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(B) The formula for the extension $l$ of a wire is given by $l = \frac{FL}{AY}$,where $F$ is the applied force,$L$ is the original length,$A$ is the c

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$3/5$

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(B) The formula for the extension $l$ of a wire is given by $l = \frac{FL}{AY}$,where $F$ is the applied force,$L$ is the original length,$A$ is the cross-sectional area,and $Y$ is the Young's modulus.Since the wires have the same length $(L)$,are stretched by the same weight $(F)$,and assuming they have the same cross-sectional area $(A)$,the extension $l$ is inversely proportional to the Young's modulus $(Y)$: $l \propto \frac{1}{Y}$.Therefore,the ratio of the increase in length of steel $(l_S)$ to copper $(l_{Cu})$ is $\frac{l_S}{l_{Cu}} = \frac{Y_{Cu}}{Y_S}$.Substituting the given values: $\frac{l_S}{l_{Cu}} = \frac{1.2 	imes 10^{11}}{2 	imes 10^{11}} = \frac{1.2}{2} = \frac{12}{20} = \frac{3}{5}$.

Steel and copper wires of the same length are stretched by the same weight one after the other. The Young's modulus of steel and copper are $2 \times 10^{11} \, N/m^2$ and $1.2 \times 10^{11} \, N/m^2$,respectively. What is the ratio of the increase in their lengths?

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