The pitch of a screw gauge is 1 mm and there are 100 divisions on circular scale. When the two studs of the screw gauge are touching each other, the `32^(nd)` division of the circulasr scale coincides with the reference line / base line / index line on the pitch scale. When a glass plate is placed in between the studs firmly, the pitch scale reads 4 divisions and the circular scale reads 16 divisions
Zero correction of the instrument is

Options

A. 0 mm
B. `+ 0.32 mm`
C. `-0.32 mm`
D. `+4 mm`

Answer

To find the zero correction of the screw gauge, we will follow these steps:

### Step 1: Understand the given data
- **Pitch of the screw gauge (P)** = 1 mm
- **Number of divisions on the circular scale (N)** = 100
- **Division on circular scale coinciding with the reference line when studs touch** = 32nd division
- **Reading on pitch scale when glass plate is placed** = 4 divisions
- **Reading on circular scale when glass plate is placed** = 16 divisions

### Step 2: Calculate the least count of the screw gauge
The least count (LC) of the screw gauge can be calculated using the formula:
\[ \text{Least Count (LC)} = \frac{\text{Pitch}}{\text{Number of divisions on circular scale}} = \frac{1 \text{ mm}}{100} = 0.01 \text{ mm} \]

### Step 3: Calculate the error due to zero correction
The error (zero correction) can be calculated using the formula:
\[ \text{Error} = \text{Number of divisions coinciding with the baseline} \times \text{Least Count} \]
Here, the number of divisions coinciding with the baseline is 32 (as given in the question).

Substituting the values:
\[ \text{Error} = 32 \times 0.01 \text{ mm} = 0.32 \text{ mm} \]

### Step 4: Determine the sign of the zero correction
Since the 32nd division coincides with the reference line when the studs are touching, it indicates a positive error. This means the screw gauge reads more than it should when the studs are in contact.

### Step 5: Write the final answer
Thus, the zero correction of the instrument is:
\[ \text{Zero Correction} = +0.32 \text{ mm} \]

### Summary of the solution:
The zero correction of the screw gauge is +0.32 mm.

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