MHT CET · Physics · Thermal Properties of Matter
Two metal rods \(\mathrm{P}\) and \(\mathrm{Q}\) have same length and same temperature difference between their ends. Their thermal conductivities are \(K_1\) and \(K_2\), also cross-sectional areas \(A_1\) and \(A_2\) respectively. If the rate of flow of heat through rod \(Q\) is three times that in rod \(P\), then
- A \(\mathrm{K}_1 \mathrm{~A}_1=3 \mathrm{~K}_2 \mathrm{~A}_2\)
- B \(3 \mathrm{~K}_1 \mathrm{~A}_1=\mathrm{K}_2 \mathrm{~A}_2\)
- C \(3 \mathrm{~K}_1 \mathrm{~A}_1=2 \mathrm{~K}_2 \mathrm{~A}_2\)
- D \(2 \mathrm{~K}_1 \mathrm{~A}_1=3 \mathrm{~K}_2 \mathrm{~A}_2\)
Answer & Solution
Correct Answer
(B) \(3 \mathrm{~K}_1 \mathrm{~A}_1=\mathrm{K}_2 \mathrm{~A}_2\)
Step-by-step Solution
Detailed explanation
Concept: the rate of heat flow through the rod is given by
\(\dot{\mathrm{Q}}=\frac{\mathrm{KA} \Delta \theta}{\mathrm{l}}\)
Where \(\mathrm{K}\) is the conductivity, \(\Delta \mathrm{Q}\) is the temperature difference across, \(\mathrm{A}\) is the area of cross-section and 1 is the length of the rod.
\(\begin{aligned}
& \therefore(\dot{\mathrm{Q}})_{\mathrm{Q}}=(\dot{\mathrm{Q}})_{\mathrm{P}} \\
& \Rightarrow \frac{\mathrm{K}_2 \mathrm{~A}_2}{\mathrm{l}}=\frac{3 \mathrm{~K}_1 \mathrm{~A}_1}{\mathrm{l}} \\
& \Rightarrow \mathrm{K}_2 \mathrm{~A}_2=3 \mathrm{~K}_1 \mathrm{~A}_1
\end{aligned}\)
\(\dot{\mathrm{Q}}=\frac{\mathrm{KA} \Delta \theta}{\mathrm{l}}\)
Where \(\mathrm{K}\) is the conductivity, \(\Delta \mathrm{Q}\) is the temperature difference across, \(\mathrm{A}\) is the area of cross-section and 1 is the length of the rod.
\(\begin{aligned}
& \therefore(\dot{\mathrm{Q}})_{\mathrm{Q}}=(\dot{\mathrm{Q}})_{\mathrm{P}} \\
& \Rightarrow \frac{\mathrm{K}_2 \mathrm{~A}_2}{\mathrm{l}}=\frac{3 \mathrm{~K}_1 \mathrm{~A}_1}{\mathrm{l}} \\
& \Rightarrow \mathrm{K}_2 \mathrm{~A}_2=3 \mathrm{~K}_1 \mathrm{~A}_1
\end{aligned}\)
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