Interface Temperature of Two Rods in Steady State

 

❓ 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.

---

🖼️ Question Image

---

✍️ Short Explanation

This problem is based on:

👉 Heat conduction
👉 Steady state heat flow
👉 Thermal resistance.

Main idea:

At equilibrium:

$$\boxed{ \text{Heat current through A} = \text{Heat current through B} }$$

Use conduction formula:

$$\boxed{ \frac{Q}{t} = \frac{KA\Delta T}{L} }$$

---

🔷 Step 1 — Let Interface Temperature be $T$ 💯

For rod A:

Hot end:

$$400\text{ K}$$

Interface:

$$T$$

Heat current:

$$H_A = \frac{K_A A_A(400-T)}{L_A}$$

---

For rod B:

$$H_B = \frac{K_B A_B(T-200)}{L_B}$$

At steady state:

$$H_A=H_B$$

---

🔷 Step 2 — Use Area Relation

Area of rod:

$$A=\pi r^2$$

Given:

$$\frac{r_A}{r_B}=2$$

Thus:

$$\frac{A_A}{A_B}=2^2=4$$

---

🔷 Step 3 — Substitute Ratios

Given:

$$\frac{K_A}{K_B}=\frac12$$
$$\frac{L_A}{L_B}=\frac12$$

Now:

$$\frac{K_AA_A}{L_A} = \frac{ \left(\frac12K_B\right)(4A_B) }{ \frac12L_B }$$
$$= 4\frac{K_BA_B}{L_B}$$

Thus:

$$H_A=4\frac{K_BA_B}{L_B}(400-T)$$

and

$$H_B=\frac{K_BA_B}{L_B}(T-200)$$

Equate:

$$4(400-T)=T-200$$

---

🔷 Step 4 — Solve

$$1600-4T=T-200$$
$$1800=5T$$
$$T=360\text{ K}$$

---

🔷 Step 5 — JEE Trap Alert 🚨

❌ Radius ratio ko directly area ratio maan lena

❌ Steady state me heat current unequal le lena

❌ Length ratio ulta use kar dena

Remember:

$$\boxed{ A\propto r^2 }$$

---

✅ Final Answer

$$\boxed{{\displaystyle 360\text{ K}}}$$

---

📚 Related Topics

• Organic Compound Percentage of Hydrogen Trick
• Coordination Compounds Hybridization and Magnetism
• Faraday Law and Nernst Equation Combined Concept