JEE Physics: Temperature at the Junction of Two Rods (Different Materials) Explained ⚙️

 

❓ Question

Two cylindrical rods A and B made of different materials are joined end-to-end in a straight line. The ratios are:

$$\frac{L_A}{L_B}=\frac{1}{2},\frac{r_A}{r_B}=2,\frac{K_A}{K_B}=\frac{1}{2}\mathrm{.}$$

The free ends of rods A and B are held at 400 K and 200 K respectively.
Find the interface temperature (temperature at the junction) when steady-state equilibrium is established.

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🖼️ Question Image

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✍️ Short Solution

In steady state the heat current $I$ (rate of heat flow) is the same through both rods. Thermal resistance of a rod of length $L$, cross-sectional area $A$, and thermal conductivity $K$ is:

$$R=\frac{L}{KA}\mathrm{.}$$

For series rods, temperature drops are proportional to their resistances. If $T_i$ is the interface temperature then:

$$\frac{400-T_i}{R_A}=\frac{T_i-200}{R_B}\mathrm{.}$$

So compute $R_A$ and $R_B$ up to a common factor using given ratios.

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Step 1 — Express areas via radii

Area $A=\pi r^2$. Given $r_A/r_B=2$ → $A_A/A_B = (r_A/r_B)^2 = 4$.

Step 2 — Use length and conductivity ratios

Let $L_B = L$ (choose a convenient reference). Then $L_A = L/2$.
Let $K_B = K$; then $K_A = K/2$.

Step 3 — Compute resistances (drop common factors)

$$R_A=\frac{L_A}{K_A{A}_A}=\frac{(L\mathrm{/}2)}{(K\mathrm{/}2)(4A_B)}\mathrm{.}$$

Simplify carefully:

$$R_A=\frac{L/2}{(K/2)\cdot4A_B}=\frac{L/2}{2K A_B}=\frac{L}{4K A_B}.$$
$$R_B=\frac{L_B}{K_B A_B}=\frac{L}{K A_B}.$$

So ratio:

$$R_A:R_B=\frac{L}{4K{A}_B}:\frac{L}{K{A}_B}=1:4.$$

Equivalently $R_A = \tfrac{1}{4}R_B$

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Step 4 — Relate temperature drops

Using $(400-T_i)/R_A = (T_i-200)/R_B$ and substituting $R_A = \tfrac{1}{4}R_B$:

$$\frac{400 - T_i}{\tfrac{1}{4}R_B} = \frac{T_i - 200}{R_B} \;\Rightarrow\;4(400 - T_i) = T_i - 200.$$

Solve:

$$1600 - 4T_i = T_i - 200 \Rightarrow 1600 + 200 = 5T_i \Rightarrow 1800 = 5T_i \Rightarrow T_i = 360\ \text{K}.$$

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🧮 Image Solution

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✅ Conclusion & Physical Check

• Interface temperature $T_i = \boxed{360\ \text{K}}$.

• Intuition check: because rod A has much lower resistance (shorter, thicker, lower conductivity), most of the temperature drop occurs across rod B (the higher resistance). So interface is closer to the hot end (400 K) — 360 K is reasonable.