There is some water in a cylindrical vessel of diameter 12 cm. A solid metallic spere of redius 4 cm is dropped in to it . Find the increase in height of the water surface if sphere is fully immersed in to the water.

Explanation

To find the increase in height of the water surface when a solid metallic sphere is immersed in a cylindrical vessel, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the dimensions of the sphere and the cylinder**:
- Diameter of the cylindrical vessel = 12 cm
- Radius of the cylindrical vessel (R) = Diameter/2 = 12 cm / 2 = 6 cm
- Radius of the sphere (r) = 4 cm

2. **Calculate the volume of the sphere**:
The formula for the volume of a sphere is given by:
\[ V_{sphere} = \frac{4}{3} \pi r^3 \]
Substituting the radius of the sphere:
\[ V_{sphere} = \frac{4}{3} \pi (4)^3 = \frac{4}{3} \pi (64) = \frac{256}{3} \pi \text{ cm}^3 \]

3. **Calculate the volume of water displaced by the sphere**:
When the sphere is fully immersed in the water, it displaces an amount of water equal to its own volume. Therefore, the volume of water displaced is also:
\[ V_{displaced} = \frac{256}{3} \pi \text{ cm}^3 \]

4. **Calculate the increase in height of the water in the cylindrical vessel**:
The volume of water in the cylinder can be expressed as:
\[ V_{cylinder} = \pi R^2 h \]
where \( R \) is the radius of the cylinder and \( h \) is the increase in height of the water. Setting the volume of water displaced equal to the volume of the cylinder, we have:
\[ \frac{256}{3} \pi = \pi (6^2) h \]
Simplifying this, we can cancel \( \pi \) from both sides:
\[ \frac{256}{3} = 36h \]

5. **Solve for \( h \)**:
\[ h = \frac{256}{3 \times 36} = \frac{256}{108} = \frac{64}{27} \text{ cm} \]

### Final Answer:
The increase in height of the water surface is \( \frac{64}{27} \) cm.

---