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The circular scale of a screw gauge has 50 divisions. Its spindle moves by 2 mm on the sleeve, when given four complete rotations calculate pitch least count - Physics

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Question

The circular scale of a screw gauge has 50 divisions. Its spindle moves by 2 mm on the sleeve, when given four complete rotations calculate

  1. pitch
  2. least count
Numerical
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Solution

Number of circular scale divisions (C.S.D.) = 50
Distance moved by screw (spindle) on sleeve = 2 mm
Number of complete rotations given = 4

(1) Pitch = `"Distance moved by screw on sleeve"/"No. of complete rotations"`

= `(2 "mm")/4`

= 0.5 mm

(2) Pitch = 0.05 cm

Least count = `"Pitch"/"No. of circular scale divisions"`

= `0.05/50`

= 0.001 cm

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Vernier Callipers
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Chapter 1: Measurements and Experimentation - Unit 4 Practice Problems 1

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Goyal Brothers Prakashan A New Approach to ICSE Physics Part 1 [English] Class 9
Chapter 1 Measurements and Experimentation
Unit 4 Practice Problems 1 | Q 1

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