Two iron shots each of diameter 6 cm are immersed in the water contained in a cylindrical vessel or radius 6 cm . The level of the water in the vessel will be raised by

Options

A. 1 cm
B. 2 cm
C. 3 cm
D. 6 cm

Answer

To solve the problem of how much the water level in the cylindrical vessel will rise when two iron shots of diameter 6 cm are immersed in it, we can follow these steps:

### Step-by-Step Solution:

1. **Determine the radius of the iron shots**:
The diameter of each iron shot is given as 6 cm. Therefore, the radius (r) of each iron shot is:
\[ r = \frac{\text{Diameter}}{2} = \frac{6 \text{ cm}}{2} = 3 \text{ cm} \]

2. **Calculate the volume of one iron shot**:
The volume (V) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
Substituting the radius of the iron shot:
\[ V = \frac{4}{3} \pi (3 \text{ cm})^3 = \frac{4}{3} \pi (27 \text{ cm}^3) = 36 \pi \text{ cm}^3 \]

3. **Calculate the total volume of two iron shots**:
Since there are two iron shots, the total volume (V_total) displaced by the two shots is:
\[ V_{\text{total}} = 2 \times 36 \pi \text{ cm}^3 = 72 \pi \text{ cm}^3 \]

4. **Determine the volume of water displaced in the cylindrical vessel**:
The volume of water displaced when the water level rises by a height (h) in a cylindrical vessel is given by:
\[ V = \pi R^2 h \]
where R is the radius of the cylindrical vessel. Given that the radius of the vessel is also 6 cm:
\[ V = \pi (6 \text{ cm})^2 h = 36 \pi h \text{ cm}^3 \]

5. **Set the volumes equal to each other**:
The volume of water displaced by the two iron shots is equal to the volume of water that raises the level in the vessel:
\[ 72 \pi = 36 \pi h \]

6. **Solve for h**:
Dividing both sides by \(36 \pi\):
\[ h = \frac{72 \pi}{36 \pi} = 2 \text{ cm} \]

### Conclusion:
The level of the water in the vessel will be raised by **2 cm**.