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Q10.9: NCERT Physics - Part 2 THERMAL PROPERTIES OF MATTER 10.9 Solution: A brass wire 1.8 m long at 27 �C is held taut with little…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often fail to take the absolute value of the thermal strain when calculating stress, leading to a negative tension which is physically impossible for a clamped wire. Remember that cooling induces compressive thermal strain, but the rigid supports generate a tensile stress equal to $Y|\alpha \Delta T|$.

A brass wire 1.8 m long at 27 �C is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of �39 �C, what is the tension developed in the wire, if its diameter is 2.0 mm? Coefficient of linear expansion of brass = 2.0 _ 10__ K__; Young's modulus of brass = 0.91 _ 10__ Pa.

Step-by-Step Verified Solution

**Answer:** Tension = 3.8×102 N3.8 \times 10^2 \text{ N} **Step-by-Step Explanation:** Thermal strain = αΔT=2.0×10−5×(−39−27)=−1.32×10−3\alpha \Delta T = 2.0 \times 10^{-5} \times (-39 - 27) = -1.32 \times 10^{-3} Since the wire is clamped, this strain is prevented, resulting in a tensile stress:
Stress=Y×∣strain∣=0.91×1011×1.32×10−3=1.20×108 Pa\text{Stress} = Y \times |\text{strain}| = 0.91 \times 10^{11} \times 1.32 \times 10^{-3} = 1.20 \times 10^8 \text{ Pa}
Area of wire A=πr2=π(1.0×10−3)2=3.14×10−6 m2A = \pi r^2 = \pi (1.0 \times 10^{-3})^2 = 3.14 \times 10^{-6} \text{ m}^2 Tension F=Stress×A=1.20×108×3.14×10−6=377 N≈3.8×102 NF = \text{Stress} \times A = 1.20 \times 10^8 \times 3.14 \times 10^{-6} = 377 \text{ N} \approx 3.8 \times 10^2 \text{ N}

All Questions in this Chapter

Q10.1 Q10.2 Q10.3 Q10.4 Q10.5 Q10.6 Q10.7 Q10.8 Q10.9 Q10.10 Q10.11 Q10.12 Q10.13 Q10.14 Q10.15 Q10.16 Q10.17 Q10.18 Q10.19 Q10.20

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