Answer :
To calculate the possible change in volume due to the measurement error, we'll use the formula for the volume of a sphere:
\[V = \frac{4}{3} \pi r^3\]
Given:
- Initial radius, \(r = 10 \, \text{mm}\)
- Measurement error, \(Δr = 0.1 \, \text{mm}\)
We'll first find the initial volume, then find the volumes with the maximum and minimum possible radii within the given error range, and finally calculate the changes in volume.
1. **Initial Volume (\(V_0\))**:
\[V_0 = \frac{4}{3} \pi (10 \, \text{mm})^3\]
\[V_0 = \frac{4}{3} \pi (1000 \, \text{mm}^3)\]
\[V_0 = \frac{4000}{3} \pi \, \text{mm}^3\]
2. **Volume with Maximum Radius (\(V_{\text{max}}\))**:
\[r_{\text{max}} = r + Δr = 10 \, \text{mm} + 0.1 \, \text{mm} = 10.1 \, \text{mm}\]
\[V_{\text{max}} = \frac{4}{3} \pi (10.1 \, \text{mm})^3\]
3. **Volume with Minimum Radius (\(V_{\text{min}}\))**:
\[r_{\text{min}} = r - Δr = 10 \, \text{mm} - 0.1 \, \text{mm} = 9.9 \, \text{mm}\]
\[V_{\text{min}} = \frac{4}{3} \pi (9.9 \, \text{mm})^3\]
Now, let's calculate these volumes:
\[V_0 \approx \frac{4000}{3} \pi \, \text{mm}^3 \approx 4188.79 \, \text{mm}^3\]
\[V_{\text{max}} \approx \frac{4}{3} \pi (10.1 \, \text{mm})^3 \approx 4241.75 \, \text{mm}^3\]
\[V_{\text{min}} \approx \frac{4}{3} \pi (9.9 \, \text{mm})^3 \approx 4136.82 \, \text{mm}^3\]
Finally, let's find the possible change in volume:
\[ΔV_{\text{max}} = V_{\text{max}} - V_0 \approx 4241.75 \, \text{mm}^3 - 4188.79 \, \text{mm}^3 \approx 52.96 \, \text{mm}^3\]
\[ΔV_{\text{min}} = V_{\text{min}} - V_0 \approx 4136.82 \, \text{mm}^3 - 4188.79 \, \text{mm}^3 \approx -51.97 \, \text{mm}^3\]
The possible change in volume due to the measurement error is approximately \(52.96 \, \text{mm}^3\) (positive or negative).
\[V = \frac{4}{3} \pi r^3\]
Given:
- Initial radius, \(r = 10 \, \text{mm}\)
- Measurement error, \(Δr = 0.1 \, \text{mm}\)
We'll first find the initial volume, then find the volumes with the maximum and minimum possible radii within the given error range, and finally calculate the changes in volume.
1. **Initial Volume (\(V_0\))**:
\[V_0 = \frac{4}{3} \pi (10 \, \text{mm})^3\]
\[V_0 = \frac{4}{3} \pi (1000 \, \text{mm}^3)\]
\[V_0 = \frac{4000}{3} \pi \, \text{mm}^3\]
2. **Volume with Maximum Radius (\(V_{\text{max}}\))**:
\[r_{\text{max}} = r + Δr = 10 \, \text{mm} + 0.1 \, \text{mm} = 10.1 \, \text{mm}\]
\[V_{\text{max}} = \frac{4}{3} \pi (10.1 \, \text{mm})^3\]
3. **Volume with Minimum Radius (\(V_{\text{min}}\))**:
\[r_{\text{min}} = r - Δr = 10 \, \text{mm} - 0.1 \, \text{mm} = 9.9 \, \text{mm}\]
\[V_{\text{min}} = \frac{4}{3} \pi (9.9 \, \text{mm})^3\]
Now, let's calculate these volumes:
\[V_0 \approx \frac{4000}{3} \pi \, \text{mm}^3 \approx 4188.79 \, \text{mm}^3\]
\[V_{\text{max}} \approx \frac{4}{3} \pi (10.1 \, \text{mm})^3 \approx 4241.75 \, \text{mm}^3\]
\[V_{\text{min}} \approx \frac{4}{3} \pi (9.9 \, \text{mm})^3 \approx 4136.82 \, \text{mm}^3\]
Finally, let's find the possible change in volume:
\[ΔV_{\text{max}} = V_{\text{max}} - V_0 \approx 4241.75 \, \text{mm}^3 - 4188.79 \, \text{mm}^3 \approx 52.96 \, \text{mm}^3\]
\[ΔV_{\text{min}} = V_{\text{min}} - V_0 \approx 4136.82 \, \text{mm}^3 - 4188.79 \, \text{mm}^3 \approx -51.97 \, \text{mm}^3\]
The possible change in volume due to the measurement error is approximately \(52.96 \, \text{mm}^3\) (positive or negative).
