Question

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

Answer: (C)

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