Question

In a screw gauge when the circular scale is given five complete rotations it moves linearly by 2.5 mm. If the circular scale has 100 divisions, the least count of screw gauge is ______ mm.

• A. \(1 \times 10^{-2}\)
• B. \(1 \times 10^{-3}\)
• C. \(5 \times 10^{-2}\)
• D. \(5 \times 10^{-3}\)

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The least count of a screw gauge is the smallest linear distance that can be accurately measured by the instrument.
It is determined by calculating the pitch of the screw first, and then dividing the pitch by the total number of divisions on the circular scale.
Step 2: Key Formula or Approach:
1. \(\text{Pitch} = \frac{\text{Linear distance moved}}{\text{Number of complete rotations}}\)
2. \(\text{Least Count (L.C.)} = \frac{\text{Pitch}}{\text{Number of divisions on circular scale}}\)
Step 3: Detailed Explanation:
Given in the problem:
Linear distance moved = \(2.5\) mm.
Number of complete rotations = \(5\).
Number of circular scale divisions = \(100\).
First, calculate the Pitch:
\(\text{Pitch} = \frac{2.5 \text{ mm}}{5} = 0.5 \text{ mm}\).
Now, calculate the Least Count:
\(\text{L.C.} = \frac{0.5 \text{ mm}}{100} = 0.005 \text{ mm}\).
Step 4: Final Answer:
Convert the decimal to scientific notation to match the options:
\(0.005 \text{ mm} = 5 \times 10^{-3} \text{ mm}\).