Answer
(a) $2F$
(b) $\frac{F}{4}$
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
(a) Since a force of $F$ is applied at each end, the total applied force is $2F$:
$2F = \frac{\Delta L ~A_0~E}{L_0}$
$F = \frac{\Delta L ~A_0~E}{2L_0}$
We can find the required force at each end when the length is $\frac{L_0}{2}$:
$2F' = \frac{\Delta L ~A_0~E}{\frac{L_0}{2}}$
$F' = 2\times \frac{\Delta L ~A_0~E}{2L_0}$
$F' = 2~F$
(b) Since the bar has half the radius, the bar has one-fourth of the area. We can find the required force at each end when the area is $\frac{A_0}{4}$:
$2F' = \frac{\Delta L ~(\frac{A_0}{4})~E}{L_0}$
$F' = \frac{1}{4}\times \frac{\Delta L ~A_0~E}{2L_0}$
$F' = \frac{F}{4}$
