There is an error of `0.2%` in measurment of the redius of a sphere. Find the percentage error in its
(i) volume, (ii) surface.

Answer

To solve the problem of finding the percentage error in the volume and surface area of a sphere due to an error in the measurement of its radius, we can follow these steps:

### Given:
- Error in radius measurement, \( \frac{dr}{r} \times 100 = 0.2\% \)

### Step 1: Find the percentage error in the volume of the sphere.

1. **Formula for Volume of a Sphere**:
\[ V = \frac{4}{3} \pi r^3 \]

2. **Differentiate Volume with respect to Radius**:
\[ \frac{dV}{dr} = 4 \pi r^2 \]

3. **Using the relationship between differential changes**:
\[ dV = 4 \pi r^2 \, dr \]

4. **Percentage error in volume**:
\[ \frac{dV}{V} = \frac{4 \pi r^2 \, dr}{\frac{4}{3} \pi r^3} \]
Simplifying this:
\[ \frac{dV}{V} = \frac{3 \, dr}{r} \]

5. **Multiply by 100 to get percentage**:
\[ \frac{dV}{V} \times 100 = 3 \times \frac{dr}{r} \times 100 \]

6. **Substituting the given error**:
\[ \frac{dr}{r} \times 100 = 0.2 \]
Therefore,
\[ \frac{dV}{V} \times 100 = 3 \times 0.2 = 0.6\% \]

### Step 2: Find the percentage error in the surface area of the sphere.

1. **Formula for Surface Area of a Sphere**:
\[ S = 4 \pi r^2 \]

2. **Differentiate Surface Area with respect to Radius**:
\[ \frac{dS}{dr} = 8 \pi r \]

3. **Using the relationship between differential changes**:
\[ dS = 8 \pi r \, dr \]

4. **Percentage error in surface area**:
\[ \frac{dS}{S} = \frac{8 \pi r \, dr}{4 \pi r^2} \]
Simplifying this:
\[ \frac{dS}{S} = \frac{2 \, dr}{r} \]

5. **Multiply by 100 to get percentage**:
\[ \frac{dS}{S} \times 100 = 2 \times \frac{dr}{r} \times 100 \]

6. **Substituting the given error**:
\[ \frac{dr}{r} \times 100 = 0.2 \]
Therefore,
\[ \frac{dS}{S} \times 100 = 2 \times 0.2 = 0.4\% \]

### Final Answers:
- (i) Percentage error in volume: **0.6%**
- (ii) Percentage error in surface area: **0.4%**

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