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(D) Given: Diameter $d = 0.015 \ m$,Length $L = 0.2 \ m$. The relation is $F = 97.2 	imes 10^6 (\Delta L)$. The formula for Young's modulus is $Y = \

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$110$

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(D) Given: Diameter $d = 0.015 \ m$,Length $L = 0.2 \ m$. The relation is $F = 97.2 	imes 10^6 (\Delta L)$. The formula for Young's modulus is $Y = \frac{F L}{A \Delta L}$. The area of cross-section $A = \pi r^2 = \pi (d/2)^2 = \frac{\pi d^2}{4}$. Substituting $A$: $Y = \frac{F L}{(\pi d^2 / 4) \Delta L} = \frac{4 F L}{\pi d^2 \Delta L}$. From the given relation,$\frac{F}{\Delta L} = 97.2 	imes 10^6 \ N/m$. Substituting the values: $Y = \frac{4 	imes (97.2 	imes 10^6) 	imes 0.2}{3.14159 	imes (0.015)^2}$. $Y = \frac{77.76 	imes 10^6}{0.00070685} \approx 110 	imes 10^9 \ Pa = 110 \ GPa$.

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