What should be the diameter of a copper wire | vedclass

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(C) The formula for Young's modulus is $Y = \frac{F L}{A \Delta L}$, where $A = \pi r^2 = \pi (d/2)^2 = \frac{\pi d^2}{4}$.Rearranging for elongation:

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$2.3$

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(C) The formula for Young's modulus is $Y = \frac{F L}{A \Delta L}$, where $A = \pi r^2 = \pi (d/2)^2 = \frac{\pi d^2}{4}$.Rearranging for elongation: $\Delta L = \frac{4 F L}{\pi d^2 Y}$.Since $F$, $L$, and $\Delta L$ are the same for both wires, we have $\frac{1}{d_c^2 Y_c} = \frac{1}{d_a^2 Y_a}$, where $c$ denotes copper and $a$ denotes aluminium.Thus, $d_c^2 Y_c = d_a^2 Y_a$.Substituting the values: $d_c^2 (12 	imes 10^{10}) = (3 	ext{ mm})^2 (7 	imes 10^{10})$.$d_c^2 = \frac{9 	imes 7}{12} 	ext{ mm}^2 = \frac{63}{12} 	ext{ mm}^2 = 5.25 	ext{ mm}^2$.$d_c = \sqrt{5.25} 	ext{ mm} \approx 2.29 	ext{ mm} \approx 2.3 	ext{ mm}$.

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