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Q8.7: NCERT Physics - Part 2 MECHANICAL PROPERTIES OF SOLIDS 8.7 Solution: Four identical hollow cylindrical columns of mild steel support …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently insert the radii in centimeters directly into $A=\pi(r_2^2-r_1^2)$, producing an area in cm$^2$ but then use SI units for force, yielding a stress that is off by a factor of $10^4$. Always convert $r_1$ and $r_2$ to meters before computing the cross‑sectional area.

Four identical hollow cylindrical columns of mild steel support a big structure of mass 50,000 kg. The inner and outer radii of each column are 30 and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column.

Step-by-Step Verified Solution

**Answer:** Compressional strain = 7.22×10−77.22 \times 10^{-7} **Step-by-Step Explanation:** Total mass = 50,00050,000 kg, so total weight = 50,000×9.8=4.9×10550,000 \times 9.8 = 4.9 \times 10^5 N. Load on each column = 4.9×105/4=1.225×1054.9 \times 10^5 / 4 = 1.225 \times 10^5 N. Area of each column = π(r22−r12)=π(0.62−0.32)=π(0.36−0.09)=π×0.27=0.848\pi(r_2^2 - r_1^2) = \pi(0.6^2 - 0.3^2) = \pi(0.36 - 0.09) = \pi \times 0.27 = 0.848 m_. Stress = F/A=1.225×105/0.848=1.445×105F/A = 1.225 \times 10^5 / 0.848 = 1.445 \times 10^5 Pa. For steel, Y=2.0×1011Y = 2.0 \times 10^{11} Pa. Strain = Stress / Y = 1.445×105/(2.0×1011)=7.22×10−71.445 \times 10^5 / (2.0 \times 10^{11}) = 7.22 \times 10^{-7}.

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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