## College Physics (4th Edition)

The correct answer is: (a) $0.50~mm$
$E = \frac{F/A}{\Delta L/L_0}$ $E$ is Young's modulus $F$ is the force $A$ is the cross-sectional area $\Delta L$ is the change in length $L_0$ is the original length We can write an expression for $\Delta L$ when a force of $F$ is applied: $\Delta L = \frac{F~L_0}{E~A_0}$ We can find $\Delta L'$ when the length is $2L_0$ and the area is $4A_0$: $\Delta L' = \frac{F~(2L_0)}{E~(4A_0)}$ $\Delta L' = \frac{1}{2}\times \frac{F~L_0}{E~A_0}$ $\Delta L' = \frac{1}{2}\times \Delta L$ $\Delta L' = \frac{1}{2}\times (1.0~mm)$ $\Delta L' = 0.50~mm$ The correct answer is: (a) $0.50~mm$