$c o p p e r s t e e l$
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.
If the ratio of the strains developed is $\frac{x}{20}$ . Find $x$ . (Y of steel $= 2 \times 10^{11} N / m^{2}$ and Y of copper = $1.3 \times 10^{11} N / m^{2})$

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Solution

Young's modulus, $Y = \frac{S t r e s s}{s t r a i n}$ where Stress, $σ = \frac{F}{A}$
As both wires have same cross sectional area $(A)$ and same tension $F$ , so stress will be equal for both the wires.
$⟹$ $S t r a i n \propto \frac{1}{Y}$
Thus, $\frac{(S t r a i n)_{s t e e l}}{(S t r a i n)_{c o p p e r}} = \frac{Y_{c o p p e r}}{Y_{s t e e l}} = \frac{1.3 \times 10^{11}}{2 \times 10^{11}} = \frac{13}{20}$
$⟹ x = 13$

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