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The circular scale of a screw gauge has 100 divisions. Its spindle moves forward by 2.5 mm when given five complete turns. Calculate pitch least count of the screw gauge. - Physics

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Question

The circular scale of a screw gauge has 100 divisions. Its spindle moves forward by 2.5 mm when given five complete turns. Calculate

  1. pitch
  2. least count of the screw gauge
Numerical

Solution

Number of circular scale divisions = 100
Distance moved by spindle (screw) = 2.5 mm
No. of complete rotations given = 5

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

= `2.5/5`

= 0.5 mm

= 0.05 cm

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

= `0.05/100` cm

= 0.0005 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 2
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