Question

A steel wire of length $4.7m$ and cross- sectional area $3.0 \times {10^{ - 5}}{m^2}$ stretches by the same amount as a copper wire of length $3.5m$and cross-sectional area of $4.0 \times {10^{ - 5}}{m^2}$ under a given load. What is the ratio of the young’s modulus of steel to that of copper?

Answer 1

We can use the formula Young’s modulus for steel
$Y = \dfrac{{F \times L}}{{A \times {\delta _1}}}$
We can recall the idea that steel wire has more young modulus than copper wire.
We can see that load is the same in this question.

Complete step by step answer:
Young’s modulus for steel ${Y_s}$
$\Rightarrow Y_s = \dfrac{{F \times {L_s}}}{{{A_s} \times {\delta _1}}}$
$F$is the force from the load
${A_S}$ Area of steel wire
${L_s}$ Length of the steel wire.
Young’s modulus for copper, ${Y_c}$
$\Rightarrow Y_c = \dfrac{{F \times {L_C}}}{{{A_C} \times {\delta _1}}}$
${A_c}$ Area of copper wire
${L_c}$ Length of the copper wire.
$\Rightarrow \therefore \dfrac{{{Y_S}}}{{{Y_C}}} = \dfrac{{{L_S} \times {A_C}}}{{{L_C} \times {A_S}}}$
$\Rightarrow \therefore \dfrac{{{Y_S}}}{{{Y_C}}} = \dfrac{{4.7 \times 4.0 \times {{10}^{ - 5}}}}{{3.5 \times 3.0 \times {{10}^{ - 5}}}}$
$\Rightarrow \dfrac{{{Y_S}}}{{{Y_C}}} = 1.79:1$

Note: We can easily express the elasticity of a material either the ratio stress to strain parameter also termed as the tensile elastic modulus or young's modulus of the material usually with the symbol ${Y_c}$. Steel has more elasticity. When we equate the both Young’s modulus cancel all common terms, and consider the power of quantities. We should be careful to answer the question, the ratio of the young’s modulus of steel to that of copper and not vice versa.