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.

4.32 $c m$

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4.38 $c m$

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4.26 $c m$

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4.36 $c m$

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Solution

The correct option is C
4.26 $c m$

$10 M S D = 1 c m$
$∴ 1 M S D = 0.1 c m = 1 m m$

Also, $10 V S D = 9 M S D$
$∴ 1 V S D = 0.9 M S D$
$1 V S D = 0.9 m m$

Least count of vernier Calliper is:
$L . C . = 1 M S D - 1 V S D$
$L . C . = 1 m m - 0.9 m m$
$L . C . = 0.1 m m$

Also given while measuring the zero error, zero of vernier is to the right of the main scale. So, zero error is:
$Z = + 6 L C$
$Z = 6 \times 0.1 m m = 0.6 m m$

Actual reading,
$M R = M S R + V S R - Z$
$M R = 43 + (2 \times 0.1) - 0.6 m m$
$M R = 4.26 c m$

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Similar questions

The least count of the main scale of a vernier callipers is $1$ mm. Its vernier scale is divided into $10$ divisions and coincide with 9 divisions of the main scale. When jaws are touching each other, the $7 t h$ division of vernier scale coincides with a division of main scale and the zero of vernier scale is lying right side of the zero of main scale. When this vernier is used to measure length of a cylinder the zero of the vernier scale between $3.1$ cm and $3.2$ cm and $4 t h$ VSD coincides with a main scale division. The length of the cylinder is: (VSD is vernier scale division)

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