Two metallic wires P and Q have same volume a | vedclass

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Question

$ \mathrm{Y}=\frac{	ext { Stress }}{	ext { Strain }}=\frac{\mathrm{F} / \mathrm{A}}{\Delta \ell / \ell}=\frac{\mathrm{F} \ell}{\mathrm{A} \Delta \el

Vedclass · Accepted Answer

$16$

Vedclass · Answer

$ \mathrm{Y}=\frac{	ext { Stress }}{	ext { Strain }}=\frac{\mathrm{F} / \mathrm{A}}{\Delta \ell / \ell}=\frac{\mathrm{F} \ell}{\mathrm{A} \Delta \ell} $

$ \Delta \ell=\frac{\mathrm{F} \ell}{\mathrm{AY}} $

$ \mathrm{V}=\mathrm{A} \ell \Rightarrow \ell=\frac{\mathrm{V}}{\mathrm{A}} $

$ \Delta \ell=\frac{\mathrm{FV}}{\mathrm{A}^2 \mathrm{Y}}$

$Y$ & $V$ is same for both the wires

$ \Delta \ell \propto \frac{\mathrm{F}}{\mathrm{A}^2} $

$ \frac{\Delta \ell_1}{\Delta \ell_2}=\frac{\mathrm{F}_1}{\mathrm{~A}_1^2} 	imes \frac{\mathrm{A}_2^2}{\mathrm{~F}_2} $

$ \Delta \ell_1=\Delta \ell_2 $

$ \mathrm{~F}_1 \mathrm{~A}_2^2=\mathrm{F}_2 \mathrm{~A}_1^2 $

$ \frac{\mathrm{F}_1}{\mathrm{~F}_2}=\frac{\mathrm{A}_1^2}{\mathrm{~A}_2^2}=\left(\frac{4}{1}ight)^2=16$

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