If the percentage error in the radius of a sphere is `a` , find the percentage error in its volume.

Options

A. `3/2 a %`
B. `2/3 a %`
C. `3 a %`
D. none of these

Explanation

To solve the problem of finding the percentage error in the volume of a sphere when the percentage error in its radius is given as \( a \% \), we can follow these steps:

### Step-by-Step Solution:

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

2. **Identify the percentage error in the radius**:
The percentage error in the radius \( r \) is given as \( a \% \). This can be expressed mathematically as:
\[ \frac{dr}{r} \times 100 = a \]
From this, we can derive:
\[ \frac{dr}{r} = \frac{a}{100} \]

3. **Differentiate the volume with respect to the radius**:
To find the error in volume, we differentiate the volume formula with respect to \( r \):
\[ dV = \frac{d}{dr} \left( \frac{4}{3} \pi r^3 \right) = 4 \pi r^2 \, dr \]

4. **Express \( dV \) in terms of \( V \)**:
We can express the differential volume \( dV \) as:
\[ dV = 4 \pi r^2 \, dr \]
The volume \( V \) can also be expressed as:
\[ V = \frac{4}{3} \pi r^3 \]

5. **Formulate the ratio \( \frac{dV}{V} \)**:
Now, we can find the ratio of the differential volume to the volume:
\[ \frac{dV}{V} = \frac{4 \pi r^2 \, dr}{\frac{4}{3} \pi r^3} \]
Simplifying this gives:
\[ \frac{dV}{V} = \frac{3 \, dr}{r} \]

6. **Calculate the percentage error in volume**:
The percentage error in volume is given by:
\[ \text{Percentage error in volume} = \frac{dV}{V} \times 100 \]
Substituting the ratio we found:
\[ \text{Percentage error in volume} = \frac{3 \, dr}{r} \times 100 \]
Since we know \( \frac{dr}{r} = \frac{a}{100} \):
\[ \text{Percentage error in volume} = 3 \left( \frac{a}{100} \right) \times 100 = 3a \]

7. **Final Result**:
Therefore, the percentage error in the volume of the sphere is:
\[ \text{Percentage error in volume} = 3a \% \]

### Summary:
If the percentage error in the radius of a sphere is \( a \% \), then the percentage error in its volume is \( 3a \% \).