The pitch of a screw gauge is 1 mm with 100 divisions on the circular scale. For a given wire, the linear scale reads 2 mm and 64th divisions on the circular scale coincides with the reference line. If the length of the wire is 3.5 cm, find the volume in `cm^(3)`?

Options

A. `19.0cm^(3)`
B. `0.19cm^(3)`
C. `76.0cm^(3)`
D. `0.76cm^(3)`

Answer

To find the volume of the wire using the given data, we will follow these steps:

### Step 1: Identify the given data
- **Pitch of the screw gauge (P)** = 1 mm
- **Number of divisions on the circular scale (N)** = 100
- **Linear scale reading (L)** = 2 mm
- **Circular scale reading (C)** = 64 divisions
- **Length of the wire (L_wire)** = 3.5 cm

### Step 2: Calculate the least count of the screw gauge
The least count (LC) can be calculated using the formula:
\[ \text{Least Count} = \frac{\text{Pitch}}{\text{Number of divisions}} = \frac{1 \text{ mm}}{100} = 0.01 \text{ mm} \]
Converting this to centimeters:
\[ \text{Least Count} = 0.01 \text{ mm} = 0.01 \times 0.1 \text{ cm} = 10^{-3} \text{ cm} \]

### Step 3: Calculate the diameter of the wire
The diameter (D) can be calculated using the formula:
\[ D = \text{Linear scale reading} + \left(\text{Circular scale reading} \times \text{Least Count}\right) \]
Substituting the values:
\[ D = 2 \text{ mm} + (64 \times 0.01 \text{ mm}) = 2 \text{ mm} + 0.64 \text{ mm} = 2.64 \text{ mm} \]
Now converting to centimeters:
\[ D = 2.64 \text{ mm} \times 0.1 \text{ cm/mm} = 0.264 \text{ cm} \]

### Step 4: Calculate the radius of the wire
The radius (r) is half of the diameter:
\[ r = \frac{D}{2} = \frac{0.264 \text{ cm}}{2} = 0.132 \text{ cm} \]

### Step 5: Calculate the volume of the wire
The volume (V) of the wire can be calculated using the formula for the volume of a cylinder:
\[ V = \pi r^2 L \]
Substituting the values:
\[ V = \pi \times (0.132 \text{ cm})^2 \times 3.5 \text{ cm} \]
Using \(\pi \approx 3.14\):
\[ V = 3.14 \times 0.017424 \text{ cm}^2 \times 3.5 \text{ cm} \approx 0.19 \text{ cm}^3 \]

### Final Answer
The volume of the wire is approximately \(0.19 \text{ cm}^3\).

---