Answer
1.19
Work Step by Step
(a) Since the volume flow rate is equal at each point, according to the equation $Q=Av$, we can tell that the upper hole must have the larger area (v is lower at the upper hole)
(b) Let's apply Bernoulli's equation to the system.
$P+\frac{1}{2}\rho v^{2}+\rho gh=constant$
$P+\frac{1}{2}\rho v_{u}^{2}+\rho gh=P+\frac{1}{2}\rho v_{l}^{2}=P+2\rho gh$
We can write,
$P+\frac{1}{2}\rho v_{u}^{2}+\rho gh=P+2\rho gh=>v_{u}=\sqrt {2gh}$
$P+\frac{1}{2}\rho v_{l}^{2}=P+2\rho gh=> v_{l}=\sqrt {4gh}$
Let's apply $A_{1}v_{1}=A_{2}v_{2}$
$\pi r_{1}^{2}\sqrt {2gh}=\pi r_{2}^{2}\sqrt {4gh}$
$\frac{r_{1}}{r_{2}}=\sqrt[4] 2=1.19$
