Question
Download Solution PDFA 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.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFCONCEPT:
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.
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