A steel rod has a radius of 50 mm and a leng | vedclass

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(C) The formula for Young's modulus $Y$ is given by $Y = \frac{FL}{A \Delta L}$.Given values: Force $F = 400 \ kN = 400 	imes 10^3 \ N$,Length $L = 2

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$2 	imes 10^{11} \ N/m^2$

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(C) The formula for Young's modulus $Y$ is given by $Y = \frac{FL}{A \Delta L}$.Given values: Force $F = 400 \ kN = 400 	imes 10^3 \ N$,Length $L = 2 \ m$,Radius $r = 50 \ mm = 50 	imes 10^{-3} \ m$,Elongation $\Delta L = 0.5 \ mm = 0.5 	imes 10^{-3} \ m$.The area of cross-section $A = \pi r^2 = \pi 	imes (50 	imes 10^{-3})^2 = \pi 	imes 2500 	imes 10^{-6} = 2.5 \pi 	imes 10^{-3} \ m^2 \approx 7.85 	imes 10^{-3} \ m^2$.Substituting these values into the formula:$Y = \frac{400 	imes 10^3 	imes 2}{(2.5 \pi 	imes 10^{-3}) 	imes (0.5 	imes 10^{-3})}$$Y = \frac{800 	imes 10^3}{1.25 \pi 	imes 10^{-6}} = \frac{800}{1.25 \pi} 	imes 10^9 \approx \frac{640}{3.14} 	imes 10^9 \approx 203.8 	imes 10^9 \approx 2 	imes 10^{11} \ N/m^2$.

$A$ steel rod has a radius of $50 \ mm$ and a length of $2 \ m$. It is stretched along its length with a force of $400 \ kN$. This causes an elongation of $0.5 \ mm$. Find the (approximate) Young's modulus of steel from this information.

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