Two exactly similar wires of steel and copper | vedclass

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(B) The formula for elongation $\Delta L$ is given by $\Delta L = \frac{FL}{AY}$,where $F$ is the force,$L$ is the original length,$A$ is the cross-se

Vedclass · Accepted Answer

$0.75 \,cm ; 1.25 \,cm$

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(B) The formula for elongation $\Delta L$ is given by $\Delta L = \frac{FL}{AY}$,where $F$ is the force,$L$ is the original length,$A$ is the cross-sectional area,and $Y$ is Young's modulus.Since $F$,$L$,and $A$ are the same for both wires,we have $\Delta L \propto \frac{1}{Y}$.Let $\Delta L_s$ be the elongation in the steel wire and $\Delta L_c$ be the elongation in the copper wire.Given $\Delta L_s + \Delta L_c = 2 \,cm$.From the proportionality,$\frac{\Delta L_s}{\Delta L_c} = \frac{Y_c}{Y_s} = \frac{12 	imes 10^{11}}{20 	imes 10^{11}} = \frac{12}{20} = 0.6$.Thus,$\Delta L_s = 0.6 \Delta L_c$.Substituting this into the total elongation equation: $0.6 \Delta L_c + \Delta L_c = 2 \,cm$.$1.6 \Delta L_c = 2 \,cm \implies \Delta L_c = \frac{2}{1.6} = 1.25 \,cm$.Then,$\Delta L_s = 2 - 1.25 = 0.75 \,cm$.Therefore,the elongation in steel wire is $0.75 \,cm$ and in copper wire is $1.25 \,cm$.

Two exactly similar wires of steel and copper are stretched by equal forces. If the total elongation is $2 \,cm$,then how much is the elongation in steel and copper wire respectively? Given,$Y_{\text{steel}} = 20 \times 10^{11} \,dyne/cm^2$,$Y_{\text{copper}} = 12 \times 10^{11} \,dyne/cm^2$.

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