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

The correct option is C $1.7$

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, $Y=\frac{\left(F/A\right)}{\left(\Delta L/L\right)}$

As applied force $F$ and extension $\Delta L$ are same for steel and copper wire
$\therefore \frac{F}{\Delta L}=\frac{Y_S{A}_S}{L_S}=\frac{Y_C{A}_C}{L_C}$

where the subscripts $C$ and $S$ refers to copper and steel respectively
$\therefore \frac{Y_S}{Y_C}=\frac{L_S}{L_C}\times \frac{A_C}{A_S}=\frac{\left(4.5m\right)(4\times {10}^{-5}{m}^2)}{\left(3.5m\right)(3\times {10}^{-5}{m}^2)}=1.7$