Question

A steel wire of length $4.7m$ and cross-sectional area $3\times {10}^{-6}{m}^2$ stretches by the same amount as a copper wire of length $3.5m$ ad cross-sectional area of $4\times {10}^{-6}{m}^2$ under a given load. The ratio of Young's modulus of steel to that of copper is

• A. $1.8$
• B. $3.6$
• C. $0.6$
• D. $8.7$

Answer 1

The correct option is A $1.8$

We know, $\Delta l=\frac{FL}{AY}$

Given, load and extension are same, thus, $\frac{Y_1}{Y_2}=\frac{\frac{L_1}{A_1}}{\frac{L_2}{A_2}}=\frac{\frac{4.7}{3}}{\frac{3.5}{4}}=1.8$