Answer
The correct answer is:
(a) $0.50~mm$
Work Step by Step
$E = \frac{F/A}{\Delta L/L_0}$
$E$ 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) $0.50~mm$
