Question

In a screw gauge the zero of main scale reference line coincides with the fifth division of the circular scale when two studs are in contact. There are 100 divisions in circular scale and pitch of screw gauge is 0.1 mm. When diameter of a sphere is measured, the reading of main scale is 5 mm and 50th division of circular scale coincides with the reference line of main scale. The diameter of sphere is _______ mm.}

• A. 5.045
• B. 5.055
• C. 5.450
• D. 5.550

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The diameter measured by a screw gauge is the sum of the Main Scale Reading (MSR) and the Circular Scale Reading (CSR) multiplied by the Least Count (LC).
A zero error must also be accounted for. If the zero of the circular scale is below the reference line (the 5th division coincides), the zero error is positive.
True Reading = Observed Reading - Zero Error.
Step 2: Key Formula or Approach:
Least Count (LC) = \(\frac{\text{Pitch}}{\text{Number of circular divisions}}\).
Observed Reading = MSR + (CSR \(\times\) LC).
Zero Error = (Coinciding division) \(\times\) LC.
Step 3: Detailed Explanation:
First, calculate the Least Count:
\[ LC = \frac{0.1 \text{ mm}}{100} = 0.001 \text{ mm} \] Determine the Zero Error:
Since the zero of the reference line coincides with the 5th division, the 0 mark of the circular scale has crossed the reference line, meaning there is a positive zero error.
\[ \text{Zero Error} = +5 \times 0.001 \text{ mm} = +0.005 \text{ mm} \] Now calculate the Observed Reading:
Main Scale Reading (MSR) = \(5 \text{ mm}\).
Circular Scale Reading (CSR) = \(50\).
\[ \text{Observed Reading} = 5 + 50 \times 0.001 = 5.050 \text{ mm} \] Finally, calculate the True Reading (Diameter of the sphere):
\[ \text{True Reading} = \text{Observed Reading} - \text{Zero Error} \] \[ \text{True Reading} = 5.050 \text{ mm} - 0.005 \text{ mm} = 5.045 \text{ mm} \] Step 4: Final Answer:
The diameter of the sphere is \(5.045\text{ mm}\).