The Young's modulus of steel is twice that of | vedclass

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(A) Let $L$ and $A$ be the length and area of cross-section of each wire respectively.In order to have the lower ends of the wires at the same level,t

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$2:1$

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(A) Let $L$ and $A$ be the length and area of cross-section of each wire respectively.In order to have the lower ends of the wires at the same level,the elongation produced in both wires must be equal.Let $W_s$ and $W_b$ be the weights added to the steel and brass wires respectively.By the definition of Young's modulus $Y = \frac{W/A}{\Delta L/L}$,the elongation produced in the steel wire is $\Delta L_s = \frac{W_s L}{Y_s A}$.The elongation produced in the brass wire is $\Delta L_b = \frac{W_b L}{Y_b A}$.Since $\Delta L_s = \Delta L_b$,we have $\frac{W_s L}{Y_s A} = \frac{W_b L}{Y_b A}$.This simplifies to $\frac{W_s}{W_b} = \frac{Y_s}{Y_b}$.Given that the Young's modulus of steel is twice that of brass,$Y_s = 2 Y_b$,so $\frac{Y_s}{Y_b} = 2$.Therefore,$\frac{W_s}{W_b} = 2$,which means the ratio is $2:1$.

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