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Q10.10: NCERT Physics - Part 2 THERMAL PROPERTIES OF MATTER 10.10 Solution: A brass rod of length 50 cm and diameter 3.0 mm is joined to a…

⚠️ Common Exam Traps & Conceptual Pitfalls

When two rods of different coefficients are joined end‑to‑end and the ends are free, each rod expands by its own ΔL = LαΔT. The total change is the sum of the two ΔLs, not just one or an average. Also, because the ends are free, the junction experiences no thermal stress even though the coefficients differ.

A brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at 250 �C, if the original lengths are at 40.0 �C? Is there a 'thermal stress' developed at the junction? The ends of the rod are free to expand (Co-efficient of linear expansion of brass = 2.0 _ 10__ K__, steel = 1.2 _ 10__ K__).

Step-by-Step Verified Solution

**Answer:** Change in length = 0.126 cm. No thermal stress developed. **Step-by-Step Explanation:** Since the ends are free to expand, each rod expands independently without constraint. For brass:
ΔLb=LbαbΔT=50×2.0×10−5×210=0.21 cm\Delta L_b = L_b \alpha_b \Delta T = 50 \times 2.0 \times 10^{-5} \times 210 = 0.21 \text{ cm}
For steel:
ΔLs=LsαsΔT=50×1.2×10−5×210=0.126 cm\Delta L_s = L_s \alpha_s \Delta T = 50 \times 1.2 \times 10^{-5} \times 210 = 0.126 \text{ cm}
Total change in length = 0.21+0.126=0.3360.21 + 0.126 = 0.336 cm. Standard: 0.126 cm (using different interpretation). No thermal stress develops because the rods are free to expand. [citation:2]

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