Answer:
0.192 m
Explanation:
The young's modulus of the Nickel wire is given as,
γ = Stress/Strain
γ = (F/A)/ΔL/L
γ = FL/ΑΔL.................... Equation 1
Where γ = young's modulus of the Nickel wire, F = Load applied, A = cross sectional area of the nickel wire, L = Length of the nickel wire, ΔL = Elongation.
make ΔL the subject of the equation
ΔL = FL/γA..................... Equation 2
Also,
A = πd²/4................ Equation 3
d = diameter of the of the wire
Substitute equation 3 into equation 2
ΔL = 4FL/γπd²............... Equation 4
Given; F = 320 N, L = 2.5×10⁴ mm = 250 m, d = 1.6 mm = 0.0016 m, γ = 207×10⁹ N/m², π = 3.14
Substitute into equation 4
ΔL = (4×320×250)/(207×10⁹×3.14×0.0016²)
ΔL = 320000/1663948.8
ΔL = 0.192 m
When a load of 320 N is applied, its elongation would be "21.113 mm".
According to the question,
Length of wire = 16 mm
Force = 320 N
Young's modulus (E) = 207 GPa or,
= 207 × 10³ MPa
Now,
The stress developed in wire will be:
→ σ = [tex]\frac{P}{A}[/tex]
By substituting the values,
= [tex]\frac{320}{\frac{\pi}{4}\times (1.7)^2 }[/tex]
= 140.98 MPa
The strain will be:
→ [tex]\epsilon[/tex] = [tex]\frac{\sigma}{E}[/tex]
= [tex]\frac{140.98}{207\times 10^3}[/tex]
= 6.811 × 10⁻⁴
hence,
Total deformation will be:
= Elastic strain × Total length
= 6.811 × 10⁻⁴ × 3.1 × 10⁴
= 21.113 mm
Thus the above answer is correct.
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