Consider the following data : $10$ MSDs = $1$ cm, $10$ VSDs = $9$ MSDs, zero of Vernier scale is to the right of the zero marking of the main scale with $6$ VSDs coinciding with MSDs and the actual reading for length measurement is $4.3$ cm with $2$ VSDs coinciding with main scale graduations. Estimate the length.

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Solution

In this case, Vernier constant = $\frac{1 m m}{10} = 0.1 m m$
Zero error = $6 \times 0.1 = + 0.6 m m$
Correction = $- 0.6 m m$
Actual length = $(4.3 + 2 \times 0.01) + C o r r e c t i o n = 4.32 - 0.06 = 4.26 c m$

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Q. Consider the following data for a vernier calliper:
10 main scale division = 1 cm.
10 vernier division matches with 9 Main scale division.

While measuring the zero error of the vernier we see that zero of vernier scale is to the right of the zero marking of the main scale with

6^{t h}

vernier division coinciding with main scale divisions.

The actual reading for length measurement is 4.3 cm with 2nd vernier division coinciding with main scale graduation.

Estimate the length.

Consider the following data: 10 main scale division = 1cm, $10^{t h}$ vernier division match with $9^{t h}$ Main scale division. While measuring the zero error of the vernier we see that zero of vernier scale is to the right of the zero marking of the main scale with $6^{t h}$ vernier division coinciding with main scale divisions. The actual reading for length measurement is 4.3 cm with 2nd vernier division coinciding with main scale graduation. Estimate the length.