Question

A steel wire of length $4.5\,m$ and cross-sectional area $3 \times {10^{ - 5}}{m^2}$ stretches by the same amount as a copper wire of length $3.5\,m$and cross-sectional area of \[4 \times {10^{ - 5}}{m^2}\] under a given load. the ratio of the young’s modulus of steel to that of copper is:
(A) $1.3$
(B) $1.5$
(C) $1.7$
(D) $1.9$

Answer 1

Hint We know here that the value of the length of the steel wire and the cross-section area is also the same as the copper wire, so we can use the young modulus principle to find the ratio of copper and steel wire. The force of both wires is also applied.

Complete step by step solution
Given by,
Copper wire, ${L_C} = 3,5m$
Area of cross- sectional of copper wire ${A_C} = 4 \times {10^{ - 5}}{m^2}$
Steel wire, ${L_S} = 4.5m$
Area of cross-sectional of steel wire \[{A_S} = 3 \times {10^{ - 5}}{m^2}\]
According to that, young modulus,
$Y = \dfrac{{\left( {F/A} \right)}}{{\Delta L/L}}$
We can by same as the copper wire,
Therefore,
$\dfrac{F}{{\Delta L}} = \dfrac{{{Y_S}{A_S}}}{{{L_S}}} = \dfrac{{{Y_C}{A_C}}}{{{L_C}}}$
Where the subscripts $C$ and $S$ refer to copper and steel respectively
Here,
$\dfrac{{{Y_S}}}{{{Y_C}}} = \dfrac{{{L_S}}}{{{L_C}}} \times \dfrac{{{A_C}}}{{{A_S}}}$
Substituting the given value in above equation,
$\dfrac{{{Y_S}}}{{{Y_C}}} = \dfrac{{\left( {4.5\,m} \right)\left( {4 \times {{10}^{ - 5}}{m^2}} \right)}}{{\left( {3.5\,m} \right)\left( {3 \times {{10}^{ - 5}}{m^2}} \right)}}$
On simplifying,
We get,
Young's modulus is often referred to as the modulus of elasticity, equal to the longitudinal stress separated by the strain.
$\dfrac{{{Y_S}}}{{{Y_C}}} = 1.7$
Hence, Thus, the ratio of the young’s modulus of steel to that of copper is $1.7$

Option C is the correct answer.

Note The elasticity modulus is the estimate of the object's stress-strain relationship. The elasticity module is the primary feature when stress is applied in the measurement of the deformation response of concrete. The range in which the stress is equal to the strain and when the outside force is removed, the material returns to its original dimensions.