Question

An aluminium wire and a steel wire of the same length and cross-section are joined end to end. The composite wire is hung from a rigid support and a load is suspended from the free end. If the increase in length of the composite wire is 1.35mm, find the ratio of a. stress in the two wires and b. strain in the two wires $(Y_{AI}=0.7 * 10^{11} N m^{-2}, Y{steel}=2*10^{11} Nm^{-2})$

Answer 1

a/c to question, An aluminium and steel wire of the same length . e.g., $L_{Al}=L_{s}$
cross section are joined end to end , so, cross sectional area of wires are same . e.g., $A_{Al}=A_{s}$
and also load applied to wires are same.e.g., $F_{Al}=F_{s}$
$Y_{AI}=0.7\times10^{11}Nm^{-2}$
$Y_{steel}=2\times10^{11} Nm^{-2}$
increase in length of composite wires, $\Delta L=\Delta L_1+\Delta L_2$ = 1.35mm



(a) stress is the ratio of force per unit area.


e.g., $\frac{stress_{Al}}{stress_{s}}=\frac{F_{Al}}{A_{Al}}\frac{A_{s}}{F_{s}}$

= $\frac{F_{Al}}{F_{s}}\frac{A_{Al}}{A_{s}}$

= 1 . 1 = 1


(b) strain in two wires, $\frac{\Delta L_{Al}}{\Delta L_{s}}=\frac{Y_{Al}}{Y_{s}}$

= 2 × 10¹¹ N/m²/0.7 × 10¹¹ N/m²

= 20/7

Answer 2

Explanation:

An aluminium wire and a steel wire of the same length and cross-section are joined end to end. The composite wire is hung from a rigid support and a load is suspended from the free end. If the increase in length of the composite wire is 1.35mm, find the ratio of a. stress in the two wires and b. strain in the two wires $(Y_{AI}=0.7 * 10^{11} N m^{-2}, Y{steel}=2*10^{11} Nm^{-2})$