Properties Of Solids And Liquids Question 390
Question: If two rods A and B of equal length L, and different areas of cross-section $ {{A} _{1}} $ and$ A _{2} $ have one end each at temperature$ {{T} _{1}} $ and $ T _{2} $ , have equal rates of flow of heat, then
Options:
A) $ A _{1}=A _{2} $
B) $ \frac{A _{1}}{A _{2}}=\frac{K _{1}}{K _{2}} $
C) $ \frac{A _{1}}{A _{2}}=\frac{K _{2}}{K _{1}} $
D) $ K _{1}=K _{2} $
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Answer:
Correct Answer: C
Solution:
[c] $ \frac{\Delta Q _{1}}{\Delta t}=\frac{\Delta Q _{2}}{\Delta t} $
$ \Rightarrow K _{1}A _{1}\frac{T _{1}-T _{2}}{L _{1}}=K _{2}A _{2}\frac{T _{1}-T _{2}}{L _{2}} $
$ ( L _{1}=L _{2} )\therefore \frac{A _{1}}{A _{2}}=\frac{K _{2}}{K _{1}} $
