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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 of $4.0 \times 10^{-5} \;m ^{2}$ under a given load. What is the ratio of the Young's modulus of steel to that of copper?
Solution
Length of the steel wire, $L_{1}=4.7 m$
Area of cross-section of the steel wire, $A_{1}=3.0 \times 10^{-5} m ^{2}$
Length of the copper wire, $L_{2}=3.5 m$
Area of cross-section of the copper wire, $A_{2}=4.0 \times 10^{-5} m ^{2}$
Change in length $=\Delta L_{1}=\Delta L_{2}=\Delta L$
Force applied in both the cases $=F$
Young's modulus of the steel wire:
$Y_{1}=\frac{F_{1}}{A_{1}} \times \frac{L_{1}}{\Delta L}$
$=\frac{F \times 4.7}{3.0 \times 10^{-5} \times \Delta L} \ldots(i)$
Young's modulus of the copper wire:
$Y_{2}=\frac{F_{2}}{A_{2}} \times \frac{L_{2}}{\Delta L_{2}}$
$=\frac{F \times 3.5}{4.0 \times 10^{-5} \times \Delta L}\dots (ii)$
Dividing ($i$) by ($ii$), we get:
$\frac{Y_{1}}{Y_{2}}=\frac{4.7 \times 4.0 \times 10^{-5}}{3.0 \times 10^{-5} \times 3.5}=1.79: 1$
The ratio of Young's modulus of steel to that of copper is $1.79: 1$
