Question

A copper wire and an aluminium wires have lengths in the ratio 3:2, diameters in the ratio 2:3 and forces applied in the ratio 4:5. Find the ratio of increase in length of the two wires. (YCu = 1.1 × 10¹¹ N m⁻², YAl = 0.7 × 10¹¹ N m⁻²)

Answer 1

it is given that
$\frac{L_{co}}{L_{al}}=\frac{3}{2}$

$\frac{d_{co}}{d_{al}}=\frac{2}{3}$

$\frac{F_{co}}{F_{al}}=\frac{4}{5}$

we know, Young's modulus = FL/A∆L
so, ∆L = FL/AY

so, ratio of increase in length of wires = $\frac{F_{co}L_{co}A_{al}Y_{al}}{F_{al}L_{al}A_{co}Y_{co}}$

= $\frac{F_{co}}{F_{Al}}\frac{L_{co}}{L_{Al}}\frac{A_{al}}{A_{co}}\frac{Y_{Al}}{Y_{co}}$

= (4/5) × (3/2) × $\frac{\pi d_{al}^2/4}{\pi d_{co}^2/4}$ × (0.7 × 10¹¹ )/(1.1 × 10¹¹)

= (4/5) × (3/2) × (3/2)² × (7/11)

= 4/5 × 27/8 × 7/11

= 189/110

Answer 2

Hii dear,

◆ Answer-
∆Lc/∆La = 1.72

◆ Explanation-
# Given-
Lc/La = 3/2
Ac/Aa = (2/3)^2 = 4/9
Fc/Fa = 4/5
Yc = 1.1×10^11 N/m^2
Ya = 0.7×10^11 N/m^2

# Solution-
For copper wire,
Yc = FcLc / Ac∆Lc
∆Lc = FcLc / AcYc

For aluminium wire,
Ya = FaLa / Aa∆La
∆La = FaLa / AaYa

Dividing ,
∆Lc/∆La = (FcLc/AcYc) / (FaLa/AaYa)
∆Lc/∆La = 4/5 × 3/2 × 9/4 × 0.7/1.1
∆Lc/∆La = 1.72

Hence, we can say that ∆Lc/∆La = 1.72 .

Hope that is useful...