A vessel is in the form of an inverted cone. Its height is 11 cm and the radius of its top, which is open, is 2.5 cm. It is filled with water up to the rim. When lead shots, each of which is a sphere of radius 0.25 cm, are dropped into the vessel `(2)/(5)` of the water flows out. Find the number of lead shots dropped into the vessel.

Answer

To solve the problem step by step, we will calculate the volume of the inverted cone and the volume of the lead shots, then find the number of lead shots that can be dropped into the vessel.

### Step 1: Calculate the volume of the inverted cone.
The formula for the volume \( V \) of a cone is given by:
\[ V = \frac{1}{3} \pi r^2 h \]
Where:
- \( r \) is the radius of the base of the cone,
- \( h \) is the height of the cone.

Given:
- Radius \( r = 2.5 \, \text{cm} \)
- Height \( h = 11 \, \text{cm} \)

Substituting the values into the formula:
\[ V = \frac{1}{3} \pi (2.5)^2 (11) \]
Calculating \( (2.5)^2 = 6.25 \):
\[ V = \frac{1}{3} \pi (6.25)(11) = \frac{1}{3} \pi (68.75) \]
\[ V = \frac{68.75}{3} \pi \, \text{cm}^3 \]

### Step 2: Calculate the volume of water that flows out.
When lead shots are dropped into the vessel, \( \frac{2}{5} \) of the water flows out. Therefore, the volume of water that flows out is:
\[ \text{Volume of water that flows out} = \frac{2}{5} \times V \]
Substituting the volume of the cone:
\[ \text{Volume of water that flows out} = \frac{2}{5} \times \frac{68.75}{3} \pi = \frac{137.5}{15} \pi \, \text{cm}^3 \]

### Step 3: Calculate the volume of one lead shot.
The volume \( V_s \) of a sphere is given by:
\[ V_s = \frac{4}{3} \pi r^3 \]
Where \( r \) is the radius of the sphere.

Given:
- Radius of each lead shot \( r = 0.25 \, \text{cm} \)

Substituting the value into the formula:
\[ V_s = \frac{4}{3} \pi (0.25)^3 \]
Calculating \( (0.25)^3 = 0.015625 \):
\[ V_s = \frac{4}{3} \pi (0.015625) = \frac{0.0625}{3} \pi \, \text{cm}^3 \]

### Step 4: Set up the equation for the number of lead shots.
Let \( n \) be the number of lead shots. The total volume of the lead shots that displaces the water is equal to the volume of water that flows out:
\[ n \times V_s = \frac{2}{5} V \]
Substituting the volumes we calculated:
\[ n \times \frac{0.0625}{3} \pi = \frac{137.5}{15} \pi \]

### Step 5: Solve for \( n \).
Cancel \( \pi \) from both sides:
\[ n \times \frac{0.0625}{3} = \frac{137.5}{15} \]
Multiplying both sides by \( 3 \):
\[ n \times 0.0625 = \frac{412.5}{15} \]
Calculating \( \frac{412.5}{15} = 27.5 \):
\[ n \times 0.0625 = 27.5 \]
Now, divide both sides by \( 0.0625 \):
\[ n = \frac{27.5}{0.0625} = 440 \]

### Final Answer:
The number of lead shots dropped into the vessel is \( n = 440 \).