A steel wire of length 4.5 m and a copper wire of length 3.5 m are stretched samt
wire of length 3.5 m are stretched same amount under a
given load. If ratio of youngs modulli of steel to that of copper is
12
then what is the ratio of Cross
sectional area of steel wire to copper wire ?
Answers
Answer:
As:Ac=0.107:1
Explanation:
Ls=4.5m ; Lc=3.5m ; Fs=Fc=F(let) ; ΔLs=ΔLc=ΔL(let) ; Ys:Yc=12:1 ; As:Ac=?
so, Ys/Yc= ((F/As)/(ΔL/Ls))/((F/Ac)/(ΔL/Lc))
=> 12/1 = (Ls/As)/(Lc/Ac) ( F and ΔL cancels out)
=> 12/1 = (4.5/As)/(3.5/Ac)
=> 12/1 = (4.5xAc)/(3.5xAs)
=>As/Ac= 4.5/(3.5x12)
= 4.5/42
= 0.107/1 <-:answer
Answer:
Y₁ : Y₂ = 9 : 5
Explanation:
Given :
For steel wire :
A₁ = 3 × 10⁻⁵ m³ and l₁ = 4.7 m
For copper wire :
A₂ = 4 × 10⁻⁵ m³ and l₂ = 3.5 m
It is said as :
Δl₁ = Δl₂ = Δl and F₁ = F₂ = F
We know :
Y₁ = F₁ l₁ / A₁ Δl₁
= > F / 3 × 10⁻⁵ × 4.7 /Δl
Also Y₂ = F₂ l₂ / A₂ Δl₂
Y₂ = F / 4 × 10⁻⁵ m³ × 3.5 / Δl
We have find ratio of Y₁ / Y₂
Y₁ : Y₂ = ( F / 3 × 10⁻⁵ × 4.7 /Δl ) / ( F / 4 × 10⁻⁵ m³ × 3.5 / Δl )
Y₁ : Y₂ = 4 × 10⁻⁵ × 4.7 / 3 × 10⁻⁵ × 3.5 )
Y₁ : Y₂ = 18.5 / 10.5 ≈ 1.8
Y₁ : Y₂ = 18 / 10 .
Y₁ : Y₂ = 9 : 5
Hence the ratio of the Young's modulus of steel to that of copper 9 : 5