Two exactly similar wires of steel and copper | vedclass

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Question

(A) The formula for Young's modulus is $Y = \frac{F L}{A l}$,where $F$ is the force,$L$ is the original length,$A$ is the cross-sectional area,and $l$

Vedclass · Accepted Answer

${l_s} = 0.75 \ cm, \ {l_c} = 1.25 \ cm$

Vedclass · Answer

(A) The formula for Young's modulus is $Y = \frac{F L}{A l}$,where $F$ is the force,$L$ is the original length,$A$ is the cross-sectional area,and $l$ is the elongation.Since the wires are similar,$F, L,$ and $A$ are the same for both. Thus,$l \propto \frac{1}{Y}$.Therefore,$\frac{l_s}{l_c} = \frac{Y_c}{Y_s} = \frac{1.2 	imes 10^{11}}{2.0 	imes 10^{11}} = \frac{1.2}{2.0} = \frac{3}{5}$.This implies $l_c = \frac{5}{3} l_s$ ... $(i)$.Given the difference in elongations is $l_c - l_s = 0.5 \ cm$ ... (ii).Substituting $(i)$ into (ii): $\frac{5}{3} l_s - l_s = 0.5 \ cm$.$\frac{2}{3} l_s = 0.5 \ cm \Rightarrow l_s = 0.5 	imes \frac{3}{2} = 0.75 \ cm$.Then,$l_c = 0.75 + 0.5 = 1.25 \ cm$.

Two exactly similar wires of steel and copper are stretched by equal forces. If the difference in their elongations is $0.5 \ cm$,find the elongation $(l)$ of each wire. Given: ${Y_s} = 2.0 \times {10^{11}} \ N/m^2$ and ${Y_c} = 1.2 \times {10^{11}} \ N/m^2$.

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