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V. Conceptual Questions · Q2

Q.Show that a screw gauge of pitch 1 mm and 100 divisions is more precise than a vernier caliper with 20 divisions on the sliding scale.

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✓ Free question

Step 1. The least count (LC) of an instrument is the smallest change it can resolve; a SMALLER least count means a MORE precise instrument.

Step 2. For the screw gauge: LC=pitchnumber of head-scale divisions=1 mm100=0.01 mm\text{LC}=\dfrac{\text{pitch}}{\text{number of head-scale divisions}}=\dfrac{1\ \text{mm}}{100}=0.01\ \text{mm}.

Step 3. For the vernier caliper (main scale divisions of 1 mm, as is standard): LC=1 main scale divisionnumber of vernier divisions=1 mm20=0.05 mm\text{LC}=\dfrac{\text{1 main scale division}}{\text{number of vernier divisions}}=\dfrac{1\ \text{mm}}{20}=0.05\ \text{mm}.

Step 4. Comparing: the screw gauge's LC (0.010.01 mm) is smaller than the vernier caliper's LC (0.050.05 mm) -- in fact exactly 5 times smaller.

Step 5. Since a smaller least count resolves a finer change, the screw gauge is the more precise of the two instruments.

✓Final answer

Screw gauge least count =1 mm/100=0.01=1\ \text{mm}/100=0.01 mm; vernier caliper least count =1 mm/20=0.05=1\ \text{mm}/20=0.05 mm. Since 0.01 mm<0.05 mm0.01\ \text{mm}<0.05\ \text{mm}, the screw gauge resolves a finer change and is therefore more precise.

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