A steel wire and a copper wire are joined end | vedclass

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(C) The formula for Young's modulus $(Y)$ is given by $Y = \frac{F \cdot L}{A \cdot \Delta L}$,where $F$ is the force,$L$ is the original length,$A$ i

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$20: 11$

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(C) The formula for Young's modulus $(Y)$ is given by $Y = \frac{F \cdot L}{A \cdot \Delta L}$,where $F$ is the force,$L$ is the original length,$A$ is the cross-sectional area,and $\Delta L$ is the elongation.Since the wires are joined end to end and subjected to the same tension $(F)$,the force $F$ is the same for both wires. Given that the cross-sectional areas $(A)$ and elongations $(\Delta L)$ are also equal,we have:$\Delta L = \frac{F \cdot L}{A \cdot Y}$Since $\Delta L$,$F$,and $A$ are constant for both wires,we can write:$\frac{L_{	ext{steel}}}{Y_{	ext{steel}}} = \frac{L_{	ext{copper}}}{Y_{	ext{copper}}}$Rearranging for the ratio of lengths:$\frac{L_{	ext{steel}}}{L_{	ext{copper}}} = \frac{Y_{	ext{steel}}}{Y_{	ext{copper}}}$Substituting the given values:$\frac{L_{	ext{steel}}}{L_{	ext{copper}}} = \frac{2.0 	imes 10^{11}}{1.1 	imes 10^{11}} = \frac{20}{11}$Thus,the ratio is $20: 11$.

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