Verified Solution • 2025–26 Session moderate

Q8.8: NCERT Physics - Part 2 MECHANICAL PROPERTIES OF SOLIDS 8.8 Solution: A piece of copper having a rectangular cross-section of 15.2 mm …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently compute the cross‑sectional area using the given dimensions in mm directly, forgetting to convert each side to metres; this makes the area $10^6$ times too large, inflating stress and yielding an absurd strain. Always change mm to m before forming $A=\ell\times w$ and then use the correct units in the stress‑strain formula.

A piece of copper having a rectangular cross-section of 15.2 mm _ 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain.

Step-by-Step Verified Solution

**Answer:** Strain = 1.38×10−31.38 \times 10^{-3} **Step-by-Step Explanation:** Area A=15.2×10−3×19.1×10−3=2.90×10−4A = 15.2 \times 10^{-3} \times 19.1 \times 10^{-3} = 2.90 \times 10^{-4} m_. Stress = F/A=44500/2.90×10−4=1.53×108F/A = 44500 / 2.90 \times 10^{-4} = 1.53 \times 10^8 Pa. For copper, Y=1.1×1011Y = 1.1 \times 10^{11} Pa. Strain = Stress / Y = 1.53×108/(1.1×1011)=1.39×10−31.53 \times 10^8 / (1.1 \times 10^{11}) = 1.39 \times 10^{-3}.

All Questions in this Chapter

Q8.1 Q8.2 Q8.3 Q8.4 Q8.5 Q8.6 Q8.7 Q8.8 Q8.9 Q8.10 Q8.11 Q8.12 Q8.13 Q8.14 Q8.15 Q8.16

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