Question

Two wires of different material have same length and area of cross-section. What is the ratio of their increase in length when forces applied are the same? (Y₁ = 0.9 × 10¹¹ N m⁻², Y₂ = 3.6 × 10¹¹ N m⁻²).

Answer 1

Given, Two wires of different material have same length and cross sectional area e.g., $L_1=L_2=L$ and $A_1=A_2=A$ and also $F_1=F_2=F$
$Y_1=0.9\times10^{11}N/m^2$,$Y_2=3.6\times10^{11}N/m^2$

we know, Young's modulus, Y = FL/A∆L

so, ∆L = FL/YA
A, F and L are constant.
e.g., $\frac{\Delta L_1}{\Delta L_2}=\frax{Y_2}{Y_1}$

= (3.6 × 10¹¹)/(0.9 × 10¹¹)

= 4/1

Hence, ratio of their increase in length = 4 : 1

Answer 2

Explanation:

therefore ratio of increase in length is 4:1 hope, it will help you