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A long cable hanging down a mine shaft stretches even before anything is attached to it, because every part of it carries the weight of the cable below. Exams ask for that stretch, and the usual wrong answer is twice the right one. This post derives the WL/2AY result, works an example with a cable and a load, and shows two other set-ups that use the same idea.
Is this on the syllabus?
The NCERT Class 11 chapter Mechanical Properties of Solids defines stress, strain and Young's modulus, Y = (F/A)/(ΔL/L), and its Table 8.1 gives steel a density of 7860 kg/m³, Y = 200 × 10⁹ N/m², an ultimate strength of 400 × 10⁶ N/m² and a yield strength of 250 × 10⁶ N/m². The Syllabus for JEE (Main) 2026 lists elastic behaviour, the stress-strain relationship, Hooke's law and Young's modulus under Unit 7, and the NEET (UG) 2026 syllabus lists the same. The chapter's formula assumes the same tension all along the wire. The own-weight case needs one extra step, which uses the integration in the JEE (Main) maths units.
Why is the tension different at each point?
Measure x upward from the free lower end of a uniform rod of length L and weight W. A cut at height x has to hold up the piece below it, whose weight is W × x/L. So the tension is zero at the bottom and rises in a straight line to W at the top.
How do you derive WL/2AY?
Take a small piece of length dx at height x. It carries tension T(x) = Wx/L, so by Hooke's law it stretches by
d(ΔL) = T(x) dx ÷ (AY) = (W/L) x dx ÷ (AY)
Add up all the pieces from x = 0 to x = L:
ΔL = (W/(LAY)) × L²/2 = WL/(2AY)
Since W = ρALg, this is also ΔL = ρgL²/(2Y). The area cancels, so a thick rod and a thin rod of the same material and length stretch by the same amount under their own weight. Doubling the length quadruples the stretch.
If a load of weight Mg also hangs from the bottom, the two effects add: ΔL = (Mg + W/2) × L/(AY).
Worked example: a steel cable in a mine shaft
A steel cable 100 m long with cross-section 2.0 cm² hangs in a shaft and carries a 500 kg cage at its lower end. Use NCERT's values for steel, ρ = 7860 kg/m³ and Y = 2.0 × 10¹¹ N/m², and g = 9.8 m/s².
Area: 2.0 cm² = 2.0 × 10⁻⁴ m².
Cable weight: W = ρALg = 7860 × 2.0 × 10⁻⁴ × 100 × 9.8 ≈ 1541 N, about 157 kg.
Stretch from the cable's own weight: WL/(2AY) = 1541 × 100 ÷ (2 × 2.0 × 10⁻⁴ × 2.0 × 10¹¹) ≈ 1.93 × 10⁻³ m. Check with ρgL²/(2Y) = 7860 × 9.8 × 100² ÷ (4.0 × 10¹¹) ≈ 1.93 mm.
Stretch from the cage: MgL/(AY) = 4900 × 100 ÷ (2.0 × 10⁻⁴ × 2.0 × 10¹¹) ≈ 1.23 × 10⁻² m, or 12.3 mm.
Total: about 14.2 mm.
Largest stress, at the top: (4900 + 1541) ÷ 2.0 × 10⁻⁴ ≈ 3.2 × 10⁷ N/m². That is well below the 250 × 10⁶ N/m² yield strength, so Hooke's law applies.
How long can a wire hang before it gives way?
The stress at the top of a hanging wire with nothing attached is W/A = ρgL. It reaches the yield strength σ_y when L = σ_y/(ρg). With NCERT's steel values:
Yield: 250 × 10⁶ ÷ (7860 × 9.8) ≈ 3.2 km.
Ultimate strength: 400 × 10⁶ ÷ (7860 × 9.8) ≈ 5.2 km.
The area cancels again, so a thicker wire cannot hang any further. That is a favourite JEE trap.
Where else does the same maths appear?
A rod pulled along a smooth floor. Pull a uniform rod with force F at one end. Every part accelerates together, so the tension falls linearly from F at the pulled end to zero at the free end, and ΔL = FL/(2AY). With F = 1000 N, L = 2.0 m, A = 1.0 × 10⁻⁴ m² and Y = 1.0 × 10¹¹ N/m² (values given in the problem), ΔL = 1.0 × 10⁻⁴ m, or 0.1 mm.
A column standing on the ground. The compressive stress grows from zero at the top to W/A at the base, and the column shortens by WL/(2AY).
Our post on stress, strain and Young's modulus covers the uniform-tension problems that come before this one.
Which mistakes cost marks?
Using W in place of W/2. This doubles the answer.
Adding the full cable weight to the load, or leaving the cable's weight out when the question gives its mass.
Unit slips. 1 cm² = 10⁻⁴ m² and 1 mm² = 10⁻⁶ m². Diameter is not radius.
Assuming a thicker wire can hang longer. The breaking length depends only on strength, density and g.
If you can follow the integral but would not have set it up yourself, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning behind each step.
A practice routine
Day 1: Derive WL/2AY on a blank page, starting from the tension at height x.
Day 2: Three problems with a rod plus a load, including one with two materials joined end to end.
Day 3: One rod pulled on a smooth floor and one breaking-length problem.
Today, redo the cable example for a 200 m cable with the same cage and check that the own-weight stretch rises to about 7.7 mm.
