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NEET PHYSICSEasy

A screw gauge has the least count of 0.01 mm0.01\text{ mm} and there are 5050 divisions in its circular scale. The pitch of the screw gauge is:

A

0.25 mm0.25\text{ mm}

B

0.5 mm0.5\text{ mm}

C

1.0 mm1.0\text{ mm}

D

0.01 mm0.01\text{ mm}

Step-by-Step Solution

  1. Recall the formula: The least count of a screw gauge is given by the formula: Least Count=PitchNumber of divisions on the circular scale\text{Least Count} = \frac{\text{Pitch}}{\text{Number of divisions on the circular scale}}.
  2. Rearrange for Pitch: Pitch=Least Count×Number of divisions on the circular scale\text{Pitch} = \text{Least Count} \times \text{Number of divisions on the circular scale}.
  3. Substitute the given values: Least count=0.01 mm\text{Least count} = 0.01\text{ mm} Number of circular scale divisions=50\text{Number of circular scale divisions} = 50 Pitch=0.01 mm×50=0.5 mm\text{Pitch} = 0.01\text{ mm} \times 50 = 0.5\text{ mm}.
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