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(B) Given: Tensile force $F = 1000 \,N$, Length $L = 1 \,m$, Change in length $\Delta L = 0.25 \,cm = 0.25 	imes 10^{-2} \,m$, Young's modulus $Y = 1

Vedclass · Accepted Answer

$2.26$

Vedclass · Answer

(B) Given: Tensile force $F = 1000 \,N$, Length $L = 1 \,m$, Change in length $\Delta L = 0.25 \,cm = 0.25 	imes 10^{-2} \,m$, Young's modulus $Y = 10^{11} \,Pa$.The formula for Young's modulus is $Y = \frac{FL}{A \Delta L}$, where $A = \pi r^2$.Rearranging for $r^2$: $r^2 = \frac{FL}{Y \Delta L \pi}$.Substituting the values: $r^2 = \frac{1000 	imes 1}{10^{11} 	imes 0.25 	imes 10^{-2} 	imes \pi} = \frac{1000}{10^9 	imes 0.25 	imes \pi} = \frac{1}{0.25 	imes \pi 	imes 10^6} = \frac{4}{\pi 	imes 10^6}$.Taking the square root: $r = \frac{2}{\sqrt{\pi} 	imes 10^3} = \frac{2}{1.77 	imes 10^3} \approx 1.13 	imes 10^{-3} \,m = 1.13 \,mm$.The diameter $d = 2r = 2 	imes 1.13 \,mm = 2.26 \,mm$.

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