Q104 / 180 · NEET 2025 · Physics

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 Figure

  • A
    $\frac{3}{2}$
  • B
    $\frac{4}{3}$
  • Correct answer
    C
    $\frac{5}{3}$
  • D
    $\frac{5}{4}$

Solution

$\begin{aligned} & \text { In steady state } \Rightarrow\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_1=\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_2=\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_3 \\ & \frac{2 \mathrm{KA}\left(3 \mathrm{~T}-\mathrm{T}_1\right)}{\mathrm{L}}=\frac{\mathrm{KA}\left(\mathrm{~T}_1-\mathrm{T}_2\right)}{\mathrm{L}}=\frac{2 \mathrm{KA}\left(\mathrm{~T}_2-\mathrm{T}\right)}{\mathrm{L}} \\ & 6 \mathrm{~T}-2 \mathrm{~T}_1=\mathrm{T}_1-\mathrm{T}_2=2 \mathrm{~T}_2-2 \mathrm{~T} \\ & 6 \mathrm{~T}-2 \mathrm{~T}_1=\mathrm{T}_1-\mathrm{T}_2 \\ & 6 \mathrm{~T}+\mathrm{T}_2=3 \mathrm{~T}_1 ....(i)\\ & \mathrm{~T}_1-\mathrm{T}_2=2 \mathrm{~T}_2-2 \mathrm{~T} \\ & 2 \mathrm{~T}-3 \mathrm{~T}_2=-\mathrm{T}_1.....(ii) \end{aligned}$ Multiply equation (2) By ' 3 ' $6 \mathrm{~T}-9 \mathrm{~T}_2=-3 \mathrm{~T}_1$....(iii) Equation (1) - (3) $\begin{aligned} & 10 \mathrm{~T}_2=6 \mathrm{~T}_1 \\ & \frac{\mathrm{~T}_1}{\mathrm{~T}_2}=\frac{5}{3} \end{aligned}$

Hence, the correct answer is option: 'C'.

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