An iron rod of length 2 , m and cross-section | vedclass

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(A) Given: Length $L = 2 \, m$,Area $A = 50 \, mm^2 = 50 	imes 10^{-6} \, m^2$,Change in length $\Delta L = 0.5 \, mm = 0.5 	imes 10^{-3} \, m$,Mass

Vedclass · Accepted Answer

$19.6 	imes 10^{10} \, N/m^2$

Vedclass · Answer

(A) Given: Length $L = 2 \, m$,Area $A = 50 \, mm^2 = 50 	imes 10^{-6} \, m^2$,Change in length $\Delta L = 0.5 \, mm = 0.5 	imes 10^{-3} \, m$,Mass $M = 250 \, kg$,Acceleration due to gravity $g = 9.8 \, m/s^2$.The formula for Young's modulus $Y$ is given by $Y = \frac{Stress}{Strain} = \frac{F/A}{\Delta L/L} = \frac{MgL}{A \Delta L}$.Substituting the values:$Y = \frac{250 	imes 9.8 	imes 2}{50 	imes 10^{-6} 	imes 0.5 	imes 10^{-3}}$$Y = \frac{4900}{25 	imes 10^{-9}}$$Y = \frac{4900}{25} 	imes 10^9 = 196 	imes 10^9 = 19.6 	imes 10^{10} \, N/m^2$.

An iron rod of length $2 \, m$ and cross-sectional area of $50 \, mm^2$ is stretched by $0.5 \, mm$ when a mass of $250 \, kg$ is hung from its lower end. The Young's modulus of the iron rod is:

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