Question

A steel wire of length $4.5\, m$ and cross-sectional area $3 × 10^{-5}\, m^2$ stretches by the same amount as a copper wire of length $3.5\, m$ and cross-sectional area of $4 × 10^{-5}\, m^2$ 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

Answer: (C)

For copper wire, $L_C = 3.5\, m,$
$A_C = 4 × 10^{-5}\, m^2$
For steel wire, $L_S = 4.5\, m,$
$A_S = 3 × 10^{-5}\, m^2$
As Young's modulus, $\Upsilon=\frac{\left(F / A\right)}{\left(\Delta L / L\right)}$
As applied force F and extension $ΔL$ are same for steel and copper wires
$\therefore \frac{F}{\Delta L}=\frac{\Upsilon_{S}A_{S}}{L_{S}}=\frac{\Upsilon_{C}A_{C}}{L_{C}}$
where the subscripts $C$ and S refers to copper and steel respectively.
$\therefore \frac{\Upsilon_{S}}{\Upsilon_{C}}=\frac{L_{S}}{L_{C}}\times\frac{A_{C}}{A_{S}}$
$=\frac{\left(4.5\,m\right)\times\left(4\times10^{-5}\,m^{2}\right)}{\left(3.5\,m\right)\times \left(3\times 10^{-5}\,m^{2}\right)}=1.7$