Physics · Mechanical Properties Of Solids · NEET
Start from Hooke's law: within the elastic limit, stress is proportional to strain, so stress = Y × strain, where Y is the constant called Young's modulus. Longitudinal stress = force per area = F/A. Longitudinal strain = change in length per original length = ΔL/L. Substitute: Y = stress/strain = (F/A)/(ΔL/L). Flip the divided fraction: Y = (F/A) × (L/ΔL) = FL/(A·ΔL). This is the full derivation used in NCERT and NEET.
The SI unit is the pascal (Pa), equal to N·m⁻². Young's modulus = stress/strain. Strain is a ratio of two lengths, so it is a pure number with no unit. Dividing stress by a unitless number keeps the unit of stress. Stress = force/area = newton/metre² = N·m⁻² = Pa. So Young's modulus carries the exact same unit as stress: pascal.
Since Young's modulus has the same dimensions as stress (pressure), its dimensional formula is [M L⁻¹ T⁻²]. You get this from stress = force/area = [M L T⁻²]/[L²] = [M L⁻¹ T⁻²]. Strain is dimensionless, so it does not change the dimensions. NEET sometimes asks you to match this dimension with pressure or energy density (both are also [M L⁻¹ T⁻²]).
Strain = change in length ÷ original length = ΔL/L. Both ΔL and L are measured in metres, so the metres cancel and you are left with a plain number. Because it is a ratio of the same physical quantity, strain is dimensionless. This is exactly why Young's modulus ends up with the unit of stress alone.
No. Young's modulus is a property of the material only, not of the shape or size. The FL and A·ΔL in the formula are just the measured quantities of one experiment; when you compute Y you always get the same value for a given material (for example, steel ≈ 2×10¹¹ Pa) no matter how long or thick the wire is. This is a common NEET trap.
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. Choose the correct answer.
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
Try the real previous-year questions from this chapter — each with the answer and a full solution.
The pascal (Pa), which equals N·m⁻². It is the same unit as stress and pressure because strain is dimensionless.
Y = stress/strain = (F/A)/(ΔL/L) = FL/(A·ΔL).
[M L⁻¹ T⁻²], the same as pressure and stress, because strain has no dimensions.
Yes. In the CGS system it is dyne·cm⁻². 1 N·m⁻² = 10 dyne·cm⁻², so 1 Pa = 10 dyne·cm⁻².
A larger Young's modulus means a bigger force is needed to produce the same strain, so steel resists stretching more than copper. Steel ≈ 2×10¹¹ Pa while copper ≈ 1.1×10¹¹ Pa.
