two exactly similar wire of steel and copper are stretched by equal force if total elongation is 1 cm find by how much is each wire elongated given Y for steel 20×10N and Y for copper is 12×10
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Given :
Ys=Y for steel 20×10¹⁰ N/M²
Yc=Y for copper is 12×10¹⁰ N/M²
ΔLs+ΔLc=1 ----------------equation 1
Ys= FL/ A ΔLs
Yc=FL/ A ΔLC
⇒Ys/Yc=[FL/ A ΔLs] /[FL/ A ΔLC]
⇒Ys/Yc=ΔLc/ΔLs
⇒ΔLc/ΔLs=Ys/Yc =20x10¹⁰/12x10¹⁰ = 5/3
As ΔLs+ΔLc=1
ΔLs[1+ΔLc/ΔLs]=1
⇒ΔLs[1+5/3]=1
⇒ΔLs[8/3]=1
⇒ΔLs=3/8=0.375cm
on substituting the value of ΔLs in equation (1), we get value of ΔLc
ΔLc=0.625cm
Ys=Y for steel 20×10¹⁰ N/M²
Yc=Y for copper is 12×10¹⁰ N/M²
ΔLs+ΔLc=1 ----------------equation 1
Ys= FL/ A ΔLs
Yc=FL/ A ΔLC
⇒Ys/Yc=[FL/ A ΔLs] /[FL/ A ΔLC]
⇒Ys/Yc=ΔLc/ΔLs
⇒ΔLc/ΔLs=Ys/Yc =20x10¹⁰/12x10¹⁰ = 5/3
As ΔLs+ΔLc=1
ΔLs[1+ΔLc/ΔLs]=1
⇒ΔLs[1+5/3]=1
⇒ΔLs[8/3]=1
⇒ΔLs=3/8=0.375cm
on substituting the value of ΔLs in equation (1), we get value of ΔLc
ΔLc=0.625cm
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