Question: 19. Three identical heat conducting rods are connected in series as shown in the figure. The rods on the sides have thermal conductivity 2 K while that in the middle has thermal conductivity K. The left end of the combination is maintained at temperature 3 T and the right end at T. The rods are thermally insulated from outside. In steady state, temperature at the left junction is T_{1} and that at the right junction is T_{2}. The ratio T_{1} / T_{2} is:
(1) \frac{3}{2}
(2) \frac{4}{3}
(3) \frac{5}{3}
(4) \frac{5}{4}
Answer: Option (3)
Explanation:
For steady heat conduction through series rods of equal length L and cross-sectional area A, the thermal resistance of a rod is R=\dfrac{L}{kA}, so resistance is inversely proportional to thermal conductivity k.
Let R_{0}=\dfrac{L}{K A} be the resistance of a rod of conductivity K.
Then the resistances are:
Left rod: R_{L}=\dfrac{R_{0}}{2},
middle rod: R_{M}=R_{0},
right rod: R_{R}=\dfrac{R_{0}}{2}.
Total resistance:
R_{\text{tot}}=R_{L}+R_{M}+R_{R}= \dfrac{R_{0}}{2}+R_{0}+\dfrac{R_{0}}{2}=2R_{0}.Temperature difference between ends:
\Delta T_{\text{total}}=3T-T=2T.Heat current (steady) is
Q=\dfrac{\Delta T_{\text{total}}}{R_{\text{tot}}}=\dfrac{2T}{2R_{0}}=\dfrac{T}{R_{0}}.Temperature drop across the left rod:
\Delta T_{L}=Q\,R_{L}=\dfrac{T}{R_{0}}\cdot\dfrac{R_{0}}{2}=\dfrac{T}{2}.Hence the left junction temperature is
T_{1}=3T-\Delta T_{L}=3T-\dfrac{T}{2}=\dfrac{5}{2}T.Temperature drop across the middle rod:
\Delta T_{M}=Q\,R_{M}=\dfrac{T}{R_{0}}\cdot R_{0}=T.So the right junction temperature is
T_{2}=T_{1}-\Delta T_{M}=\dfrac{5}{2}T - T=\dfrac{3}{2}T.Therefore the required ratio is
\dfrac{T_{1}}{T_{2}}=\dfrac{\tfrac{5}{2}T}{\tfrac{3}{2}T}=\dfrac{5}{3}.Thus T_{1}/T_{2}=\dfrac{5}{3},
i.e., Option (3).