Question

A steel wire of length 4.7 m and cross-sectional area $3 \times 10^{-5} m ^{2}$ stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of $4.0 \times 10^{-5} m ^{2}$ under a given load. The ratio of Young's modulus of steel to that of copper is

• A. 1.2
• B. 1.4
• C. 1.6
• D. 1.8

Answer 1

Answer: (D)

$Y=\frac{F l}{A \cdot \Delta l}$ or $\Delta l=\frac{F \cdot l}{A \cdot Y}$
$\Rightarrow \Delta l=\frac{F \cdot l_{S}}{A_{S} \cdot Y_{S}}$
$=\frac{F \cdot l_{C}}{A_{C} \cdot Y_{C}}$
$\Rightarrow \frac{Y_{S}}{Y_{C}}=\frac{l_{S} \cdot A_{C}}{A_{S} \cdot l_{C}}$
$=\frac{(4.7 m ) \times\left(4 \times 10^{-5} m ^{2}\right)}{\left(3 \times 10^{-5} m ^{2}\right) \times(3.5 m )}=1.8$