The following four wires of length L and radi | vedclass

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(A) The extension $\Delta \ell$ of a wire is given by the formula $\Delta \ell = \frac{F L}{A Y}$,where $F$ is the tension,$L$ is the length,$A$ is th

Vedclass · Accepted Answer

$L = 100 \ cm, r = 0.2 \ mm$

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(A) The extension $\Delta \ell$ of a wire is given by the formula $\Delta \ell = \frac{F L}{A Y}$,where $F$ is the tension,$L$ is the length,$A$ is the cross-sectional area,and $Y$ is Young's modulus.Since $A = \pi r^2$,we have $\Delta \ell = \frac{F L}{\pi r^2 Y}$.Given that $F$ and $Y$ are constant for all wires,the extension $\Delta \ell$ is proportional to $\frac{L}{r^2}$.Calculating the ratio $\frac{L}{r^2}$ for each option:$A: \frac{100}{(0.2)^2} = \frac{100}{0.04} = 2500 \ cm/mm^2$$B: \frac{200}{(0.4)^2} = \frac{200}{0.16} = 1250 \ cm/mm^2$$C: \frac{300}{(0.6)^2} = \frac{300}{0.36} \approx 833.33 \ cm/mm^2$$D: \frac{400}{(0.8)^2} = \frac{400}{0.64} = 625 \ cm/mm^2$Comparing these values,the ratio is largest for option $A$.

The following four wires of length $L$ and radius $r$ are made of the same material. Which of these will have the largest extension,when the same tension is applied?

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