Two uniform rods AB and BC have Young's modul | vedclass

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(B) Let the lengths of rods $AB$ and $BC$ be $L_1$ and $L_2$ respectively. Let their Young's moduli be $Y_1 = 1.2 	imes 10^{11} \, N/m^2$ and $Y_2 =

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$1.2$

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(B) Let the lengths of rods $AB$ and $BC$ be $L_1$ and $L_2$ respectively. Let their Young's moduli be $Y_1 = 1.2 	imes 10^{11} \, N/m^2$ and $Y_2 = 1.5 	imes 10^{11} \, N/m^2$. Let their coefficients of linear expansion be $\alpha_1 = 1.5 	imes 10^{-5} /^{\circ}C$ and $\alpha_2$ respectively.For no shift of the junction $B$ at any temperature change $\Delta 	heta$,the thermal expansion of rod $AB$ must be equal to the thermal compression of rod $BC$ (or vice versa).Thermal expansion $\Delta L_1 = \alpha_1 L_1 \Delta 	heta$.Thermal stress induced in rod $BC$ is $\sigma = Y_2 \frac{\Delta L_2}{L_2}$. The force $F = \sigma A = Y_2 A \frac{\Delta L_2}{L_2}$.Since the junction does not shift,the force exerted by both rods must be equal: $F = \frac{Y_1 A \Delta L_1}{L_1} = \frac{Y_2 A \Delta L_2}{L_2}$.Given $\Delta L_1 = \Delta L_2$ for no shift,we have $Y_1 \alpha_1 \Delta 	heta = Y_2 \alpha_2 \Delta 	heta$.Thus,$Y_1 \alpha_1 = Y_2 \alpha_2$.Substituting the values: $(1.2 	imes 10^{11}) 	imes (1.5 	imes 10^{-5}) = (1.5 	imes 10^{11}) 	imes \alpha_2$.$\alpha_2 = \frac{1.2 	imes 10^{11} 	imes 1.5 	imes 10^{-5}}{1.5 	imes 10^{11}} = 1.2 	imes 10^{-5} /^{\circ}C$.

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