If the error in the measurement of radius of a sphere in `2%` then the error in the determination of volume of the spahere will be

Options

A. `8%`
B. `2%`
C. `4%`
D. `6%`

Explanation

To solve the problem of finding the error in the determination of the volume of a sphere when the error in the measurement of its radius is given as 2%, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula for 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. **Identify the Error in Radius**:
We are given that the error in the measurement of the radius \( r \) is 2%. This can be expressed mathematically as:
\[ \frac{\Delta r}{r} \times 100\% = 2\% \]
where \( \Delta r \) is the absolute error in the radius.

3. **Use the Error Propagation Formula**:
To find the error in volume \( V \), we can use the error propagation formula. For a function \( y = x^n \), the relative error in \( y \) can be approximated as:
\[ \frac{\Delta y}{y} \approx n \frac{\Delta x}{x} \]
In our case, since \( V = \frac{4}{3} \pi r^3 \), we have \( n = 3 \).

4. **Calculate the Relative Error in Volume**:
Using the error propagation formula:
\[ \frac{\Delta V}{V} \approx 3 \frac{\Delta r}{r} \]
Substituting the known error in radius:
\[ \frac{\Delta V}{V} \approx 3 \times \frac{2}{100} = 6\% \]

5. **Conclusion**:
Therefore, the error in the determination of the volume of the sphere is:
\[ \Delta V \text{ (percentage)} = 6\% \]

### Final Answer:
The error in the determination of the volume of the sphere will be **6%**.