A steel and a brass wire, each of length 50 , | vedclass

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(B) Let $Y_s$ and $Y_b$ be the Young's modulus of steel and brass respectively. Given $Y_s = 2 	imes 10^{12} \,dynes/cm^{2}$ and $Y_b = 1 	imes 10^{

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$5$

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(B) Let $Y_s$ and $Y_b$ be the Young's modulus of steel and brass respectively. Given $Y_s = 2 	imes 10^{12} \,dynes/cm^{2}$ and $Y_b = 1 	imes 10^{12} \,dynes/cm^{2}$.Let $L = 50 \,cm$ be the length, $A = 0.005 \,cm^{2}$ be the cross-sectional area, and $\Delta L = 0.1 \,cm$ be the extension for both wires.The tension $T$ in a wire is given by $T = \frac{Y A \Delta L}{L}$.Since $A, \Delta L,$ and $L$ are the same for both wires, the tension $T$ is directly proportional to the Young's modulus $Y$ $(T \propto Y)$.Let $T_s$ be the tension in the steel wire and $T_b$ be the tension in the brass wire.$T_s = \frac{Y_s A \Delta L}{L}$ and $T_b = \frac{Y_b A \Delta L}{L}$.Taking torque about the point where the load $W$ is applied at a distance $x$ from the steel wire:$T_s \cdot x = T_b \cdot (15 - x)$.Substituting the proportionality $T_s \propto Y_s$ and $T_b \propto Y_b$:$Y_s \cdot x = Y_b \cdot (15 - x)$.$(2 	imes 10^{12}) \cdot x = (1 	imes 10^{12}) \cdot (15 - x)$.$2x = 15 - x$.$3x = 15$.$x = 5 \,cm$.

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