Question 20: The maximum elongation of a steel wire of 1 m length if the elastic limit of steel and its Young’s modulus, respectively, are 8 \times 10^{8} \mathrm{~N} \mathrm{~m}^{-2} and 2 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2} is :
(1) 4 mm
(2) 0.4 mm
(3) 40 mm
(4) 8 mm
Answer: Option (1)
Explanation:
Young’s modulus is defined as the ratio of stress to strain within the elastic limit.
\mathrm{Y}=\frac{\text{Stress}}{\text{Strain}}Stress at the elastic limit is given as 8 \times 10^{8} \ \mathrm{N\,m^{-2}}
and Young’s modulus is 2 \times 10^{11} \ \mathrm{N\,m^{-2}}.
Therefore, the maximum strain is
\text{Strain}=\frac{8 \times 10^{8}}{2 \times 10^{11}} \text{Strain}=4 \times 10^{-3}Strain is also given by the ratio of change in length to original length.
\text{Strain}=\frac{\Delta L}{L}Here, L=1 \ \mathrm{m},
so the maximum elongation is
\Delta L = 4 \times 10^{-3} \times 1 \Delta L = 4 \times 10^{-3} \ \mathrm{m}Converting into millimetres,
\Delta L = 4 \ \mathrm{mm}Hence, the maximum elongation of the wire is 4 mm.