Question

A copper wire and an aluminium wire 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 the increase in length of the two wires.
(Y = 1.1 x 10 Nm?, Y = 0.70 x 10 Nm?)
1) 110: 189
2) 180: 110
3) 189: 110
4) 80:11​

Answer 1

Answer:

X = Length increase = L*Force/(Area)(Y)

Y is Young's modulus

X(copper)/X(aluminum) = (3/2)*(4/5)/[(2/3)^2*(1.1/.7]

= 1.718

which happens to equal 189/110 ratio = 189:110

Answer 2

Answer:

Explanation:

it is given that

we know, Young's modulus = FL/A∆L

so, ∆L = FL/AY

so, ratio of increase in length of wires =

=

= (4/5) × (3/2) ×  × (0.7 × 10¹¹ )/(1.1 × 10¹¹)

= (4/5) × (3/2) × (3/2)² × (7/11)

= 4/5 × 27/8 × 7/11

= 189/110