In a cylindrical vessel of diameter 24 cm filled up with sufficient quantity of water, a solid spherical ball of raidus 6 cm is completely immersed. Then the increase in height of water level is :

Options

A. a)1.5 cm
B. b)2 cm
C. c)3 cm
D. d)4 cm

Answer

To solve the problem of finding the increase in height of water level when a solid spherical ball is immersed in a cylindrical vessel, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the dimensions of the cylinder and sphere:**
- The diameter of the cylindrical vessel is given as 24 cm. Therefore, the radius (r_cylinder) of the cylinder is:
\[ r_{\text{cylinder}} = \frac{24}{2} = 12 \text{ cm} \]
- The radius of the solid spherical ball is given as 6 cm:
\[ r_{\text{sphere}} = 6 \text{ cm} \]

2. **Calculate the volume of the sphere:**
- The formula for the volume (V) of a sphere is:
\[ V_{\text{sphere}} = \frac{4}{3} \pi r_{\text{sphere}}^3 \]
- Substituting the radius of the sphere:
\[ V_{\text{sphere}} = \frac{4}{3} \pi (6)^3 = \frac{4}{3} \pi (216) = \frac{864}{3} \pi = 288 \pi \text{ cm}^3 \]

3. **Calculate the volume of water displaced in the cylinder:**
- When the sphere is immersed in the water, it displaces an amount of water equal to its own volume. Therefore, the volume of water displaced is:
\[ V_{\text{displaced}} = V_{\text{sphere}} = 288 \pi \text{ cm}^3 \]

4. **Relate the volume of displaced water to the height increase in the cylinder:**
- The volume of a cylinder is given by the formula:
\[ V_{\text{cylinder}} = \pi r_{\text{cylinder}}^2 h \]
- We need to find the height increase (h) when the volume of water displaced is equal to the volume of the cylinder:
\[ V_{\text{displaced}} = V_{\text{cylinder}} \implies 288 \pi = \pi (12)^2 h \]
- Simplifying this, we can cancel \(\pi\) from both sides:
\[ 288 = 144 h \]
- Solving for h:
\[ h = \frac{288}{144} = 2 \text{ cm} \]

### Final Answer:
The increase in height of the water level is **2 cm**.