In a complete rotation, spindle of a screw gauge advances by `(1)/(2)mm`. There are `50` divisions on circular scale. The main scale has `(1)/(2)mm` marks to (is graduated to `(1)/(2)mm`) If a wire is put between the jaws, `3` main scale divisions are clearly visible, and `20th` division of circular scale coincides with reference line. Find diameter of wire in correct significant figures.

Options

A. (a)1.7 mm
B. (b)1.70 mm
C. (c)1.58mm
D. (d)1.30 mm

Explanation

To find the diameter of the wire using the screw gauge measurements, we can follow these steps:

### Step 1: Calculate the Least Count (LC) of the Screw Gauge
The least count is defined as the smallest measurement that can be read on the instrument. It can be calculated using the formula:

\[ LC = \frac{\text{Value of one complete rotation}}{\text{Number of divisions on circular scale}} \]

Given:
- Value of one complete rotation = \( \frac{1}{2} \) mm
- Number of divisions on circular scale = 50

So,

\[ LC = \frac{\frac{1}{2} \text{ mm}}{50} = \frac{0.5 \text{ mm}}{50} = 0.01 \text{ mm} \]

### Step 2: Read the Main Scale (MSD) and Circular Scale (CSD) Values
From the problem:
- The main scale reading (MSD) is given as 3 divisions.
- Each main scale division is \( \frac{1}{2} \) mm.

Thus,

\[ \text{MSD} = 3 \times \frac{1}{2} \text{ mm} = \frac{3}{2} \text{ mm} = 1.5 \text{ mm} \]

Next, the circular scale reading (CSD) is the 20th division. Each division on the circular scale corresponds to the least count:

\[ \text{CSD} = 20 \times LC = 20 \times 0.01 \text{ mm} = 0.2 \text{ mm} \]

### Step 3: Calculate the Diameter of the Wire
The diameter of the wire can be calculated using the formula:

\[ \text{Diameter} (D) = \text{MSD} + \text{CSD} \]

Substituting the values we found:

\[ D = 1.5 \text{ mm} + 0.2 \text{ mm} = 1.7 \text{ mm} \]

### Final Answer
The diameter of the wire is \( 1.7 \text{ mm} \).

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