A metal wire of length $2.0\ \text{m}$ and diameter $0.8\ \text{mm}$ has a mass of $4.0\ \text{kg}$ hung from its lower end, producing an extension of $0.5\ \text{mm}$. Taking $g = 9.8\ \text{m/s}^2$, calculate (a) the tensile stress in the wire, (b) the longitudinal strain, and (c) the Young's modulus of the material of the wire.

Given: $L = 2.0\ \text{m}$, diameter $d = 0.8\ \text{mm} = 8\times10^{-4}\ \text{m}$, mass $m = 4.0\ \text{kg}$, $g = 9.8\ \text{m/s}^2$, extension $\Delta L = 0.5\ \text{mm} = 5\times10^{-4}\ \text{m}$. Cross-sectional area, with radius $r = d/2 = 4\times10^{-4}\ \text{m}$: $$A = \pi r^2 = \pi \times (4\times10^{-4})^2 = \pi \times 1.6\times10^{-7} \approx 5.03\times10^{-7}\ \text{m}^2$$ Weight (the stretching force): $F = mg = 4.0 \times 9.8 = 39.2\ \text{N}$. **(a) Tensile stress:** $$\text{Stress} = \frac{F}{A} = \frac{39.2}{5.03\times10^{-7}} \approx 7.8\times10^7\ \text{Pa}$$ **(b) Longitudinal strain:** $$\text{Strain} = \frac{\Delta L}{L} = \frac{5\times10^{-4}}{2.0} = 2.5\times10^{-4}$$ **(c) Young's modulus:** $$Y = \frac{\text{Stress}}{\text{Strain}} = \frac{7.8\times10^7}{2.5\times10^{-4}} \approx 3.1\times10^{11}\ \text{Pa}$$ > [!ANSWER] > Stress $\approx 7.8\times10^7\ \text{Pa}$; strain $= 2.5\times10^{-4}$; $Y \approx 3.1\times10^{11}\ \text{Pa}$.

A metal wire of length 2.0m and diameter 0.8mm has a mass of 4.0kg hung from its lower end, producing an exten

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2.0\ \text{m}

(a) the tensile stress in the wire,

(b) the longitudinal strain, and

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Concept understanding — Young's Modulus

Young's modulus $Y$ , named after the physicist Thomas Young, is the modulus of elasticity associated with a change in length of an object such as a wire, rod or beam — the elasticity of length. It applies to solids only. For a wire of length $L$ and radius $r$ , loaded with mass $M$ at its free end and producing an elongation $l$ , the longitudinal stress is $Mg/\pi r^2$ and the longitudinal strain is $l/L$ , so

Y = \dfrac{\text{Longitudinal stress}}{\text{Longitudinal strain}} = \dfrac{MgL}{\pi r^2 l}

Its SI unit is $\text{N/m}^2$ and its dimensional formula is $[L^{-1}M^1T^{-2}]$ .

Young's modulus measures how strongly a solid resists being stretched or compressed along its length: a large $Y$ (steel, $21\times10^{10}$ Pa) means the material barely stretches under a given tensile stress, while a small $Y$ (lead, $1.5\times10^{10}$ Pa) means it stretches comparatively much more for the same stress — this makes Young's modulus one of the most practically useful single numbers for characterising a solid material under tension or compression.

Young's modulus can be measured experimentally by suspending a long, thin, uniform wire vertically from a rigid support, measuring its original length and (using a screw gauge) its radius, then hanging a series of known loads from its free end and measuring the resulting elongation for each load (using a vernier arrangement or a reference marker and micrometer, to keep the measurement precise for the very small elongations involved). Plotting load against elongation gives a straight line within the elastic limit, whose slope, combined with the wire's original length and cross-sectional area, gives $Y = MgL/\pi r^2 l$ directly from the measured quantities.

Young's modulus is drawn straight from the NCERT Class 11 Physics chapter on Mechanical Properties of Solids, and its experimental-determination questions are a CBSE board-exam and JEE Main favourite. This is exactly the kind of concept behind searches for "Young's modulus formula and SI unit" or "Young's modulus experiment class 11 physics".

Given: $L = 2.0\ \text{m}$ , diameter $d = 0.8\ \text{mm} = 8\times10^{-4}\ \text{m}$ , mass $m = 4.0\ \text{kg}$ , $g = 9.8\ \text{m/s}^2$ , extension $\Delta L = 0.5\ \text{mm} = 5\times10^{-4}\ \text{m}$ .

Cross-sectional area, with radius $r = d/2 = 4\times10^{-4}\ \text{m}$ :

A = \pi r^2 = \pi \times (4\times10^{-4})^2 = \pi \times 1.6\times10^{-7} \approx 5.03\times10^{-7}\ \text{m}^2

Weight (the stretching force): $F = mg = 4.0 \times 9.8 = 39.2\ \text{N}$ .

Tensile stress: $\text{Stress} = \frac{F}{A} = \frac{39.2}{5.03\times10^{-7}} \approx 7.8\times10^7\ \text{Pa}$
Longitudinal strain: $\text{Strain} = \frac{\Delta L}{L} = \frac{5\times10^{-4}}{2.0} = 2.5\times10^{-4}$
Young's modulus: $Y = \frac{\text{Stress}}{\text{Strain}} = \frac{7.8\times10^7}{2.5\times10^{-4}} \approx 3.1\times10^{11}\ \text{Pa}$
✓Final answer
$\approx 7.8\times10^7\ \text{Pa}$

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Q.A metal wire of length 2.0 m2.0\ \text{m}2.0 m and diameter 0.8 mm0.8\ \text{mm}0.8 mm has a mass of 4.0 kg4.0\ \text{kg}4.0 kg hung from its lower end, producing an extension of 0.5 mm0.5\ \text{mm}0.5 mm. Taking g=9.8 m/s2g = 9.8\ \text{m/s}^2g=9.8 m/s2, calculate

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Q.A metal wire of length $2.0\ \text{m}$ and diameter $0.8\ \text{mm}$ has a mass of $4.0\ \text{kg}$ hung from its lower end, producing an extension of $0.5\ \text{mm}$ . Taking $g = 9.8\ \text{m/s}^2$ , calculate

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