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(D) The Young's modulus $Y$ is given by $Y = \frac{F L}{A \Delta x}$,where $L$ is the length,$A$ is the area,and $\Delta x$ is the extension.Since the

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$9F$

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(D) The Young's modulus $Y$ is given by $Y = \frac{F L}{A \Delta x}$,where $L$ is the length,$A$ is the area,and $\Delta x$ is the extension.Since the volume $V = A L$ is constant,we can write $L = \frac{V}{A}$.Substituting $L$ into the formula,we get $F = \frac{Y A \Delta x}{L} = \frac{Y A \Delta x}{(V/A)} = \frac{Y A^2 \Delta x}{V}$.Since $Y, V,$ and $\Delta x$ are the same for both wires,we have $F \propto A^2$.Therefore,$\frac{F_2}{F_1} = \left(\frac{A_2}{A_1}\right)^2 = \left(\frac{3A}{A}\right)^2 = 3^2 = 9$.Thus,$F_2 = 9 F_1 = 9F$.

Two wires are made of the same material and have the same volume. However,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?

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