A cylindrical vessel of base radius 14 cm is filled with water to some height. If a rectangular solid of dimensions 22 cm `xx` 7 cm `xx` 5 cm is immersed in it what is the rise in the water level ?

Options

A. `0.5` cm
B. `1.25` cm
C. 1 cm
D. `1.5` cm

Answer

To find the rise in the water level when a rectangular solid is immersed in a cylindrical vessel, we can follow these steps:

### Step 1: Calculate the Volume of the Rectangular Solid
The volume \( V \) of a rectangular solid (cuboid) can be calculated using the formula:
\[ V = \text{length} \times \text{breadth} \times \text{height} \]
Given dimensions are:
- Length = 22 cm
- Breadth = 7 cm
- Height = 5 cm

Substituting the values:
\[ V = 22 \, \text{cm} \times 7 \, \text{cm} \times 5 \, \text{cm} = 770 \, \text{cm}^3 \]

### Step 2: Calculate the Base Area of the Cylindrical Vessel
The base area \( A \) of the cylindrical vessel can be calculated using the formula:
\[ A = \pi r^2 \]
Given the radius \( r = 14 \, \text{cm} \):
\[ A = \pi \times (14 \, \text{cm})^2 = \pi \times 196 \, \text{cm}^2 \approx 615.75 \, \text{cm}^2 \quad (\text{using } \pi \approx 3.14) \]

### Step 3: Relate the Volume of the Solid to the Rise in Water Level
The volume of water displaced by the immersed solid will be equal to the volume of the solid. This volume will also cause a rise in the water level in the cylindrical vessel.

Let \( h \) be the rise in water level. The volume of the cylindrical section that corresponds to this rise can be expressed as:
\[ \text{Volume} = \text{Base Area} \times \text{Rise in Height} = A \times h \]
Setting the volume of the rectangular solid equal to the volume of the displaced water:
\[ 770 \, \text{cm}^3 = 615.75 \, \text{cm}^2 \times h \]

### Step 4: Solve for \( h \)
Rearranging the equation to solve for \( h \):
\[ h = \frac{770 \, \text{cm}^3}{615.75 \, \text{cm}^2} \approx 1.25 \, \text{cm} \]

### Conclusion
The rise in the water level when the rectangular solid is immersed in the cylindrical vessel is approximately **1.25 cm**.

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