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For an inviscid flow that is an incompressible flow and an irrotational flow, write $\mathbf{u}=\nabla \varphi$. The Euler equations for an inviscid fluid, without gravity, integrate spatially to the Unsteady Bernoulli equationA time-dependent addition to $\varphi$ can absorb $C(t)$.

In the ideal uniform tube-plug model, let $u(t)$ be the axial speed toward the outlet. Incompressibility and fixed cross-section make the speed the same at both ends. With $\varphi =u(t)x$, the kinetic terms cancel when the Unsteady Bernoulli equation is subtracted between entrance and exit, givingConservation of volume in the spherical balloon gives $4\pi R^2\dot{R}=-\pi a^{2}u$, or $u=-4R^2\dot{R}/a^2$. Differentiating and eliminating $u$ yieldsThis is the constant-tension balloon discharge equation; entrance losses and a reservoir-to-tube kinetic-head correction are excluded by this ideal plug model.

Multiplication by $2R\dot{R}$ turns the equation intoThe printed time formula additionally assumes the water is initially at rest, $\dot{R}(0)=0$, although only $R_0$ is explicitly specified in the question. Under that assumption $C=K{R}_0^2$. On the shrinking branch,Set $r=R_0\sin \vartheta$ and $\theta =\arcsin (R/R_0)$. The integral becomes $R_0^2{K}^{-1/2}{\int }_{\theta }^{\pi /2}{\sin }^2\vartheta d\vartheta$, hencePutting $R=0$, or $\theta =0$, gives the ideal emptying timeFor an independently prescribed initial rate $v_0=\dot{R}(0)$, the first integral instead has $C=K{R}_0^2+R_0^4{v}_0^2$, so radius alone does not determine that emptying time. This explicitly identifies the extra initial datum needed for the displayed answer.