Question

A steel wire of length 4.5 m and cross-sectional area $3 \times 10^{- 5}\text{ }\text{m}^{2}$ stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area$4 \times 10^{- 5}\text{ }\text{m}^{2}$ of 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

Answer: (C)

1.7

Answer 2

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

As applied force F and extension $\Delta L$are same for steel and copper wires

$\therefore\frac{F}{\Delta L} = \frac{Y_{S}A_{S}}{L_{S}} = \frac{Y_{C}A_{C}}{L_{C}}$

Where the subscripts C and S refers to copper and steel respectively.

$\therefore\frac{Y_{S}}{Y_{C}} = \frac{L_{S}}{L_{C}} \times \frac{A_{C}}{A_{S}} = \frac{(4.5m) \times (4 \times 10^{- 5}m^{2})}{(3.5m) \times (3 \times 10^{- 5}m^{2})} = 1.7$