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 = Y_{c} × (ΔL_{c}/L_{c}) = Y_{s} × (ΔL_{s}/L_{s}) where

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

ΔL_{c} /ΔL_{s} = (Y_{s}/Y_{c}) × (L_{c} /L_{s})

Given Lc = 2.2 m, Ls = 1.6 m, Yc = 1.1 × 1011 Nm^{–2}, and Y_{s} = 2.0 × 10^{11} Nm^{–2}.

ΔL_{c}/ΔL_{s} = (2.0 × 10^{11}/1.1 × 10^{11}) × (2.2/1.6) = 2.5.

The total elongation is given to be

ΔL_{c} + ΔL_{s} = 7.0 × 10^{-4} m

Solving the above equations,

ΔL_{c} = 5.0 × 10^{-4} m, and ΔL_{s} = 2.0 × 10^{-4} m.

Therefore

W = (A × Y_{c} × ΔL_{c})/L_{c}

= π (1.5 × 10^{-3})^{2} × [(5.0 × 10^{-4} × 1.1 × 10^{11})/2.2]

= 1.8 × 10^{2 }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 × 10

** 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/πr^{2}

=100 10^{3} N/3.14 10^{2} m^{2}

= 3.18 × 10^{8} N m^{–2}

The elongation,

ΔL = (F/A) Y

= (3.18 10^{8} N m^{-2} 1m)/ (2 10^{11} 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

** Answer**:- Length of the steel wire, L

Area of cross-section of the steel wire, A_{1} = 3.0 × 10^{–5}m^{2}

Length of the copper wire, L_{2 }= 3.5 m

Area of cross-section of the copper wire, A_{2 }= 4.0 × 10^{–5}m^{2}

Change in length = ΔL_{1} = ΔL_{2} = ΔL

Force applied in both the cases = F

Young’s modulus of the steel wire:

Y_{1} = (F_{1}/A_{1}) (L_{1}/ ΔL)

=F x 4.7/ (3.0 × 10^{–5}x ΔL) … (i)

Young’s modulus of the copper wire:

Y_{2} = (F_{2}/A_{2}) (L_{2}/ ΔL)

= F x3.5/ (4.0 × 10^{–5}x ΔL) … (ii)

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

Y_{1}/Y_{2} = (4.7x4.0 × 10^{–5})/ (3.0 × 10^{–5}x3.5)

=1.79:1

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

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