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(B) The formula for Young's Modulus $Y$ is given by $Y = \frac{F \cdot L}{A \cdot \Delta l}$,where $F$ is the force,$L$ is the original length,$A$ is

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$16 	imes 10^3 \ N$

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(B) The formula for Young's Modulus $Y$ is given by $Y = \frac{F \cdot L}{A \cdot \Delta l}$,where $F$ is the force,$L$ is the original length,$A$ is the cross-sectional area,and $\Delta l$ is the change in length.Rearranging for force,we get $F = \frac{Y \cdot A \cdot \Delta l}{L}$.Since $A = \pi r^2 = \pi (D/2)^2 = \frac{\pi D^2}{4}$,we have $F \propto D^2$ (where $D$ is the diameter,and $Y, L, \Delta l$ are constant).Given the initial force $F_1 = 10^3 \ N$ and initial diameter $D_1 = D$,the new diameter is $D_2 = 4D$.Therefore,the ratio of forces is $\frac{F_2}{F_1} = \left(\frac{D_2}{D_1}ight)^2 = (4)^2 = 16$.Thus,$F_2 = 16 	imes F_1 = 16 	imes 10^3 \ N$.

$A$ force of $10^3 \ N$ stretches the length of a hanging wire by $1 \ mm$. The force required to stretch a wire of the same material and length,but having four times the diameter,by $1 \ mm$ is:

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