screw gauge of pitch 0.1 cm and 50 divison on circular scale measure thickness of object which of following measurement is possible for thickness

Options

A. 2.123 cm
B. 2.124 cm
C. 2.125 cm
D. 2.127 cm

Answer

To solve the problem of determining which measurement of thickness is possible using a screw gauge with a pitch of 0.1 cm and 50 divisions on the circular scale, we can follow these steps:

### Step 1: Understand the Given Data
- **Pitch of the screw gauge (P)** = 0.1 cm
- **Number of divisions on the circular scale (n)** = 50

### Step 2: Calculate the Least Count (LC)
The least count of a screw gauge can be calculated using the formula:
\[ \text{Least Count (LC)} = \frac{\text{Pitch}}{\text{Number of divisions on circular scale}} = \frac{P}{n} \]
Substituting the values:
\[ LC = \frac{0.1 \text{ cm}}{50} = 0.002 \text{ cm} \]

### Step 3: Analyze the Given Options
We have four options for the thickness of the object:
1. 2.123 cm
2. 2.124 cm
3. 2.125 cm
4. 2.2 cm

### Step 4: Determine Which Measurements Are Possible
To find out which of these measurements can be achieved with the least count of 0.002 cm, we need to check if the difference between each option and the nearest whole number is a multiple of the least count (0.002 cm).

- **Option 1: 2.123 cm**
- Difference from 2.12 cm = 0.003 cm (not a multiple of 0.002)

- **Option 2: 2.124 cm**
- Difference from 2.12 cm = 0.004 cm (multiple of 0.002)

- **Option 3: 2.125 cm**
- Difference from 2.12 cm = 0.005 cm (not a multiple of 0.002)

- **Option 4: 2.2 cm**
- Difference from 2.12 cm = 0.08 cm (not a multiple of 0.002)

### Step 5: Conclusion
The only option that is a multiple of the least count (0.002 cm) is:
- **Option 2: 2.124 cm**

Thus, the possible measurement for the thickness of the object is **2.124 cm**.

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