathrm{l}_{1}}\)
It is given that \(\mathrm{K}_{\mathrm{Cu}}=9 \mathrm{~K}_{\mathrm{s}}\). So, if \(\mathrm{K}_{\mathrm{s}}=\mathrm{K}_{1}=\mathrm{K}\), then \(\mathrm{K}_{\mathrm{Cu}}=\mathrm{K}_{2}=9 \mathrm{~K}\)
\(\Rightarrow \theta=\frac{9 \mathrm{~K} \times 100 \times 6+\mathrm{K} \times 0 \times 18}{9 \mathrm{~K} \times 6+\mathrm{K} \times 18}\)
\(=\frac{5400 \mathrm{~K}}{72 \mathrm{~K}}=75^{\circ} \mathrm{C}\)
\(\theta=\frac{\mathrm{K}_{1} \theta_{1} \mathrm{l}_{2}+\mathrm{K}_{2} \theta_{2} \mathrm{l}_{1}}{\mathrm{~K}_{1} \mathrm{l}_{2}+\mathrm{K}_{2} \mathrm{l}_{1}}\)
It is given that \(\mathrm{K}_{\mathrm{Cu}}=9 \mathrm{~K}_{\mathrm{s}}\). So, if \(\mathrm{K}_{\mathrm{s}}=\mathrm{K}_{1}=\mathrm{K}\), then \(\mathrm{K}_{\mathrm{Cu}}=\mathrm{K}_{2}=9 \mathrm{~K}\)
\(\Rightarrow \theta=\frac{9 \mathrm{~K} \times 100 \times 6+\mathrm{K} \times 0 \times 18}{9 \mathrm{~K} \times 6+\mathrm{K} \times 18}\)
\(=\frac{5400 \mathrm{~K}}{72 \mathrm{~K}}=75^{\circ} \mathrm{C}\)
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