Define Variables and Relationships

Let s be the side length of the cubic cavity and d be the diameter of the spherical cavity.

According to the problem statement, s = d.

Let r be the radius of the spherical cavity. Since the diameter is twice the radius, d = 2r.

Therefore, the side length of the cube can be expressed in terms of the sphere's radius as s = 2r.

Calculate Volumes

The volume of the cubic cavity ($V_{cube}$) is calculated as $s^3$. Substituting $s = 2r$:

$V_{cube} = (2r)^3 = 8r^3$

The volume of the spherical cavity ($V_{sphere}$) is calculated using the formula:

$V_{sphere} = \frac{4}{3}\pi r^3$

Determine Liquid Volumes

The cubic cavity is half-filled with liquid. The volume of liquid in the cubic cavity ($V_{liquid\_cube}$) is:

$V_{liquid\_cube} = \frac{1}{2} V_{cube} = \frac{1}{2} (8r^3) = 4r^3$

The spherical cavity is completely filled with liquid. The volume of liquid in the spherical cavity ($V_{liquid\_sphere}$) is:

$V_{liquid\_sphere} = V_{sphere} = \frac{4}{3}\pi r^3$

Calculate the Ratio

The question asks for the ratio of the volume of liquid in the cubic cavity to that in the spherical cavity:

Ratio $= \frac{V_{liquid\_cube}}{V_{liquid\_sphere}} = \frac{4r^3}{\frac{4}{3}\pi r^3}$

Simplify the expression by canceling out $r^3$ and rearranging the constants:

Ratio $= \frac{4}{\frac{4}{3}\pi} = \frac{4 \times 3}{4 \times \pi} = \frac{3}{\pi}$

Approximate the Ratio

Using the approximate value of $\pi \approx 3.14159$:

Ratio $\approx \frac{3}{3.14159} \approx 0.955$

This calculated ratio is approximately equal to 1.

Final Answer

The approximate ratio of the volume of liquid in the cubic cavity to that in the spherical cavity is $1:1$.