Question

The pitch and the number of circular scale divisions in a screw gauge with least count 0.02 mm are respectively

• A. 1 mm and 100
• B. 0.5 mm and 50
• C. 1 mm and 50
• D. 0.5 mm and 100
• E. 1 mm and 200

Hint:

To quickly check, you can rearrange the formula to \( \text{Pitch} = LC \times N \). Only option (C) satisfies \( 1 = 0.02 \times 50 \).

Answer 1

The Correct Option is C

Concept: A screw gauge is a precision instrument used for measuring small dimensions. Its precision is determined by its Least Count (LC).
• Pitch: The linear distance moved by the screw in one complete rotation of the circular scale.
• Least Count Formula: \( LC = \frac{\text{Pitch}}{\text{Total number of circular scale divisions (N)}} \).

Step 1: Evaluate the options against the Least Count formula.
We are given \( LC = 0.02 \text{ mm} \). We must find the pair of Pitch and \( N \) that satisfies the relationship \( \frac{\text{Pitch}}{N} = 0.02 \).

Step 2: Calculate the ratio for each option.

• (A): \( 1 / 100 = 0.01 \text{ mm} \) (Incorrect)
• (B): \( 0.5 / 50 = 0.01 \text{ mm} \) (Incorrect)
• (C): \( 1 / 50 = 0.02 \text{ mm} \) (Correct)
• (D): \( 0.5 / 100 = 0.005 \text{ mm} \) (Incorrect)
• (E): \( 1 / 200 = 0.005 \text{ mm} \) (Incorrect)