A student measued the diameter of a small steel ball using a screw gauge of least count `1.001 cm`. The main scale reading is `5 mm` and zero of circular scale division coincides with 25 divisions above the reference level. If screw gauge has a zero erroof `-0.004 cm`, the correct diameter of the ball is

Options

A. `0.053 cm`
B. `0.529 cm`
C. `0.525 cm`
D. `0.521 cm`

Explanation

To find the correct diameter of the small steel ball using the given measurements and corrections, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Measurements**:
- Main Scale Reading (MSR) = 5 mm = 0.5 cm (since 1 cm = 10 mm)
- Circular Scale Reading (CSR) = 25 divisions
- Least Count (LC) = 1.001 cm
- Zero Error (ZE) = -0.004 cm

2. **Calculate the Total Reading**:
The total reading from the screw gauge can be calculated using the formula:
\[ \text{Total Reading} = \text{MSR} + (\text{CSR} \times \text{LC}) \]
Substituting the values:
\[ \text{Total Reading} = 0.5 \, \text{cm} + (25 \times 0.001 \, \text{cm}) \]
\[ \text{Total Reading} = 0.5 \, \text{cm} + 0.025 \, \text{cm} \]
\[ \text{Total Reading} = 0.525 \, \text{cm} \]

3. **Adjust for Zero Error**:
Since the zero error is negative, we need to add it to the total reading:
\[ \text{Correct Diameter} = \text{Total Reading} - \text{Zero Error} \]
Substituting the values:
\[ \text{Correct Diameter} = 0.525 \, \text{cm} - (-0.004 \, \text{cm}) \]
\[ \text{Correct Diameter} = 0.525 \, \text{cm} + 0.004 \, \text{cm} \]
\[ \text{Correct Diameter} = 0.529 \, \text{cm} \]

4. **Final Result**:
The correct diameter of the ball is **0.529 cm**.