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PhyEasyMain 2026 • 4th Apr - Shift 1

Least Count Calculation of a Screw Gauge

In a screw gauge when the circular scale is given five complete rotations it moves linearly by 2.5 mm2.5\text{ mm}. If the circular scale has 100100 divisions, the least count of screw gauge is ______ mm\text{mm}.

Options

A

1×10−21 \times 10^{-2}

B

1×10−31 \times 10^{-3}

C

5×10−25 \times 10^{-2}

D

5×10−35 \times 10^{-3}

Correct

Topics & Concepts

Units and MeasurementsMeasuring Instruments

Step-by-Step Solution

To find the least count of the screw gauge, we first need to determine its pitch.

The pitch of a screw gauge is defined as the distance moved linearly by the spindle per complete rotation of the circular scale: Pitch=Linear distance movedNumber of complete rotations\text{Pitch} = \frac{\text{Linear distance moved}}{\text{Number of complete rotations}}

Given:

  • Linear distance moved = 2.5 mm2.5 \text{ mm}
  • Number of complete rotations = 55

Pitch=2.5 mm5=0.5 mm\text{Pitch} = \frac{2.5 \text{ mm}}{5} = 0.5 \text{ mm}

The least count (LC) of the screw gauge is given by: Least Count=PitchTotal number of divisions on the circular scale\text{Least Count} = \frac{\text{Pitch}}{\text{Total number of divisions on the circular scale}}

Given:

  • Total number of divisions on the circular scale = 100100

Least Count=0.5 mm100=0.005 mm=5×10−3 mm\text{Least Count} = \frac{0.5 \text{ mm}}{100} = 0.005 \text{ mm} = 5 \times 10^{-3} \text{ mm}

Thus, the least count of the screw gauge is 5×10−3 mm5 \times 10^{-3} \text{ mm}, which corresponds to option D.

Least Count Calculation of a Screw Gauge | Physics PYQ Solution - JEE Challenger