A cylindrical beaker, whose base has a radius of 15 cm, is filled with water up to a height of 20 cm. A heavy iron spherical ball of radius 10 cm is dropped to submerge completely in water in the beaker. Find the increase in the level of water.

Answer

To solve the problem of finding the increase in the water level when a spherical ball is submerged in a cylindrical beaker, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Radius of the cylindrical beaker (r_cylinder) = 15 cm
- Height of water in the beaker (h_initial) = 20 cm
- Radius of the spherical ball (r_ball) = 10 cm

2. **Calculate the Volume of the Spherical Ball:**
The volume \( V \) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
Substituting the radius of the ball:
\[ V_{ball} = \frac{4}{3} \pi (10)^3 = \frac{4}{3} \pi (1000) = \frac{4000}{3} \pi \, \text{cm}^3 \]

3. **Set Up the Volume of Water Displaced:**
When the spherical ball is submerged, it displaces an equivalent volume of water. The volume of water that rises in the cylindrical beaker can be expressed as:
\[ V_{water} = \pi r_{cylinder}^2 h \]
where \( h \) is the increase in water level. Substituting the radius of the cylinder:
\[ V_{water} = \pi (15)^2 h = 225 \pi h \, \text{cm}^3 \]

4. **Equate the Two Volumes:**
Since the volume of water displaced by the ball equals the volume of water that rises in the beaker, we can set the two volumes equal to each other:
\[ V_{ball} = V_{water} \]
\[ \frac{4000}{3} \pi = 225 \pi h \]

5. **Cancel \(\pi\) from Both Sides:**
Dividing both sides by \(\pi\):
\[ \frac{4000}{3} = 225 h \]

6. **Solve for \( h \):**
Rearranging the equation to solve for \( h \):
\[ h = \frac{4000}{3 \times 225} \]
Simplifying:
\[ h = \frac{4000}{675} = \frac{4000 \div 25}{675 \div 25} = \frac{160}{27} \approx 5.92 \, \text{cm} \]

### Final Answer:
The increase in the level of water is approximately **5.92 cm**.