A sphere of diameter 26.4 cm is half-filled with acid. The acid is drained into a cylindrical beaker of diameter 16 cm. Find the depth of the acid in the beaker.

First, let's find the volume of the sphere that is half-filled with acid.

The radius of the sphere is half of the diameter, so the radius of the sphere is 26.4/2 = 13.2 cm.

The volume of a sphere is given by the formula V = (4/3)πr^3, where r is the radius.

So, the volume of the sphere that is half-filled with acid is V = (4/3)π(13.2)^3 = 35947.2 cm^3.

Since the sphere is half-filled with acid, the volume of the acid is half of the total volume, so the volume of the acid is 35947.2/2 = 17973.6 cm^3.

Now, let's find the depth of the acid in the cylindrical beaker.

The radius of the cylindrical beaker is 16/2 = 8 cm.

The volume of a cylinder is given by the formula V = πr^2h, where r is the radius and h is the height (depth).

We know that the volume of the acid is 17973.6 cm^3 and the radius of the cylindrical beaker is 8 cm.

So, 17973.6 = π(8)^2h
17973.6 = 64πh
h = 17973.6 / 64π
h ≈ 71.07 cm

Therefore, the depth of the acid in the beaker is approximately 71.07 cm.

Volume of the sphere:

```
V_sphere = (4/3)πr³ = (4/3)π(13.2 cm)³ = 9113.8 cm³
```

Volume of the acid:

```
V_acid = V_sphere / 2 = 9113.8 cm³ / 2 = 4556.9 cm³
```

Volume of the cylinder:

```
V_cylinder = πr²h
```

where r is the radius of the cylinder and h is its height (depth of the acid).

Equating the volumes:

```
V_acid = V_cylinder
4556.9 cm³ = π(8 cm)²h
```

Solving for h:

```
h = 4556.9 cm³ / (π(8 cm)²)
h = 28.5 cm
```

Therefore, the depth of the acid in the beaker is 28.5 cm.