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 $4.0\times 10^{-5}{m}^2$ under a given load. The ratio of the Young's modulus of steel to that of copper is

• A. 1.2
• B. 1.8
• C. 1.5
• D. 1.19

Answer 1

Answer: (B)

Given, for steel wire
Length $\left(l_1\right)=4.7m$
Area of cross-section $\left(A_1\right)=3.0\times 10^{-5}{m}^2$
For copper wire
Length$(l_2)=3.5m$
Area of cross-section $\left(A_2\right)=4.0\times 10^{-5}{m}^2$
Let $F$ be the given load under which steel and copper wires be stretched by the same amount $\Delta l$.
Young's modulus $(Y)=\frac{F/A}{\Delta l/l}=\frac{F\times l}{A\times \Delta l}$
For steel, $Y_s=\frac{F\times l_1}{A_1\times \Delta l}$ ........(i)
For copper, $Y_c=\frac{F\times l_2}{A_2\times \Delta l}$ ..........(ii)
Dividing Eq. (i) by Eq. (ii), we get
$\frac{Y_s}{Y_c}=\frac{F\times l_1}{A_1\times \Delta l}\times \frac{A_2\times \Delta l}{F\times l_2}$
$=\frac{l_1}{l_2}\times \frac{A_2}{A_1}$
$=\frac{4.7}{3.5}\times \frac{4.0\times 10^{-5}}{3.0\times 10^{-5}}$
$=\frac{18.8}{10.5}=1.79=1.8$