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Q8.5: NCERT Physics - Part 2 MECHANICAL PROPERTIES OF SOLIDS 8.5 Solution: Two wires of diameter 0.25 cm, one made of steel and the other…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently treat the two wires as if the same weight acts on each, adding the 4 kg and 6 kg loads to both or to the wrong wire; instead, draw a free‑body diagram: the steel wire supports the combined weight (4 kg + 6 kg) while the brass wire carries only the lower 6 kg, then use $F=mg$ for each respective wire.

Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in Fig. 8.11. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m. Compute the elongations of the steel and the brass wires.

Step-by-Step Verified Solution

**Answer:** Steel elongation = 1.49×10−41.49 \times 10^{-4} m, Brass elongation = 1.3×10−41.3 \times 10^{-4} m **Step-by-Step Explanation:** Given: Diameter d=0.25d = 0.25 cm =0.0025= 0.0025 m, so radius r=0.00125r = 0.00125 m. Area A=πr2=4.91×10−6A = \pi r^2 = 4.91 \times 10^{-6} m_. For steel: L1=1.5L_1 = 1.5 m, Y1=2.0×1011Y_1 = 2.0 \times 10^{11} Pa. Force F1=(4+6)×9.8=98F_1 = (4 + 6) \times 9.8 = 98 N.
ΔL1=F1L1AY1=98×1.54.91×10−6×2.0×1011=1.49×10−4 m\Delta L_1 = \frac{F_1 L_1}{A Y_1} = \frac{98 \times 1.5}{4.91 \times 10^{-6} \times 2.0 \times 10^{11}} = 1.49 \times 10^{-4} \text{ m}
For brass: L2=1.0L_2 = 1.0 m, Y2=0.91×1011Y_2 = 0.91 \times 10^{11} Pa. Force F2=6×9.8=58.8F_2 = 6 \times 9.8 = 58.8 N.
ΔL2=F2L2AY2=58.8×1.04.91×10−6×0.91×1011=1.3×10−4 m\Delta L_2 = \frac{F_2 L_2}{A Y_2} = \frac{58.8 \times 1.0}{4.91 \times 10^{-6} \times 0.91 \times 10^{11}} = 1.3 \times 10^{-4} \text{ m}

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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