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When the two jaws of a Vernier caliper are brought into contact, the zero mark of the Vernier scale is found to be to the left of the zero mark of the main scale. It is observed that the 6^ th division of the Vernier scale coincides exactly with a main scale division. If 1 main scale division is equal to 1 mm and the Vernier scale has 10 divisions, the zero error of the instrument is

Options

  1. A0.06 cm positive zero error
  2. B0.06 cm negative zero error
  3. C0.04 cm negative zero error
  4. D0.04 cm positive zero error

Correct answer

C. 0.04 cm negative zero error

Step-by-step solution

The least count (LC) of the Vernier caliper is given by LC = 1 MSD Total VSD . Substituting the given values, LC = 1 mm 10 = 0.1 mm = 0.01 cm . Since the zero of the Vernier scale lies to the left of the main scale zero, the instrument has a negative zero error. For a negative zero error, the magnitude is calculated by subtracting the coinciding Vernier division from the total number of Vernier divisions. Zero error = - ( Total VSD - n) LC , where n is the coinciding division. Given n = 6 and total VSD = 10 , Zero

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