A steel wire and a copper wire of equal length and equal cross-sectional area are joined end to end and the combination is subjected to a tension. Find the ratio of the stresses developed in steel and copper wire
$Y$ of steel is $2 \times 10^{11} {N/m}^{2}$ and $Y$ of copper is $1.3 \times 10^{11} {N/m}^{2}$ .

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Solution

Tension is same in both steel and copper,
$F_{s t} = F_{c u}$
Area of steel = $A_{s t}$
Area of copper = $A_{c u}$
${Stress}_{s t} = \frac{F_{s t}}{A_{s t}}$ .....(1)
${Stress}_{c u} = \frac{F_{c u}}{A_{c u}}$ ....(2)
Divide (1) by (2) we get
$\frac{{Stress}_{s t}}{{Stress}_{c u}}$ = 1

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Q. A steel wire and a copper wire of equal length and equal cross-sectional area are joined end to end and the combination is subjected to a tension. Find the ratio of the stresses developed in steel and copper wire

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2 \times 10^{11} {N/m}^{2}

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1.3 \times 10^{11} {N/m}^{2}

A steel wire and a copper wire of equal length and equal cross-sectional area are joined end to end and the combination is subjected to a tension. Find the ratio of (a) the stresses developed in the two wires and (b) the strains developed. Y of steel = $2 \times 10^{11} N m^{- 2}$ . Y of copper = $1.3 \times 1011 N m^{- 2}$ .