The radius of the base of a right circular cone is doubled keeping its height fixed. The volume of the cone will be :

Options

A. three times of the previous volume
B. four time of the previous volume
C. `sqrt(2)` times of the previous volume
D. double of the previous volume

Answer

To solve the problem, we need to determine how the volume of a right circular cone changes when the radius of its base is doubled while keeping the height fixed.

### Step-by-Step Solution:

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

2. **Define the Original Cone**:
Let the original radius be \( r_1 \) and the height be \( h \). Thus, the volume of the original cone \( V_1 \) is:
\[ V_1 = \frac{1}{3} \pi r_1^2 h \]

3. **Define the New Cone**:
When the radius is doubled, the new radius \( r_2 \) becomes:
\[ r_2 = 2r_1 \]
The height remains the same, so it is still \( h \). The volume of the new cone \( V_2 \) is:
\[ V_2 = \frac{1}{3} \pi (r_2)^2 h = \frac{1}{3} \pi (2r_1)^2 h \]

4. **Calculate the Volume of the New Cone**:
Substituting \( r_2 \) into the volume formula:
\[ V_2 = \frac{1}{3} \pi (4r_1^2) h = \frac{4}{3} \pi r_1^2 h \]

5. **Determine the Ratio of the Volumes**:
Now we can find the ratio of the new volume \( V_2 \) to the original volume \( V_1 \):
\[ \text{Ratio} = \frac{V_2}{V_1} = \frac{\frac{4}{3} \pi r_1^2 h}{\frac{1}{3} \pi r_1^2 h} \]
The \( \frac{1}{3} \pi r_1^2 h \) cancels out:
\[ \text{Ratio} = \frac{4}{1} = 4 \]

6. **Conclusion**:
Therefore, the volume of the new cone is 4 times the volume of the original cone.

### Final Answer:
The volume of the cone will be **4 times** the original volume.

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