Question

In a Screw Gauge, fifth division of the circular scale coincides with the reference line when the ratchet is closed. There are 50 divisions on the circular scale, and the main scale moves by 0.5 mm on a complete rotation. For a particular observation the reading on the main scale is 5 mm and the 20th division of the circular scale coincides with reference line. Calculate the true reading.

• A. 5.20 mm
• B. 5.25 mm
• C. 5.15 mm
• D. 5.00 mm

Hint:

Always subtract the zero error algebraically. If the zero error is positive (divisions above reference), subtract the magnitude. If negative, you end up adding the magnitude.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The measurement with a screw gauge involves determining the Least Count (LC), identifying the zero error when the jaws are closed, and calculating the final reading by subtracting the zero error from the observed reading.
Step 2: Key Formula or Approach:
1. Least Count (LC) = \(\frac{\text{Pitch}}{\text{Number of circular scale divisions}}\)
2. Zero Error = \((\text{Coinciding division}) \times \text{LC}\)
3. Observed Reading = \(\text{Main Scale Reading (MSR)} + (\text{Circular Scale Reading (CSR)} \times \text{LC})\)
4. True Reading = \(\text{Observed Reading} - \text{Zero Error}\)
Step 3: Detailed Explanation:
Given:
Pitch = 0.5 mm
Number of divisions = 50
\[ \text{LC} = \frac{0.5 \text{ mm}}{50} = 0.01 \text{ mm} \]
When closed, the 5th division coincides. Since it is above zero, it is a positive zero error.
\[ \text{Zero Error} = +5 \times 0.01 \text{ mm} = +0.05 \text{ mm} \]
For the observation:
MSR = 5 mm
CSR = 20
\[ \text{Observed Reading} = 5 \text{ mm} + (20 \times 0.01 \text{ mm}) = 5.20 \text{ mm} \]
Calculating the true reading:
\[ \text{True Reading} = 5.20 \text{ mm} - 0.05 \text{ mm} = 5.15 \text{ mm} \]
Step 4: Final Answer:
The true reading of the observation is 5.15 mm.