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 distance moved by the screw when it is rotated through one division of the circular scale. It is calculated by dividing the pitch by the total number of divisions on the circular scale.

Step 2: Key Formula or Approach:
1. $\text{Pitch} = \frac{\text{Distance moved}}{\text{Number of full rotations}}$
2. $\text{Least Count (L.C.)} = \frac{\text{Pitch}}{\text{Number of circular scale divisions}}$

Step 3: Detailed Explanation:
1. First, calculate the pitch: \[ \text{Pitch} = \frac{2.5 \text{ mm}}{5} = 0.5 \text{ mm} \] 2. Now, calculate the least count: \[ \text{L.C.} = \frac{0.5 \text{ mm}}{100} = 0.005 \text{ mm} \] 3. Expressing in scientific notation: \[ 0.005 = 5 \times 10^{-3} \text{ mm} \]

Step 4: Final Answer:
The least count of the screw gauge is \(5 \times 10^{-3}\) mm.