Young's modulus

Young's modulus is one of the three elastic moduli named explicitly in the CBSE Class 11 unit Mechanical Properties of Solids, alongside the bulk modulus (for volume changes) and the shear modulus (which this syllabus treats only qualitatively). You are expected to define $Y$ from stress and strain, derive and apply $Y = FL/(A\,\Delta L)$ to numericals on wires, rods, and columns under tension or compression, read $Y$ off the slope of a stress-strain curve, and connect it to the stored elastic energy per unit volume. This is by far the most heavily numerically tested modulus in the chapter — a large share of board and competitive-exam problems on this unit are, at heart, Young's-modulus problems in disguise.

Start from the two ideas that build every elastic-modulus formula: stress is the internal restoring force per unit area that a deformed body develops, and strain is the fractional deformation that produced it. For a rod or wire pulled along its length by a force $F$ applied over a cross-sectional area $A$, the longitudinal stress is

where: $\sigma$ (sigma) is longitudinal stress in pascals, $F$ is the applied force in newtons acting perpendicular to and centred on the cross-section, and $A$ is the cross-sectional area in square metres. Physically, stress is "how hard each square metre of the material's cross-section is being pulled" — the same force spread over a thinner wire produces a larger stress, which is why thin wires snap more easily than thick ones under the same load. The corresponding longitudinal strain is the fractional change in length,

where: $\varepsilon$ (epsilon) is longitudinal strain, dimensionless (a pure number, often quoted as a percentage), $\Delta L$ is the change in length in metres, and $L$ is the original, unstretched length in metres. Strain measures "how much longer the wire got, relative to how long it started" — a wire that stretches by 1 mm out of an original 1 m has the same strain as a wire that stretches by 1 cm out of an original 10 m. Robert Hooke's observation, valid for small deformations within the elastic limit, is that stress is directly proportional to strain: $\sigma \propto \varepsilon$. The constant of proportionality for longitudinal deformation is precisely Young's modulus:

where: $Y$ is Young's modulus in pascals, and every symbol on the right is as defined above. Since strain is dimensionless, $Y$ carries the same dimensions as stress, $[\text{ML}^{-1}\text{T}^{-2}]$, and the same SI unit, the pascal — for solids this is such a large number that gigapascals (GPa, $10^9$ Pa) are the practical unit. A larger $Y$ means a stiffer material: it takes a larger stress to produce the same strain, or equivalently, a given stress produces a smaller strain. Steel ($Y \approx 2.0\times10^{11}\ \text{Pa}$) is roughly three times stiffer than aluminium ($Y \approx 7\times10^{10}\ \text{Pa}$) and around 1500 times stiffer than rubber, which is why a steel cable and a rubber band respond so differently to the same pull. Graphically, if you plot stress on the vertical axis against strain on the horizontal axis for a test wire being loaded gradually, the initial portion of the curve is a straight line through the origin — this is the region where Hooke's law holds, and $Y$ is exactly the slope of that line. Beyond a point called the elastic limit, the curve bends away from the straight line; the material still recovers its shape when unloaded up to this point, but no longer obeys Hooke's law, and the CBSE syllabus keeps the numerical treatment restricted to the linear, Hooke's-law region.

A rearrangement of the same law that you will use constantly in problems is the extension a wire develops under a known load:

This shows that extension is directly proportional to the applied force and the original length, and inversely proportional to the cross-sectional area and Young's modulus. A longer or thinner wire of the same material stretches more under an identical load than a shorter or thicker one — a fact that matters directly whenever you compare two wires in a numerical. Finally, stretching a solid stores elastic potential energy in it, exactly as compressing a spring does. The energy stored per unit volume is

where: $u$ is elastic potential energy per unit volume in joules per cubic metre. This is the area under the straight-line portion of the stress-strain graph up to the working point — physically, it is the work done per unit volume in stretching the material against its internal restoring force, and it is exactly why a stretched elastic solid, like a stretched bow or spring, can do work when released.

Board level - elongation of a hanging steel wire

A steel wire of length $3.0\ \text{m}$ and diameter $1.0\ \text{mm}$ hangs a load of mass $20\ \text{kg}$ from a rigid support. Given Young's modulus of steel $Y = 2.0\times10^{11}\ \text{Pa}$ and $g = 9.8\ \text{m/s}^2$, find the elongation of the wire.

Step 1. Find the force applied. The load hangs in equilibrium, so the tension in the wire equals the weight: $F = mg = 20 \times 9.8 = 196\ \text{N}$.

Step 2. Find the cross-sectional area from the diameter. The radius is $r = 0.5\ \text{mm} = 0.5\times10^{-3}\ \text{m}$, so $A = \pi r^2 = \pi(0.5\times10^{-3})^2 = 7.854\times10^{-7}\ \text{m}^2$.

