A steel rod has a radius of 10 ,mm and a leng | vedclass

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(B) We know that, Young's modulus $(Y)$ is given by the formula:$Y = \frac{	ext{Stress}}{	ext{Strain}} = \frac{F/A}{\Delta l/l}$Rearranging for the

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$\frac{4}{\pi} \,mm$

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(B) We know that, Young's modulus $(Y)$ is given by the formula:$Y = \frac{	ext{Stress}}{	ext{Strain}} = \frac{F/A}{\Delta l/l}$Rearranging for the change in length $(\Delta l)$:$\Delta l = \frac{F \cdot l}{A \cdot Y}$Given values:Force $(F)$ = $80 \,kN = 80 	imes 10^3 \,N$Length $(l)$ = $1 \,m$Radius $(r)$ = $10 \,mm = 10 	imes 10^{-3} \,m = 10^{-2} \,m$Area $(A)$ = $\pi r^2 = \pi 	imes (10^{-2} \,m)^2 = \pi 	imes 10^{-4} \,m^2$Young's modulus $(Y)$ = $2 	imes 10^{11} \,N/m^2$Substituting these values into the formula:$\Delta l = \frac{80 	imes 10^3 	imes 1}{(\pi 	imes 10^{-4}) 	imes (2 	imes 10^{11})}$$\Delta l = \frac{80 	imes 10^3}{\pi 	imes 2 	imes 10^7}$$\Delta l = \frac{40}{\pi} 	imes 10^{-4} \,m = \frac{4}{\pi} 	imes 10^{-3} \,m$Since $10^{-3} \,m = 1 \,mm$, we get:$\Delta l = \frac{4}{\pi} \,mm$

$A$ steel rod has a radius of $10 \,mm$ and a length of $1 \,m$. $A$ $80 \,kN$ force stretches it along its length. If the Young's modulus of the rod is $2 \times 10^{11} \,N/m^2$, then the change in length is

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