A wire of diameter 7 mm and length 1 m is stretched within the elastic limit by the 77 kN pull. If the elongation of the wire for this force is noted as 2 mm, then find Young's modulus of elasticity for the material of the wire.

CONCEPT:

Stress:

• Stress is the ratio of the load or force to the cross-sectional area of the material to which the load is applied.
  • The standard unit of stress is N/m².

Strain:

• Strain is a measure of the deformation of the material as a result of the force applied.
  • The strain is a unitless quantity.

Hooke's law:

• Hooke's law states that within the elastic limit the stress applied on a body is directly proportional to the strain produced.

⇒ Stress ∝ Strain

⇒  Stress = E × Strain (Where E = modulus of elasticity)

\( ⇒ σ=\frac{F(N)}{A(m^2)}\)

\( ⇒ strain=\frac{dl}{l}\)    

where σ = stress, F = applied force, A = cross-sectional area, dl = change in length, l = initial length and E = young's modulus of elasticity

CALCULATION:

Given F = 77 kN = 77 × 10³ N, d = 7 mm = 7 × 10^-3 m, l = 1 m and dl = 2 mm = 2 × 10^-3 m

• The cross-sectional area of the wire is given as,

\(⇒ A=\frac{\pi}{4}d^2\)

\(⇒ A=\frac{22}{7×4}×7^2\times10^{-6}\)

\(⇒ A=\frac{77\times10^{-6}}{2}\) m²

• We know that the stress is given as,

\( ⇒ σ=\frac{F(N)}{A(m^2)}\)       ----(1)

• We know that the strain is given as,

\( ⇒ strain=\frac{dl}{l}\)     -----(2)

By Hooke's law,

⇒ σ = E × Strain         -----(3)

By equation 1, equation 2, and equation 3,

\( ⇒ \frac{F}{A}=E\times\frac{dl}{l}\)

\( ⇒ \frac{77\times10^3\times2}{77\times10^{-6}}=E\times\frac{2\times10^{-3}}{1}\)

⇒ E = 10¹² Pa

Hence, option 1 is correct.