Transmission Of Heat Question 125
Question: Five rods of same dimensions are arranged as shown in the figure. They have thermal conductivities K1, K2, K3, K4 and K5. When points A and B are maintained at different temperatures, no heat flows through the central rod if
[KCET 2002]
Options:
A) $ K _{1}=K _{4},\text{and},\ K _{2}=K _{3} $
B) $ K _{1}K _{4}=K _{2}K _{3} $
C) $ K _{1}K _{2}=K _{3}K _{4} $
D) $ \frac{K _{1}}{K _{4}}=\frac{K _{2}}{K _{3}} $
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Answer:
Correct Answer: B
Solution:
For no current flow between C and D
$ {( \frac{Q}{t} )}_AC=( \frac{Q}{t} )_CB $
Therefore $ \frac{K _{1}A({\theta _{A}}-{\theta _{C}})}{l}=\frac{K _{2}A({\theta _{C}}-{\theta _{B}})}{l} $
Therefore $ \frac{{\theta _{A}}-{\theta _{C}}}{{\theta _{C}}-{\theta _{B}}}=\frac{K _{2}}{K _{1}} $ …(i)
Also $ (\frac{Q}{t})_AD=(\frac{Q}{t})_DB $
Therefore $ \frac{K _{3}A({\theta _{A}}-{\theta _{D}})}{l}=\frac{K _{4}A({\theta _{D}}-{\theta _{B}})}{l} $
Therefore $ \frac{{\theta _{A}}-{\theta _{D}}}{{\theta _{D}}-{\theta _{B}}}=\frac{K _{4}}{K _{3}} $ …(ii)
It is given that $ {\theta _{C}}={\theta _{D}}, $
hence from equation (i) and (ii) we get
$ \frac{K _{2}}{K _{1}}=\frac{K _{4}}{K _{3}} $
Therefore $ K _{1}K _{4}=K _{2}K _{3} $
BETA
