The diameter of a sphere is measured as 1.71 cm using an instrument with a least count 0.01 cm What is the percentage error in its surface area?

Explanation

To find the percentage error in the surface area of a sphere given the diameter and its least count, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given values**:
- Diameter of the sphere, \( D = 1.71 \, \text{cm} \)
- Least count of the measuring instrument, \( \delta D = 0.01 \, \text{cm} \)

2. **Calculate the radius of the sphere**:
\[ r = \frac{D}{2} = \frac{1.71 \, \text{cm}}{2} = 0.855 \, \text{cm} \]

3. **Write the formula for the surface area of a sphere**:
\[ A = 4\pi r^2 \]

4. **Determine the formula for the error in surface area**:
The error in surface area can be related to the error in diameter using the formula:
\[ \frac{\Delta A}{A} = 2 \frac{\Delta D}{D} \]
where \( \Delta A \) is the error in surface area and \( \Delta D \) is the error in diameter.

5. **Substitute the values into the error formula**:
- We know \( \Delta D = 0.01 \, \text{cm} \) and \( D = 1.71 \, \text{cm} \).
\[ \frac{\Delta A}{A} = 2 \cdot \frac{0.01 \, \text{cm}}{1.71 \, \text{cm}} \]

6. **Calculate the value**:
\[ \frac{\Delta A}{A} = 2 \cdot \frac{0.01}{1.71} \approx 0.01169 \]

7. **Convert to percentage**:
\[ \text{Percentage error} = \frac{\Delta A}{A} \times 100 \approx 0.01169 \times 100 \approx 1.169\% \]

8. **Round to three significant figures**:
\[ \text{Percentage error} \approx 1.17\% \]

### Final Answer:
The percentage error in the surface area of the sphere is approximately **1.17%**.