Question

A steel wire of length $4.7 m$ and cross-sectional area $3.0 \times 10^{-5} m ^{2}$ stretches by the same amount as a copper wire of length $3.5 m$ 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 $\left(l_{1}\right)=4.7 \,m$
Area of cross-section $\left(A_{1}\right)=3.0 \times 10^{-5} m ^{2}$
For copper wire
Length$(l_{2})=3.5\, m$
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}} $
$=\frac{18.8}{10.5}=1.79=1.8$