JEE Challenger
More from Units and Measurements
PhyMedAdv 2015 • Paper 1

Vernier Callipers and Screw Gauge Least Count Analysis

Consider a Vernier callipers setup where 1 cm1\text{ cm} on the main scale is split into 88 equal divisions. A screw gauge with 100100 circular scale divisions is also given. In the Vernier callipers, 55 divisions on the Vernier scale match 44 main scale divisions. In the screw gauge, one full turn of its circular scale shifts it by 22 divisions on its linear scale. Select the correct statement(s):

Options

A

If the screw gauge pitch is two times the Vernier callipers least count, its least count equals 0.01 mm0.01\text{ mm}.

B

If the screw gauge pitch is two times the Vernier callipers least count, its least count equals 0.005 mm0.005\text{ mm}.

Correct
C

If the linear scale least count of the screw gauge is double the Vernier callipers least count, the screw gauge least count is 0.01 mm0.01\text{ mm}.

Correct
D

If the linear scale least count of the screw gauge is double the Vernier callipers least count, the screw gauge least count is 0.005 mm0.005\text{ mm}.

Topics & Concepts

Units and MeasurementsMeasuring Instruments

Step-by-Step Solution

To determine the correct statements, let us systematically calculate the least counts of both the Vernier callipers and the screw gauge.

1. Least Count of the Vernier Callipers (LCVC\text{LC}_{\text{VC}})

  • Main Scale Division (1 MSD1\text{ MSD}): Given that 1 cm1\text{ cm} on the main scale is divided into 88 equal divisions: 1 MSD=18 cm=0.125 cm=1.25 mm1\text{ MSD} = \frac{1}{8}\text{ cm} = 0.125\text{ cm} = 1.25\text{ mm}

  • Vernier Scale Division (1 VSD1\text{ VSD}): Given that 55 divisions of the Vernier scale coincide with 44 main scale divisions: 5 VSD=4 MSD  ⟹  1 VSD=45 MSD5\text{ VSD} = 4\text{ MSD} \implies 1\text{ VSD} = \frac{4}{5}\text{ MSD}

  • Least Count (LCVC\text{LC}_{\text{VC}}): LCVC=1 MSD−1 VSD=1 MSD−45 MSD=15 MSD\text{LC}_{\text{VC}} = 1\text{ MSD} - 1\text{ VSD} = 1\text{ MSD} - \frac{4}{5}\text{ MSD} = \frac{1}{5}\text{ MSD} LCVC=15×1.25 mm=0.25 mm\text{LC}_{\text{VC}} = \frac{1}{5} \times 1.25\text{ mm} = 0.25\text{ mm}

2. Screw Gauge Formulae

  • Number of circular scale divisions (NN) = 100100.
  • One complete rotation moves the circular scale by 22 divisions on the linear scale (LSD\text{LSD}). Pitch=2×LSD\text{Pitch} = 2 \times \text{LSD}
  • The least count of the screw gauge (LCSG\text{LC}_{\text{SG}}) is given by: LCSG=PitchN=Pitch100\text{LC}_{\text{SG}} = \frac{\text{Pitch}}{N} = \frac{\text{Pitch}}{100}

3. Analysis of Options (A) and (B)

  • Condition: Pitch of the screw gauge is twice the least count of the Vernier callipers. Pitch=2×LCVC=2×0.25 mm=0.5 mm\text{Pitch} = 2 \times \text{LC}_{\text{VC}} = 2 \times 0.25\text{ mm} = 0.5\text{ mm}

  • Calculating Least Count of Screw Gauge (LCSG\text{LC}_{\text{SG}}): LCSG=0.5 mm100=0.005 mm\text{LC}_{\text{SG}} = \frac{0.5\text{ mm}}{100} = 0.005\text{ mm}

Hence, Option (B) is correct, and Option (A) is incorrect.

4. Analysis of Options (C) and (D)

  • Condition: The linear scale least count (LSD\text{LSD}) of the screw gauge is double the Vernier callipers least count. LSD=2×LCVC=2×0.25 mm=0.5 mm\text{LSD} = 2 \times \text{LC}_{\text{VC}} = 2 \times 0.25\text{ mm} = 0.5\text{ mm}

  • Calculating Pitch: Pitch=2×LSD=2×0.5 mm=1.0 mm\text{Pitch} = 2 \times \text{LSD} = 2 \times 0.5\text{ mm} = 1.0\text{ mm}

  • Calculating Least Count of Screw Gauge (LCSG\text{LC}_{\text{SG}}): LCSG=Pitch100=1.0 mm100=0.01 mm\text{LC}_{\text{SG}} = \frac{\text{Pitch}}{100} = \frac{1.0\text{ mm}}{100} = 0.01\text{ mm}

Hence, Option (C) is correct, and Option (D) is incorrect.

Conclusion

The correct options are B and C.

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