The radii of the circular ends of bucket in the form of frustum are 28 cm and 7 cm and the height is 45 cm. The capacity of the bucket is

Options

A. `48150cm^(3)`
B. `48510cm^(3)`
C. `48105cm^(3)`
D. `48205cm^(3)`

Answer

To find the capacity of the bucket in the form of a frustum, we need to calculate the volume using the formula for the volume of a frustum of a cone. The formula is:

\[ V = \frac{1}{3} \pi h (R_1^2 + R_1 R_2 + R_2^2) \]

where:
- \( R_1 \) is the radius of the larger circular end,
- \( R_2 \) is the radius of the smaller circular end,
- \( h \) is the height of the frustum.

### Step-by-Step Solution:

1. **Identify the given values**:
- \( R_1 = 28 \) cm (radius of the larger end)
- \( R_2 = 7 \) cm (radius of the smaller end)
- \( h = 45 \) cm (height of the frustum)

2. **Substitute the values into the formula**:
\[ V = \frac{1}{3} \pi h (R_1^2 + R_1 R_2 + R_2^2) \]
Substituting the values:
\[ V = \frac{1}{3} \pi (45) (28^2 + 28 \times 7 + 7^2) \]

3. **Calculate \( R_1^2 \), \( R_1 R_2 \), and \( R_2^2 \)**:
- \( R_1^2 = 28^2 = 784 \)
- \( R_1 R_2 = 28 \times 7 = 196 \)
- \( R_2^2 = 7^2 = 49 \)

4. **Add these values**:
\[ R_1^2 + R_1 R_2 + R_2^2 = 784 + 196 + 49 = 1029 \]

5. **Substitute back into the volume formula**:
\[ V = \frac{1}{3} \pi (45) (1029) \]

6. **Use \( \pi \approx \frac{22}{7} \)**:
\[ V = \frac{1}{3} \times \frac{22}{7} \times 45 \times 1029 \]

7. **Calculate \( \frac{1}{3} \times 45 = 15 \)**:
\[ V = 15 \times \frac{22}{7} \times 1029 \]

8. **Calculate \( 15 \times 1029 = 15435 \)**:
\[ V = \frac{22}{7} \times 15435 \]

9. **Multiply \( 22 \times 15435 = 339570 \)**:
\[ V = \frac{339570}{7} \]

10. **Divide \( 339570 \div 7 = 48510 \)**:
\[ V = 48510 \text{ cm}^3 \]

### Final Answer:
The capacity of the bucket is \( 48510 \text{ cm}^3 \).