Answer
$19\%$
Work Step by Step
We know that
$Q=\frac{\Delta V}{\Delta t}$
$\implies Q=\frac{(P_1-P_2)\pi r^4}{8nL}$
$\implies Q\propto r^4$
$\frac{Q}{r^4}=constant$
Now, the above equation can be written for the two conditions
$\frac{Q_1}{r_1^4}=\frac{Q_2}{r_2^4}$
$\implies \frac{Q_2}{Q_1}=(\frac{r_2}{r_1})^4$
This simplifies to:
$\frac{r_2}{r_1}=(\frac{Q_2}{Q_1})^{\frac{1}{4}}$
If $Q_2=2Q_1$ then
$\frac{r_2}{r_1}=(\frac{Q_2}{Q_1})^{\frac{1}{4}}$
$\frac{r_2}{r_1}=2^{\frac{1}{4}}$
$\frac{r_2}{r_1}=1.19$
$\frac{d_1}{d_2}=1.19$
$\implies d_1=1.19d_2$
Thus, the increase in diameter$=1-1.19=0.19=19\%$
