Physics · Mechanical Properties Of Solids · NEET
Start from the definition of Young's modulus: Y = stress / strain = (F/A) / (ΔL/L). Rearranging for ΔL gives ΔL = FL / (A·Y). So the formula is not a new law — it is just Young's modulus written with elongation as the subject. Whenever a numerical gives you F, L, A and Y, plug straight in.
More. In ΔL = FL/(AY), length L is in the numerator, so a longer wire elongates more for the same force. Careful though: the strain (ΔL/L) is the same for both because the extra length cancels. NEET often tests this — the wire stretches more in absolute cm, but the fractional stretch (strain) stays equal.
Thicker wire stretches less. Area A is in the denominator, so if you double the area, elongation halves. If a wire is described by its radius or diameter, remember A = πr², so doubling the radius makes area 4 times larger and elongation 4 times smaller. This diameter-to-area step is a frequent slip.
Elongation ΔL is the actual increase in length, measured in metres (or mm). Strain is ΔL/L, a pure ratio with no unit. Formulas that ask for 'how much it stretches' want ΔL; formulas involving Young's modulus, stress or energy density usually want strain. Mixing them up changes your answer by a factor of L.
Use the elastic limit (maximum safe stress). Maximum elongation ΔL_max = (elastic-limit stress × L) / Y. Here you replace F/A with the given maximum stress directly, so ΔL_max = σ_max·L / Y. NEET 2024 asked exactly this with σ = 8×10⁸ and Y = 2×10¹¹, giving 4 mm.
Two wires are made of the same material and have the same volume. The first wire has cross-sectional area A and the second wire has cross-sectional area 3A. If the length of the first wire is increased by Δl on applying a force F, how much force is needed to stretch the second wire by the same amount?
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 × 10⁸ N m⁻² and 2 × 10¹¹ N m⁻², is
Match List I with List II. List I: A. Young's Modulus, B. Compressibility, C. Bulk Modulus, D. Poisson's Ratio. List II: I. (Δd/d)/(ΔL/L), II. FL/(A·ΔL), III. -(1/V)(ΔV/P), IV. -V·P/ΔV.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
ΔL = FL / (A·Y), where F is the stretching force (load), L the original length, A the cross-sectional area, and Y the Young's modulus of the material.
Elongation ΔL is a length, so its SI unit is the metre (m). In NEET numericals answers often come out in millimetres (mm), so convert carefully: 1 mm = 10⁻³ m.
Partly. It depends on the material through Young's modulus Y, but also on the shape (length L and area A) and the load F. Two wires of the same material can have very different elongations if their length or thickness differs.
The stretching force equals the weight, F = mg. So ΔL = mgL/(AY). For a mass M hanging from a wire, substitute F = Mg into the elongation formula.
Area A is in the denominator of ΔL = FL/(AY). A larger area spreads the same force over more material, lowering the stress, so the wire stretches less. Doubling the radius makes area four times larger, cutting elongation to one-fourth.
