A cylindrical vessel of diameter 4 cm is partly filled with water. 300 lead balls are dropped in it. The rise in water level is 0.8 cm. The diameter of each ball is ______.

Options

A. 0.8 cm
B. 0.4 cm
C. 0.2 cm
D. 0.5 cm

Answer

To find the diameter of each lead ball dropped into the cylindrical vessel, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Diameter of the cylindrical vessel = 4 cm
- Radius of the cylindrical vessel (r) = Diameter / 2 = 4 cm / 2 = 2 cm
- Rise in water level (h) = 0.8 cm
- Number of lead balls = 300

2. **Calculate the Volume of Water Displaced:**
The volume of water displaced when the lead balls are dropped can be calculated using the formula for the volume of a cylinder:
\[ \text{Volume} = \pi r^2 h \]
Substituting the values:
\[ \text{Volume} = \frac{22}{7} \times (2 \text{ cm})^2 \times 0.8 \text{ cm} \]
\[ \text{Volume} = \frac{22}{7} \times 4 \text{ cm}^2 \times 0.8 \text{ cm} \]
\[ \text{Volume} = \frac{22 \times 4 \times 0.8}{7} \text{ cm}^3 \]
\[ \text{Volume} = \frac{70.4}{7} \text{ cm}^3 \approx 10.06 \text{ cm}^3 \]

3. **Relate the Volume of Water Displaced to the Volume of the Lead Balls:**
The volume of water displaced is equal to the total volume of the 300 lead balls. The volume of one lead ball (which is spherical) can be calculated using the formula:
\[ \text{Volume of one ball} = \frac{4}{3} \pi r^3 \]
Therefore, the total volume of 300 balls is:
\[ \text{Total Volume} = 300 \times \frac{4}{3} \pi r^3 \]

4. **Set the Two Volumes Equal:**
Since the volume of water displaced is equal to the total volume of the balls:
\[ 300 \times \frac{4}{3} \pi r^3 = 10.06 \]

5. **Substituting \(\pi\) and Solving for \(r^3\):**
Substitute \(\pi\) with \(\frac{22}{7}\):
\[ 300 \times \frac{4}{3} \times \frac{22}{7} \times r^3 = 10.06 \]
Rearranging gives:
\[ r^3 = \frac{10.06 \times 7}{300 \times \frac{4}{3} \times 22} \]

6. **Calculating \(r^3\):**
\[ r^3 = \frac{70.42}{300 \times \frac{88}{3}} = \frac{70.42 \times 3}{300 \times 88} \]
\[ r^3 = \frac{211.26}{26400} \approx 0.008 \]

7. **Finding \(r\):**
Taking the cube root:
\[ r = \sqrt[3]{0.008} = 0.2 \text{ cm} \]

8. **Calculating the Diameter:**
The diameter \(d\) of each lead ball is:
\[ d = 2r = 2 \times 0.2 \text{ cm} = 0.4 \text{ cm} \]

### Final Answer:
The diameter of each ball is **0.4 cm**.