Four wires of the same material are stretched | vedclass

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(A) The elongation $\Delta L$ is given by the formula $\Delta L = \frac{FL}{AY} = \frac{FL}{\pi r^2 Y}$.Since the wires are of the same material and s

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$L=100 \ cm, r=1 \ mm$

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(A) The elongation $\Delta L$ is given by the formula $\Delta L = \frac{FL}{AY} = \frac{FL}{\pi r^2 Y}$.Since the wires are of the same material and stretched by the same load,$F$ and $Y$ are constant.Therefore,$\Delta L \propto \frac{L}{r^2}$.Calculating the proportionality factor $\frac{L}{r^2}$ for each case:For $A$: $\frac{100}{1^2} = 100$.For $B$: $\frac{200}{3^2} = \frac{200}{9} \approx 22.22$.For $C$: $\frac{300}{3^2} = \frac{300}{9} \approx 33.33$.For $D$: $\frac{400}{4^2} = \frac{400}{16} = 25$.Comparing the values,the wire with dimensions $L=100 \ cm$ and $r=1 \ mm$ has the largest value of $\frac{L}{r^2}$,hence it will elongate the most.

Four wires of the same material are stretched by the same load. Which one of them will elongate the most if their dimensions are as follows?

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