Step 3. Apply the extension formula $\Delta L = \dfrac{FL}{AY}$ with $F = 196\ \text{N}$, $L = 3.0\ \text{m}$, $A = 7.854\times10^{-7}\ \text{m}^2$, $Y = 2.0\times10^{11}\ \text{Pa}$.

Step 4. Compute numerator and denominator separately. Numerator: $196 \times 3.0 = 588$. Denominator: $7.854\times10^{-7} \times 2.0\times10^{11} = 1.571\times10^{5}$. So $\Delta L = \dfrac{588}{1.571\times10^{5}} = 3.74\times10^{-3}\ \text{m}$.

The wire stretches by about 3.74 mm under the 20 kg load — small compared to its 3 m length, consistent with steel being a very stiff material.

Competitive level - comparing two wires and the stored energy

Two wires, A and B, are made of the same steel ($Y = 2.0\times10^{11}\ \text{Pa}$) and have the same original length $L = 3.0\ \text{m}$, but wire A has twice the diameter of wire B (so wire B is the wire from the board-level example above, of diameter 1.0 mm, while wire A has diameter 2.0 mm). Both are loaded with the same $196\ \text{N}$ force used above. Find the ratio of their elongations, and the elastic potential energy density stored in wire B.

Step 1. Relate area to diameter. Since $A = \pi d^2/4$, doubling the diameter makes the area four times as large: $A_A = 4A_B$, so $A_A = 4 \times 7.854\times10^{-7} = 3.142\times10^{-6}\ \text{m}^2$.

Step 2. Use $\Delta L = \dfrac{FL}{AY}$ with identical $F$, $L$, $Y$ for both wires, so the ratio depends only on area: $\dfrac{\Delta L_A}{\Delta L_B} = \dfrac{A_B}{A_A} = \dfrac{1}{4}$, giving $\Delta L_A : \Delta L_B = 1 : 4$. Numerically, $\Delta L_A = 3.74\ \text{mm}/4 = 0.935\ \text{mm}$, while $\Delta L_B = 3.74\ \text{mm}$ as found above.

Step 3. Find the strain in wire B: $\varepsilon_B = \dfrac{\Delta L_B}{L} = \dfrac{3.74\times10^{-3}}{3.0} = 1.247\times10^{-3}$.

Step 4. Apply $u = \tfrac12 Y\varepsilon^2$ to wire B: $u_B = \tfrac12(2.0\times10^{11})(1.247\times10^{-3})^2 = \tfrac12(2.0\times10^{11})(1.555\times10^{-6}) = 1.56\times10^{5}\ \text{J/m}^3$.

The thicker wire (A) stretches only one-quarter as much as the thinner wire (B) under the same force, $\Delta L_A : \Delta L_B = 1:4$, and wire B stores about $1.56\times10^{5}\ \text{J/m}^3$ of elastic potential energy. Timing note: the area ratio ($d\to A$ scaling as $d^2$) is the fast route to the elongation ratio and should take well under a minute; save the slower per-wire arithmetic for when the question explicitly asks for the numeric elongation or the stored energy, not just the ratio.

• Using the diameter directly in $A = \pi r^2$ instead of converting to radius first — a very common arithmetic slip that makes the computed area four times too large and the elongation four times too small.
• Forgetting to convert millimetres to metres (or square millimetres to square metres) before substituting into an SI formula, producing answers off by several orders of magnitude.
• Assuming Young's modulus itself changes with the dimensions of the wire — it does not; $Y$ is a property of the material only, never of a particular sample's length or area. Only the elongation, not $Y$, depends on the wire's dimensions.
• Confusing "stress" (which depends on area) with "force" — two wires under the same force but different areas experience different stress and hence different strain, even though both have the same $Y$ if they are the same material.
• Writing $u = \tfrac12 F\Delta L$ (an energy in joules, not a density) when the question asks for energy per unit volume — remember to divide by $AL$, or equivalently just use $u = \tfrac12 Y\varepsilon^2$ directly.

• State the defining formula $Y = FL/(A\,\Delta L)$ for Young's modulus and correctly identify every symbol and its SI unit.
• Compute a wire's cross-sectional area from a given diameter, converting units to SI before substituting into any modulus formula.
• Calculate elongation, force, area, length, or Young's modulus given the other quantities in $Y = FL/(A\,\Delta L)$, including comparison problems between two wires.
• Read Young's modulus off the slope of the linear region of a stress-strain graph and explain why the slope, not the curve overall, gives $Y$.
• Compute the elastic potential energy density $u = \tfrac12 Y\varepsilon^2$ stored in a stretched wire or rod.
• Explain, in words, why $Y$ is a property of the material alone while elongation depends on the specific wire's length and area.

Ask the tutor: Stuck on a multi-wire Young's modulus problem or unsure whether to use $\Delta L = FL/(AY)$ or the energy-density formula — ask the AI tutor to walk through your unit conversions and area calculation step by step.