Question
The given sphere has a radius of 2 inches.
A sphere with radius of 2 inches.
What happens to the volume if the radius is doubled?
The volume of the new sphere is
times larger than the volume of the original sphere.
A sphere with radius of 2 inches.
What happens to the volume if the radius is doubled?
The volume of the new sphere is
times larger than the volume of the original sphere.
Answers
GPT-4o mini
To find out how the volume changes when the radius of a sphere is doubled, we can start with the formula for the volume of a sphere:
\[
V = \frac{4}{3} \pi r^3
\]
For the original sphere with a radius of 2 inches:
\[
V_{\text{original}} = \frac{4}{3} \pi (2)^3 = \frac{4}{3} \pi (8) = \frac{32}{3} \pi \text{ cubic inches}
\]
If the radius is doubled, the new radius becomes:
\[
r_{\text{new}} = 2 \times 2 = 4 \text{ inches}
\]
Now we calculate the volume of the new sphere:
\[
V_{\text{new}} = \frac{4}{3} \pi (4)^3 = \frac{4}{3} \pi (64) = \frac{256}{3} \pi \text{ cubic inches}
\]
Next, we need to determine how many times larger the volume of the new sphere is compared to the original sphere. We can find the ratio of the new volume to the original volume:
\[
\text{Ratio} = \frac{V_{\text{new}}}{V_{\text{original}}} = \frac{\frac{256}{3} \pi}{\frac{32}{3} \pi}
\]
The \(\frac{3}{\pi}\) cancels out:
\[
\text{Ratio} = \frac{256}{32} = 8
\]
Thus, the volume of the new sphere is **8 times larger** than the volume of the original sphere.
Therefore, the final answer is:
The volume of the new sphere is **8** times larger than the volume of the original sphere.
\[
V = \frac{4}{3} \pi r^3
\]
For the original sphere with a radius of 2 inches:
\[
V_{\text{original}} = \frac{4}{3} \pi (2)^3 = \frac{4}{3} \pi (8) = \frac{32}{3} \pi \text{ cubic inches}
\]
If the radius is doubled, the new radius becomes:
\[
r_{\text{new}} = 2 \times 2 = 4 \text{ inches}
\]
Now we calculate the volume of the new sphere:
\[
V_{\text{new}} = \frac{4}{3} \pi (4)^3 = \frac{4}{3} \pi (64) = \frac{256}{3} \pi \text{ cubic inches}
\]
Next, we need to determine how many times larger the volume of the new sphere is compared to the original sphere. We can find the ratio of the new volume to the original volume:
\[
\text{Ratio} = \frac{V_{\text{new}}}{V_{\text{original}}} = \frac{\frac{256}{3} \pi}{\frac{32}{3} \pi}
\]
The \(\frac{3}{\pi}\) cancels out:
\[
\text{Ratio} = \frac{256}{32} = 8
\]
Thus, the volume of the new sphere is **8 times larger** than the volume of the original sphere.
Therefore, the final answer is:
The volume of the new sphere is **8** times larger than the volume of the original sphere.