Two metallic wires P and Q have the same volu | vedclass

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(A) The Young's modulus $Y$ is given by $Y = \frac{	ext{Stress}}{	ext{Strain}} = \frac{F/A}{\Delta l/l} = \frac{Fl}{A\Delta l}$.Rearranging for exte

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$16$

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(A) The Young's modulus $Y$ is given by $Y = \frac{	ext{Stress}}{	ext{Strain}} = \frac{F/A}{\Delta l/l} = \frac{Fl}{A\Delta l}$.Rearranging for extension,we get $\Delta l = \frac{Fl}{AY}$.Since volume $V = A 	imes l$,we can write $l = \frac{V}{A}$.Substituting $l$ into the extension formula: $\Delta l = \frac{F(V/A)}{AY} = \frac{FV}{A^2Y}$.Since $Y$ and $V$ are the same for both wires,$\Delta l \propto \frac{F}{A^2}$.Given $\Delta l_1 = \Delta l_2$,we have $\frac{F_1}{A_1^2} = \frac{F_2}{A_2^2}$.Therefore,$\frac{F_1}{F_2} = \frac{A_1^2}{A_2^2} = \left(\frac{A_1}{A_2}ight)^2$.Given the ratio of areas $\frac{A_1}{A_2} = \frac{4}{1}$,we get $\frac{F_1}{F_2} = (4)^2 = 16$.

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