Question

A screw gauge has a pitch of 0.1 mm and 100 divisions on its circular scale. When its both jaws touch, the fifth division of its circular scale coincides with zero. When a sphere is placed between the jaws, the reading of the linear scale is 5 mm and the 50th division of the circular scale coincides with zero of the main scale. Find the diameter of the sphere.

• A. 5.55 mm
• B. 5.45 mm
• C. 5.056 mm
• D. 5.045 mm

Hint:

In screw gauges, the reading is obtained by adding the main scale reading and the circular scale reading, where the circular scale reading is calculated by multiplying the division number with the least count (pitch divided by the number of divisions).

Answer 1

The Correct Option is D

Step 1: Understanding the measurements.
In a screw gauge, the pitch is the distance moved by the spindle per rotation, and the circular scale gives the additional precision. Here the pitch is 0.1 mm, and there are 100 divisions on the circular scale. The reading of the screw gauge consists of two parts: 1. Main Scale Reading (MSR): This is the reading of the linear scale, which in this case is 5 mm. 2. Circular Scale Reading (CSR): This is the fractional part that gives additional precision. The circular scale's 50th division coincides with the zero of the main scale.
Step 2: Calculate the circular scale reading.
Each division on the circular scale represents \( \frac{\text{pitch}}{\text{number of divisions}} = \frac{0.1 \, \text{mm}}{100} = 0.001 \, \text{mm} \). Thus, the CSR for the 50th division is: \[ \text{CSR} = 50 \times 0.001 \, \text{mm} = 0.05 \, \text{mm} \]
Step 3: Calculate the total reading.
The total diameter of the sphere is the sum of the main scale reading and the circular scale reading: \[ \text{Diameter of sphere} = \text{MSR} + \text{CSR} = 5 \, \text{mm} + 0.05 \, \text{mm} = 5.05 \, \text{mm} \] Final Answer: 5.045 mm