A solid right circular cone of radius 4 cm and height 7 cm is put inside a cylindrical vessel of radius 5 cm and height 8 cm. How much water in cubic cm will be required to fill the cylindrical vessel completely ?

Options

A. 1022.48 cubic cm
B. 1533.72 cubic cm
C. 511.24 cubic cm
D. 255.62 cubic cm

Answer

To find out how much water in cubic centimeters will be required to fill the cylindrical vessel completely, we need to calculate the volume of the cylinder and subtract the volume of the cone that is placed inside it.

### Step-by-Step Solution:

1. **Calculate the Volume of the Cylinder:**
The formula for the volume \( V \) of a cylinder is given by:
\[ V = \pi r^2 h \]
where \( r \) is the radius and \( h \) is the height.

For the cylindrical vessel:
- Radius \( r = 5 \) cm
- Height \( h = 8 \) cm

Substituting the values:
\[ V_{cylinder} = \pi (5)^2 (8) = \pi (25)(8) = 200\pi \, \text{cm}^3 \]

2. **Calculate the Volume of the Cone:**
The formula for the volume \( V \) of a cone is given by:
\[ V = \frac{1}{3} \pi r^2 h \]
where \( r \) is the radius and \( h \) is the height.

For the cone:
- Radius \( r = 4 \) cm
- Height \( h = 7 \) cm

Substituting the values:
\[ V_{cone} = \frac{1}{3} \pi (4)^2 (7) = \frac{1}{3} \pi (16)(7) = \frac{112}{3}\pi \, \text{cm}^3 \]

3. **Calculate the Remaining Volume in the Cylinder:**
The remaining volume in the cylinder, which represents the volume of water needed to fill it completely, is given by:
\[ V_{remaining} = V_{cylinder} - V_{cone} \]
Substituting the volumes we calculated:
\[ V_{remaining} = 200\pi - \frac{112}{3}\pi \]

To subtract these, we need a common denominator:
\[ V_{remaining} = \left(200 - \frac{112}{3}\right)\pi = \left(\frac{600}{3} - \frac{112}{3}\right)\pi = \frac{488}{3}\pi \, \text{cm}^3 \]

4. **Calculate the Numerical Value:**
Using \( \pi \approx 3.14 \):
\[ V_{remaining} \approx \frac{488}{3} \times 3.14 \approx 511.34 \, \text{cm}^3 \]

Thus, the amount of water required to fill the cylindrical vessel completely is approximately **511.34 cm³**.