= Solution
For constant fluid density $\rho$, steady <Euler equations for an inviscid fluid> under a <conservative force> per unit mass $-\nabla\Phi$ read $(\mathbf u\cdot\nabla)\mathbf u=-\nabla p/\rho-\nabla\Phi$. Here $\mathbf u$ is velocity, $p$ pressure and $\Phi$ potential per unit mass. Dotting with $\mathbf u$ gives
$$
\mathbf u\cdot\nabla\left(\frac12|\mathbf u|^2+\frac p\rho+\Phi\right)=0,
\qquad \boxed{\frac12|\mathbf u|^2+\frac p\rho+\Phi=\text{constant along each streamline}.}
$$
This is the <Bernoulli equation>. The constants can differ between <streamlines>. If density varies, the same derivation for a <barotropic fluid> replaces $p/\rho$ by $\int^p dp'/\rho(p')$; constant density is the appropriate water approximation here.
For the draining cone, the radius at height $h$ is $h$ and the horizontal area is $\pi h^2$. Both the free surface and the jet are at atmospheric pressure; their gravitational potential difference is approximately $gh$. The speed of the descending surface is smaller than the jet speed by the area ratio $\epsilon^2/h^2$, so it can be neglected when $h\gg\epsilon$. The <Bernoulli equation> then gives <Torricelli's law>, $U\simeq\sqrt{2gh}$. <Conservation of mass> gives
$$
\pi h^2\dot h\simeq-\pi\epsilon^2\sqrt{2gh},\qquad
h(t)^{5/2}\simeq h_0^{5/2}-\frac52\epsilon^2\sqrt{2g}\,t.
$$
Hence the leading <drainage time of a conical tank> is
$$
\boxed{T\simeq\frac{2h_0^{5/2}}{5\epsilon^2\sqrt{2g}}=\left(\frac{2h_0^5}{25\epsilon^4g}\right)^{1/2}.}
$$
Removing the tip places the aperture at height of order $\epsilon$, and the quasi-steady calculation fails in the last layer of that depth. Integrating down only to $h=O(\epsilon)$ changes the displayed leading time by relative order $(\epsilon/h_0)^{5/2}$; replacing the actual head $h-\epsilon$ by $h$ introduces a larger, but still vanishing, relative correction of order $\epsilon/h_0$. The formula is the ideal inviscid estimate under the question's stated approximation, not an exact law for the final transient or a real contracted jet.
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