= Solution
In steady state the incoming <volumetric flow rate> equals the discharge. <Bernoulli's principle> between a nearly stationary free surface and the atmospheric outlet gives the ideal <Torricelli's law> <speed> $\sqrt{2gh}$. The specified <discharge coefficient> multiplies the opening-flow rate, so
$$
Q=\frac{10\times10^{-3}}{60}=1.6667\times10^{-4}\,\mathrm{m^3\,s^{-1}},\qquad
Q=C_d A\sqrt{2gh}.
$$
Thus, with $A=10^{-4}\,\mathrm{m^2}$ and $g=9.81\,\mathrm{m\,s^{-2}}$,
$$
\boxed{h=\frac1{2g}\left(\frac{Q}{0.62A}\right)^2=0.368\,\mathrm m.}
$$
Neglect surface-speed and capillary corrections. The coefficient includes contraction and loss effects; it should not be applied a second time to the calculated height.
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