The measure of radius of a sphere is (4.22 +2%) cm. The percentage error in volume of the sphere is

Options

A. `315pm 6%`
B. `315 pm 2% `
C. `315 pm14%`
D. `315 pm8%`

Answer

To find the percentage error in the volume of a sphere given the radius and its percentage error, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Radius \( r = 4.22 \, \text{cm} \)
- Percentage error in radius \( \delta r = 2\% \)

2. **Calculate the Volume of the Sphere:**
The formula for the volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]
Plugging in the radius:
\[ V = \frac{4}{3} \pi (4.22)^3 \]

3. **Calculate \( (4.22)^3 \):**
\[ (4.22)^3 = 4.22 \times 4.22 \times 4.22 \approx 75.05 \, \text{cm}^3 \]
Now, substituting this back into the volume formula:
\[ V \approx \frac{4}{3} \times 3.14 \times 75.05 \approx 314.63 \, \text{cm}^3 \]
Rounding this gives:
\[ V \approx 315 \, \text{cm}^3 \]

4. **Determine the Percentage Error in Volume:**
The percentage error in volume can be calculated using the formula:
\[ \text{Percentage error in volume} = 3 \times \left( \frac{\delta r}{r} \right) \times 100 \]
Here, \( \delta r = 2\% \):
\[ \text{Percentage error in volume} = 3 \times 2\% = 6\% \]

5. **Final Result:**
Therefore, the percentage error in the volume of the sphere is:
\[ \text{Percentage error in volume} = 6\% \]

### Conclusion:
The volume of the sphere is approximately \( 315 \, \text{cm}^3 \) with a percentage error of \( 6\% \).

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