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A micrometre screw gauge has a negative zero error of 7 divisions. While measuring the diameter of a wire the reading on the main scale is 2 divisions

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Question

A micrometre screw gauge has a negative zero error of 7 divisions. While measuring the diameter of a wire the reading on the main scale is 2 divisions and 79th circular scale division coincides with baseline.
If the number of divisions on the main scale is 10 to a centimetre and circular scale has 100 divisions, calculate

  1. pitch
  2. observed diameter
  3. least count
  4. corrected diameter.
Numerical
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Solution

(1) The number of divisions on the main scale are 10 to a centimetre

⇒ Pitch = `"Unit"/"No. of divisions in unit"=(1 "cm")/10` = 0.1 cm

(2) No. of circular scale divisions = 100

∴ Least count (L.C.) = `"Pitch"/"No. of circular scale divisions"`

= `0.1/100` cm

= 0.001 cm

(3) Main scale reading = 2 division
⇒ Main scale reading = 2 × Pitch = 0.2 cm
Circular scale reading = 79 division
∴ Observed diametre = M.S. reading + L.C. × C.S. reading
= 0.2 + 0.001 × 79
= 0.2 + 0.079 cm
= 0.279 cm

(4) Negative zero error = 7 division
∴ Correct = −(−7 × L.C.)
= −(−7 × 0.001) cm
= +0.007 cm
Correct diametre = Observed diametre + Correction
= 0.279 + 0.007
= 0.286 cm

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

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Goyal Brothers Prakashan A New Approach to ICSE Physics [English] Class 9
Chapter 1 Measurements and Experimentation
Unit 4 Practice Problems 4 | Q 2

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