Theoretical discharge
$Q_t = A \sqrt{2gH}$
$Q_t = \frac{1}{4}\pi (0.0125^2) \sqrt{2(9.81)(5.5)}$
$Q_t = 0.001275 ~ \text{m}^3\text{/s}$
Actual discharge
$Q_a = \dfrac{\text{Volume}}{\text{time}}$
$Q_a = \dfrac{0.45}{9.5(60)}$
$Q_a = 0.000789 ~ \text{m}^3\text{/s}$
Coefficient of discharge
$C = \dfrac{Q_a}{Q_t}$
$C = \dfrac{0.000789}{0.001275}$
$C = 0.62$ ← answer
