Answer
(a) increased
(b) $\sqrt 2$
Work Step by Step
(a) We know that
$R=\rho(\frac{L}{A})$
$\implies R=\rho \frac{L}{\frac{\pi D^2}{4}}$
$\implies R=\rho\frac{L}{\frac{\pi(2r)^2}{4}}$
$\implies R=\rho\frac{L}{\pi r^2}$
This can be rearranged as:
$r^2=\frac{\rho L}{\pi R}$
$\implies r=\sqrt{\frac{\rho L}{\pi R}}$
Given that the length is doubled
$\implies r_1=\sqrt{\frac{\rho(2L)}{\pi R}}$
$\implies r_1=\sqrt{2}\sqrt{\frac{\rho L}{\pi R}}$
$\implies r_1=\sqrt{2}r$
Thus, the radius must be increased.
(b) As $r_1=\sqrt{2}r$, this shows the radius must be increased by a factor of $\sqrt 2$.
