A brass rod of length 2,m and cross-sectional | vedclass

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(D) Given:Length of brass rod,$\ell_{B} = 2\,m$Cross-sectional area of brass rod,$A_{B} = 2.0\,cm^2 = 2.0 	imes 10^{-4}\,m^2$Young's modulus of brass

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(D) Given:Length of brass rod,$\ell_{B} = 2\,m$Cross-sectional area of brass rod,$A_{B} = 2.0\,cm^2 = 2.0 	imes 10^{-4}\,m^2$Young's modulus of brass,$Y_{B} = 1.0 	imes 10^{11}\,N/m^2$Length of steel rod,$\ell_{S} = L$Cross-sectional area of steel rod,$A_{S} = 1.0\,cm^2 = 1.0 	imes 10^{-4}\,m^2$Young's modulus of steel,$Y_{S} = 2.0 	imes 10^{11}\,N/m^2$Applied force,$F = 5 	imes 10^4\,N$Condition for equal elongation: $\Delta \ell_{B} = \Delta \ell_{S}$Using the formula for elongation,$\Delta \ell = \frac{FL}{AY}$,we have:$\frac{F \ell_{B}}{A_{B} Y_{B}} = \frac{F \ell_{S}}{A_{S} Y_{S}}$Canceling $F$ from both sides:$\frac{\ell_{B}}{A_{B} Y_{B}} = \frac{L}{A_{S} Y_{S}}$Rearranging for $L$:$L = \ell_{B} 	imes \frac{A_{S} Y_{S}}{A_{B} Y_{B}}$Substituting the values:$L = 2 	imes \frac{1.0 	imes 10^{-4} 	imes 2.0 	imes 10^{11}}{2.0 	imes 10^{-4} 	imes 1.0 	imes 10^{11}}$$L = 2 	imes \frac{2.0 	imes 10^7}{2.0 	imes 10^7} = 2\,m$Thus,the length of the steel rod is $2\,m$.

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$A$ rod is fixed between two points at $20\,^oC$. The coefficient of linear expansion of the material of the rod is $1.1 \times 10^{-5}/\,^oC$ and Young's modulus is $1.2 \times 10^{11}\,N/m^2$. Find the stress developed in the rod if the temperature of the rod becomes $10\,^oC$.

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