Question

In a screw gauge, when the circular scale is given five complete rotations, it moves linearly by \(2.5\,\text{mm\). If the circular scale has 100 divisions, the least count of the screw gauge is:}

• A. \(1\times10^{-2}\) mm
• B. \(5\times10^{-3}\) mm
• C. \(1\times10^{-3}\) mm
• D. \(4\times10^{-2}\) mm

Answer 1

The Correct Option is B

Concept:
Least count of a screw gauge is given by \[ \text{Least Count} = \frac{\text{Pitch}}{\text{Number of divisions on circular scale}} \] Pitch is the linear distance moved by the screw in one complete rotation. Step 1: Find the pitch. Given 5 rotations move the screw by \(2.5\,\text{mm}\): \[ \text{Pitch}=\frac{2.5}{5} \] \[ \text{Pitch}=0.5\,\text{mm} \] \[ =5\times10^{-4}\,\text{m} \] Step 2: Calculate the least count. Number of divisions on circular scale \(=100\) \[ \text{L.C.}=\frac{\text{Pitch}}{\text{Number of divisions}} \] \[ =\frac{5\times10^{-4}}{100}\,\text{m} \] \[ =5\times10^{-6}\,\text{m} \] Step 3: Convert into millimetres. \[ 5\times10^{-6}\,\text{m}=5\times10^{-3}\,\text{mm} \]