Class 11 Physics Mechanical Properties of Solids Youngs Modulus

Young’s Modulus

• Young’s modulus is derived from the name of the scientist who defined it.
• It is the ratio of longitudinal stress to longitudinal strain.
• It is denoted by Y.
• Mathematically:
• Y= longitudinal stress/ longitudinal strain = σ/ ε
• = (F/A)/ (ΔL/L)
• Y=FL/ΔL
• If Young’s modulus is more, to produce a small change in length more force required.
• S.I. Unit is N m–2 or Pascal (Pa).
• Metals have comparatively greater Young’s Modulus. To change the length of metals, greater force is required.

Problem: - A copper wire of length 2.2 m and a steel wire of length 1.6 m, both of

diameter 3.0 mm, are connected end to end. When stretched by a load, the net

elongation is found to be 0.70 mm. Obtain the load applied.

Answer:- The copper and steel wires are under a tensile stress because they have the same tension (equal to the load W) and the same area of cross-section A.

Stress = strain × Young’s modulus. Therefore

W/A = Yc × (ΔLc/Lc) = Ys × (ΔLs/Ls) where

The subscripts c and s refer to copper and stainless steel respectively. Or,

ΔLc /ΔLs = (Ys/Yc) × (Lc /Ls)

Given Lc = 2.2 m, Ls = 1.6 m, Yc = 1.1 × 1011 Nm–2, and Ys = 2.0 × 1011 Nm–2.

ΔLc/ΔLs = (2.0 × 1011/1.1 × 1011) × (2.2/1.6) = 2.5.

The total elongation is given to be

ΔLc + ΔLs = 7.0 × 10-4 m

Solving the above equations,

ΔLc = 5.0 × 10-4 m, and ΔLs = 2.0 × 10-4 m.

Therefore

W = (A × Yc × ΔLc)/Lc

= π (1.5 × 10-3)2 × [(5.0 × 10-4 × 1.1 × 1011)/2.2]

= 1.8 × 102 N

Problem: A structural steel rod has a radius of 10 mm and a length of 1.0 m. A 100 kN force stretches it along its length. Calculate (a) stress, (b) elongation, and (c) strain on the rod. Young’s modulus, of structural steel is 2.0 × 1011 N m-2.

Answer:  We assume that the rod is held by a clamp at one end, and the force F is applied at the other end, parallel to the length of the rod.

Then the stress on the rod is given by

Stress =F/A = F/πr2

=100 103 N/3.14 102 m2

= 3.18 × 108 N m–2

The elongation,

ΔL = (F/A) Y

= (3.18 108 N m-2 1m)/ (2 1011 N m-2)

= 1.59 × 10–3 m

= 1.59 mm

The strain is given by

Strain = ΔL/L

= (1.59 × 10–3 m)/ (1m)

= 1.59 × 10–3

= 0.16 %

Young’s Modulus: Application

• In industrial constructions steel is preferred over copper. The reason behind this is steel is more elastic than copper.
• If there is slight deformation in steel due to contraction and expansion it will come back to its original position.
• Steel is preferred over copper to construct bridges.

Problem:- A steel wire of length 4.7 m and cross-sectional area 3.0 × 10-5 m2 stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of  4.0 × 10–5m2, under a given load. What is the ratio of the Young’s modulus of steel to that of copper?

Answer:- Length of the steel wire, L1 = 4.7 m

Area of cross-section of the steel wire, A1 = 3.0 × 10–5m2

Length of the copper wire, L2 = 3.5 m

Area of cross-section of the copper wire, A2 = 4.0 × 10–5m2

Change in length = ΔL1 = ΔL2 = ΔL

Force applied in both the cases = F

Young’s modulus of the steel wire:

Y1 = (F1/A1) (L1/ ΔL)

=F x 4.7/ (3.0 × 10–5x ΔL)    … (i)

Young’s modulus of the copper wire:

Y2 = (F2/A2) (L2/ ΔL)

= F x3.5/ (4.0 × 10–5x ΔL)    … (ii)

Dividing (i) by (ii), we get:

Y1/Y2 = (4.7x4.0 × 10–5)/ (3.0 × 10–5x3.5)

=1.79:1

The ratio of Young’s modulus of steel to that of copper is 1.79: 1.

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