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(D) Let $V$ be the volume of the wires. Since $V = A 	imes L$,and both wires have the same volume and material,we have $A_1 L_1 = A_2 L_2$.Given $A_1

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$9F$

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(D) Let $V$ be the volume of the wires. Since $V = A 	imes L$,and both wires have the same volume and material,we have $A_1 L_1 = A_2 L_2$.Given $A_1 = A$ and $A_2 = 3A$,we get $A L_1 = 3A L_2$,which implies $L_1 = 3 L_2$.The Young's modulus $Y$ is given by $Y = \frac{F L}{A \Delta x}$.For wire $1$: $Y = \frac{F L_1}{A \Delta x} \implies F = \frac{Y A \Delta x}{L_1}$.For wire $2$: $Y = \frac{F' L_2}{3A \Delta x} \implies F' = \frac{Y (3A) \Delta x}{L_2}$.Substitute $L_2 = \frac{L_1}{3}$ into the equation for $F'$:$F' = \frac{Y (3A) \Delta x}{L_1 / 3} = 9 \left( \frac{Y A \Delta x}{L_1} ight)$.Since $F = \frac{Y A \Delta x}{L_1}$,we get $F' = 9F$.

Two wires are made of the same material and have the same volume. Wire $1$ has cross-sectional area $A$ and wire $2$ has cross-sectional area $3A$. If the length of wire $1$ increases by $\Delta x$ on applying force $F$,how much force is needed to stretch wire $2$ by the same amount $\Delta x$?

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