An open cylindrical vessel of internal diameter 7 cm and height 8 cm stands on a horizontal table. Inside this is placed a solid metallic right circular cone, the diameter of whose base is `3 (1)/(2) cm` and height 8 cm. Find the volume of water required to fill the vessel.
If this cone is replaced by another cone, whose height is `1 (3)/(4)cm` and the radius of whose base is 2 cm, find the drop in the water level.

Answer

To solve the problem step-by-step, we will first calculate the volume of the cylindrical vessel, then the volume of the first cone, and finally the volume of water required to fill the vessel. After that, we will calculate the drop in the water level when the first cone is replaced by the second cone.

### Step 1: Calculate the volume of the cylindrical vessel

1. **Find the radius of the cylinder**:
- Given the diameter of the cylinder = 7 cm
- Radius \( r = \frac{7}{2} = 3.5 \) cm

2. **Use the formula for the volume of a cylinder**:
\[ V_{\text{cylinder}} = \pi r^2 h \]
Where \( h = 8 \) cm.

3. **Substitute the values**:
\[ V_{\text{cylinder}} = \frac{22}{7} \times (3.5)^2 \times 8 \]
\[ = \frac{22}{7} \times 12.25 \times 8 \]
\[ = \frac{22 \times 12.25 \times 8}{7} \]
\[ = \frac{2200}{7} \approx 314.29 \text{ cm}^3 \]

### Step 2: Calculate the volume of the first cone

1. **Find the radius of the cone**:
- Given the diameter of the cone = \( 3 \frac{1}{2} \) cm = \( 3.5 \) cm
- Radius \( r = \frac{3.5}{2} = 1.75 \) cm

2. **Use the formula for the volume of a cone**:
\[ V_{\text{cone}} = \frac{1}{3} \pi r^2 h \]
Where \( h = 8 \) cm.

3. **Substitute the values**:
\[ V_{\text{cone}} = \frac{1}{3} \times \frac{22}{7} \times (1.75)^2 \times 8 \]
\[ = \frac{1}{3} \times \frac{22}{7} \times 3.0625 \times 8 \]
\[ = \frac{1}{3} \times \frac{22 \times 3.0625 \times 8}{7} \]
\[ = \frac{1}{3} \times \frac{528.125}{7} \approx 25.24 \text{ cm}^3 \]

### Step 3: Calculate the volume of water required to fill the vessel

1. **Calculate the volume of water needed**:
\[ V_{\text{water}} = V_{\text{cylinder}} - V_{\text{cone}} \]
\[ = 314.29 - 25.24 \approx 289.05 \text{ cm}^3 \]

### Step 4: Calculate the drop in the water level when the second cone is placed

1. **Find the radius and height of the second cone**:
- Radius = 2 cm
- Height = \( 1 \frac{3}{4} \) cm = \( 1.75 \) cm

2. **Calculate the volume of the second cone**:
\[ V_{\text{new cone}} = \frac{1}{3} \pi r^2 h \]
\[ = \frac{1}{3} \times \frac{22}{7} \times (2)^2 \times 1.75 \]
\[ = \frac{1}{3} \times \frac{22}{7} \times 4 \times 1.75 \]
\[ = \frac{1}{3} \times \frac{22 \times 7}{7} \approx 15.43 \text{ cm}^3 \]

3. **Calculate the drop in water level**:
- Volume of water displaced = Volume of the first cone - Volume of the new cone
\[ V_{\text{drop}} = V_{\text{cone}} - V_{\text{new cone}} = 25.24 - 15.43 \approx 9.81 \text{ cm}^3 \]

4. **Using the volume of the cylinder to find the drop in height**:
\[ V_{\text{drop}} = \pi r^2 h \]
\[ h = \frac{V_{\text{drop}}}{\pi r^2} \]
\[ = \frac{9.81}{\frac{22}{7} \times (3.5)^2} \]
\[ = \frac{9.81}{\frac{22}{7} \times 12.25} \]
\[ \approx 0.14 \text{ cm} \]

### Final Answers:
- Volume of water required to fill the vessel: **289.05 cm³**
- Drop in water level when the second cone is placed: **0.14 cm**