If the radius of the base is doubled , keeping the height constant , what is the ratio of the volume of the larger cone to the smaller cone ?

Options

A. `2:1`
B. `3:1`
C. `4:1`
D. `4:3`

Answer

To solve the problem of finding the ratio of the volume of the larger cone to the smaller cone when the radius of the base is doubled while keeping the height constant, we can follow these steps:

### Step-by-Step Solution:

1. **Define the Variables:**
- Let the radius of the smaller cone be \( r_1 \).
- Let the height of the smaller cone be \( h_1 \).
- The volume \( V_1 \) of the smaller cone can be calculated using the formula for the volume of a cone:
\[ V_1 = \frac{1}{3} \pi r_1^2 h_1 \]

2. **Determine the Dimensions of the Larger Cone:**
- If the radius of the base is doubled, then the radius of the larger cone \( r_2 \) will be:
\[ r_2 = 2r_1 \]
- The height of the larger cone remains the same as the smaller cone, so:
\[ h_2 = h_1 \]

3. **Calculate the Volume of the Larger Cone:**
- The volume \( V_2 \) of the larger cone can be calculated using the same volume formula:
\[ V_2 = \frac{1}{3} \pi r_2^2 h_2 \]
- Substituting \( r_2 \) and \( h_2 \):
\[ V_2 = \frac{1}{3} \pi (2r_1)^2 h_1 \]
- Simplifying this gives:
\[ V_2 = \frac{1}{3} \pi (4r_1^2) h_1 = \frac{4}{3} \pi r_1^2 h_1 \]

4. **Find the Ratio of the Volumes:**
- Now, we need to find the ratio of the volume of the larger cone to the volume of the smaller cone:
\[ \text{Ratio} = \frac{V_2}{V_1} = \frac{\frac{4}{3} \pi r_1^2 h_1}{\frac{1}{3} \pi r_1^2 h_1} \]
- Simplifying this ratio:
\[ \text{Ratio} = \frac{4}{3} \cdot \frac{3}{1} = 4 \]

5. **Conclusion:**
- The ratio of the volume of the larger cone to the smaller cone is:
\[ \text{Ratio} = 4:1 \]