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# A steel wire of length 4.7 m and cross-sectional area 3.0 × 10-5 m2 stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 × 10–5 m2 under a given load. What is the ratio of the Young’s modulus of steel to that of co

Q1  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?

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Let the Young's Modulus of steel and copper be YS and YC respectively.

Length of the steel wire l= 4.7 m

Length of the copper wire l= 4.7 m

The cross-sectional area of the steel wire AS = $3.0 \times 10^{-5} m^2$

The cross-sectional area of the Copper wire AC = $4.0 \times 10^{-5} m^2$

Let the load and the change in the length be F and $\Delta l$ respectively

$\\Y=\frac{Fl}{A\Delta l}\\ \frac{F}{\Delta l}=\frac{AY}{l}$

Since F and $\Delta l$ is the same for both wires we have

$\\\frac{A_{S}Y_{S}}{l_{S}}=\frac{A_{C}Y_{C}}{l_{C}}\\ \frac{Y_{S}}{Y_{C}}=\frac{A_{C}l_{S}}{A_{S}l_{C}}\\ \frac{Y_{S}}{Y_{C}}=\frac{4.0\times 10^{-5}\times 4.7}{3.0\times 10^{-5}\times3.5}\\ \frac{Y_{S}}{Y_{C}}=1.79$

The ratio of Young’s modulus of steel to that of copper is 1.79.

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