The following four wires are made of the same | vedclass

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(A) The Young's modulus $Y$ is given by $Y = \frac{F/A}{\Delta \ell / \ell}$.Rearranging for extension $\Delta \ell$,we get $\Delta \ell = \frac{F \el

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Length $50\,cm$ and diameter $0.5\,mm$

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(A) The Young's modulus $Y$ is given by $Y = \frac{F/A}{\Delta \ell / \ell}$.Rearranging for extension $\Delta \ell$,we get $\Delta \ell = \frac{F \ell}{AY}$.Since the area $A = \pi (d/2)^2 = \frac{\pi d^2}{4}$,we have $\Delta \ell = \frac{4F \ell}{\pi d^2 Y}$.Given that $F$ and $Y$ are constant for all wires,$\Delta \ell \propto \frac{\ell}{d^2}$.For option $A$: $\frac{\ell}{d^2} = \frac{50}{(0.5)^2} = \frac{50}{0.25} = 200$.For option $B$: $\frac{\ell}{d^2} = \frac{100}{(1)^2} = 100$.For option $C$: $\frac{\ell}{d^2} = \frac{200}{(2)^2} = \frac{200}{4} = 50$.For option $D$: $\frac{\ell}{d^2} = \frac{300}{(3)^2} = \frac{300}{9} \approx 33.3$.Comparing the values,option $A$ has the largest ratio,hence the largest extension.

The following four wires are made of the same material. Which of these will have the largest extension when the same tension is applied?

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