Three identical heat conducting rods are connected in series as shown in the figure. The rods on the sides have thermal conductivity 2K while that in the middle has thermal conductivity K. The left end of the combination is maintained at temperature 3T and the right end at T. The rods are thermally insulated from outside. In steady state, temperature at the left junction is T1 and that at the right junction is T₂. The ratio T₂/T₂ is

                                                    

**Step 1 – Understand the configuration**

The three rods are placed in series. The two outer rods have thermal conductivity $2K$ and the middle rod has conductivity $K$. All three rods are identical in length (let the common length be $L$). The left end of the whole assembly is kept at temperature $3T$ and the right end at temperature $T$. Because the sides are insulated, a steady‑state heat flow $q$ is the same through each rod.

**Step 2 – Write the thermal resistance of each rod**

Thermal resistance of a rod is $R=\backslash dfrac\{L\}\{kA\}$. Since the cross‑sectional area $A$ is the same for all rods, we can work with the resistance per unit area (ignore A):

R\_\{\backslash text\{side\}\}=\backslash dfrac\{L\}\{2K\},\backslash qquad\; R\_\{\backslash text\{middle\}\}=\backslash dfrac\{L\}\{K\}=2\backslash ,\backslash dfrac\{L\}\{2K\}\backslash ,.

**Step 3 – Total resistance of the series**

R\_\{\backslash text\{total\}\}=R\_\{\backslash text\{side\}\}+R\_\{\backslash text\{middle\}\}+R\_\{\backslash text\{side\}\}\; =\backslash dfrac\{L\}\{2K\}+2\backslash dfrac\{L\}\{2K\}+\backslash dfrac\{L\}\{2K\}\; =4\backslash ,\backslash dfrac\{L\}\{2K\}=2\backslash ,\backslash dfrac\{L\}\{K\}\backslash ,.

**Step 4 – Heat flux through the system**

The temperature difference between the ends is

\backslash Delta\; T\; =\; 3T\; -\; T\; =\; 2T\backslash ,.

Hence the steady heat flux is

q\; =\; \backslash dfrac\{\backslash Delta\; T\}\{R\_\{\backslash text\{total\}\}\}\; =\; \backslash dfrac\{2T\}\{2L/K\}=\; \backslash dfrac\{K\backslash ,T\}\{L\}\backslash ,.

**Step 5 – Temperature drop across each outer rod**

The temperature drop across an outer rod is

\backslash Delta\; T\_\{\backslash text\{side\}\}\; =\; q\backslash ,R\_\{\backslash text\{side\}\}\; =\; \backslash dfrac\{K\backslash ,T\}\{L\}\backslash times\backslash dfrac\{L\}\{2K\}=\; \backslash dfrac\{T\}\{2\}\backslash ,.

Therefore

• Temperature at the left junction (after the first rod) is

T\_1\; =\; 3T\; -\; \backslash Delta\; T\_\{\backslash text\{side\}\}\; =\; 3T\; -\; \backslash dfrac\{T\}\{2\}=\; \backslash dfrac\{5T\}\{2\}\backslash ,.

• Temperature at the right junction (before the last rod) is

T\_2\; =\; T\; +\; \backslash Delta\; T\_\{\backslash text\{side\}\}\; =\; T\; +\; \backslash dfrac\{T\}\{2\}=\; \backslash dfrac\{3T\}\{2\}\backslash ,.

**Step 6 – Form the required ratio**

The problem asks for the ratio $T\_1/T\_2$ (the image’s notation “T₂/T₂” is a typographical error). Using the values found above:

\backslash dfrac\{T\_1\}\{T\_2\}=\; \backslash dfrac\{\backslash dfrac\{5T\}\{2\}\}\{\backslash dfrac\{3T\}\{2\}\}\; =\; \backslash dfrac\{5\}\{3\}\backslash ,.

Thus the correct ratio is **5 / 3**.