To find the angle of minimum deviation for the liquid in the equilateral hollow prism, we can use the formula for the angle of minimum deviation based on the refractive index $n$ and the prism angle $A$:

$n = \frac{\sin \frac{A + \delta_m}{2}}{\sin \frac{A}{2}}$

In this case, the refractive index $n = \sqrt{2}$, and since the prism is equilateral, the prism angle $A = 60^\circ$. Now, we can substitute these values into the formula and solve for the angle of minimum deviation $\delta_m$:

$\sqrt{2} = \frac{\sin \frac{60^\circ + \delta_m}{2}}{\sin \frac{60^\circ}{2}}$

First, let's find the sine of half the prism angle:

$\sin \frac{60^\circ}{2} = \sin 30^\circ = \frac{1}{2}$

Now, we can substitute this value into the formula:

$\sqrt{2} = \frac{\sin \frac{60^\circ + \delta_m}{2}}{\frac{1}{2}}$

To isolate the sine term, we multiply both sides by $\frac{1}{2}$:

$\frac{\sqrt{2}}{2} = \sin \frac{60^\circ + \delta_m}{2}$

Now, we can find the angle inside the sine function:

$\frac{60^\circ + \delta_m}{2} = \sin^{-1} \frac{\sqrt{2}}{2} = 45^\circ$

Finally, we can solve for the angle of minimum deviation $\delta_m$:

$60^\circ + \delta_m = 2 \cdot 45^\circ$

$\delta_m = 90^\circ - 60^\circ = 30^\circ$

The angle of minimum deviation for the liquid in the equilateral hollow prism is $30^\circ$.