If the radius of a sphere is doubled, its volume becomes

Options

A. a.double
B. b. four times
C. c.six times
D. d.eight times

Answer

To solve the problem, we need to determine how the volume of a sphere changes when its radius is doubled.

### Step-by-Step Solution:

1. **Understand the formula for the volume of a sphere**:
The volume \( V \) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
where \( r \) is the radius of the sphere.

2. **Calculate the volume of the original sphere**:
Let the original radius be \( r \). The volume \( V_1 \) of the sphere with radius \( r \) is:
\[ V_1 = \frac{4}{3} \pi r^3 \]

3. **Determine the new radius when it is doubled**:
If the radius is doubled, the new radius \( r' \) becomes:
\[ r' = 2r \]

4. **Calculate the volume of the new sphere**:
The volume \( V_2 \) of the sphere with the new radius \( r' \) is:
\[ V_2 = \frac{4}{3} \pi (r')^3 = \frac{4}{3} \pi (2r)^3 \]
Simplifying this, we have:
\[ V_2 = \frac{4}{3} \pi (2^3 r^3) = \frac{4}{3} \pi (8 r^3) = \frac{32}{3} \pi r^3 \]

5. **Compare the volumes**:
Now we compare the original volume \( V_1 \) and the new volume \( V_2 \):
\[ V_1 = \frac{4}{3} \pi r^3 \]
\[ V_2 = \frac{32}{3} \pi r^3 \]

6. **Determine the factor of increase**:
To find out how many times the new volume is compared to the original volume, we can set up the ratio:
\[ \frac{V_2}{V_1} = \frac{\frac{32}{3} \pi r^3}{\frac{4}{3} \pi r^3} \]
Simplifying this gives:
\[ \frac{V_2}{V_1} = \frac{32}{4} = 8 \]

### Conclusion:
Thus, when the radius of the sphere is doubled, its volume becomes **8 times** the original volume.

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