A cylindrical vessel of radius 4 cm. contains water. A solid sphere of radius 3 cm is dipped into the water until it is completely immersed. The water level in the vessel will rise by

Options

A. 3.5 cm
B. 2.25 cm
C. 2 cm
D. 3.8 cm

Answer

To solve the problem of how much the water level in the cylindrical vessel rises when a solid sphere is immersed in it, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Radius of the cylindrical vessel (R) = 4 cm
- Radius of the solid sphere (r) = 3 cm

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

3. **Relate the Volume of the Sphere to the Increase in Water Level:**
When the sphere is immersed in the water, it displaces an amount of water equal to its own volume. The increase in volume of water in the cylindrical vessel can be expressed as:
\[ \text{Increase in Volume} = \pi R^2 h \]
where \( h \) is the height the water level rises.

4. **Set the Volume of the Sphere Equal to the Increase in Volume of the Cylinder:**
\[ \pi R^2 h = \frac{4}{3} \pi r^3 \]
Substituting the known values:
\[ \pi (4)^2 h = 36 \pi \]

5. **Simplify the Equation:**
Cancel out \( \pi \) from both sides:
\[ 16h = 36 \]

6. **Solve for \( h \):**
\[ h = \frac{36}{16} = 2.25 \text{ cm} \]

### Final Answer:
The water level in the vessel will rise by **2.25 cm**.

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