The distance moved by the screw of a screw gauge is `2 mm` in four rotations and there are `50` divisions on its cap. When nothing is put between its jaws, `20th` divisions of circular scale coincides with reference line, and zero of linear scale is hidden from circular scale when two jaws touch each other or zero circular scale is laying above the reference line. When plate is placed between the jaws, main scale reads `2` divisions and circular scale reads `20` divisions. Thickness of plate is.

Options

A. 1.1 mm
B. 1.2 mm
C. 1.4 mm
D. 1.5 mm

Explanation

To find the thickness of the plate using the screw gauge, we will follow these steps:

### Step 1: Calculate the Pitch of the Screw Gauge
The pitch of the screw gauge is defined as the distance moved by the screw in one complete rotation. Given that the distance moved by the screw in 4 rotations is 2 mm, we can calculate the pitch as follows:

\[ \text{Pitch} = \frac{\text{Distance moved}}{\text{Number of rotations}} = \frac{2 \text{ mm}}{4} = 0.5 \text{ mm} \]

### Step 2: Calculate the Least Count of the Screw Gauge
The least count is the smallest measurement that can be read on the screw gauge, calculated as the pitch divided by the number of divisions on the circular scale. Given that there are 50 divisions on the circular scale:

\[ \text{Least Count} = \frac{\text{Pitch}}{\text{Number of divisions}} = \frac{0.5 \text{ mm}}{50} = 0.01 \text{ mm} = \frac{1}{100} \text{ mm} \]

### Step 3: Determine the Zero Error
When nothing is placed between the jaws, the 20th division of the circular scale coincides with the reference line, indicating a zero error. Since the zero of the linear scale is hidden, we have a negative error. The number of divisions from 0 to 20 is 20, and the remaining divisions to complete 50 is:

\[ \text{Number of divisions} = 50 - 20 = 30 \]

The error can be calculated as:

\[ \text{Error} = \text{Number of divisions} \times \text{Least Count} = 30 \times 0.01 \text{ mm} = 0.3 \text{ mm} \]

### Step 4: Measure the Thickness of the Plate
When the plate is placed between the jaws, the main scale reads 2 divisions, and the circular scale reads 20 divisions. The observed length can be calculated as follows:

1. **Main Scale Reading (MSR)**:
\[ \text{MSR} = 2 \text{ divisions} \times \text{Pitch} = 2 \times 0.5 \text{ mm} = 1 \text{ mm} \]

2. **Circular Scale Reading (CSR)**:
\[ \text{CSR} = 20 \text{ divisions} \times \text{Least Count} = 20 \times 0.01 \text{ mm} = 0.2 \text{ mm} \]

3. **Total Observed Length (L_observed)**:
\[ L_{\text{observed}} = \text{MSR} + \text{CSR} = 1 \text{ mm} + 0.2 \text{ mm} = 1.2 \text{ mm} \]

### Step 5: Calculate the Actual Thickness of the Plate
To find the actual thickness of the plate, we need to subtract the zero error from the observed length:

\[ L_{\text{actual}} = L_{\text{observed}} - \text{Error} = 1.2 \text{ mm} - (-0.3 \text{ mm}) = 1.2 \text{ mm} + 0.3 \text{ mm} = 1.5 \text{ mm} \]

### Final Answer
The thickness of the plate is **1.5 mm**.

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