I am a vernier calipers. I have a task to measure the length of a pencil as shown in the figure. The vernier constant of my vernier scale is 0.05 times the value of one division of main-scale.

My Vernier Scale has n-divisions and coincides with (n – 4) divisions of main scale. While measuring the length of pencil, the $\frac{n}{5}$ th division of Vernier-Scale exactly coincides with a division of main-scale. Neglect any zero error.
Now choose the correct option(s)

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The vernier constant of Vernier scale is 0.005 cm

The length of Pencil is 8.48 cm

The value of n is 40

The length of Pencil is 8.44 cm

answer is B, D.

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Detailed Solution

$\begin{array}{l} L . C . = 0.05 t i m e s 1 M S D = 0.05 \times 0.1 c m = 0.005 c m = 0.05 m m \\ L e t 1 V . S . D . = x \\ ∴ L . C . = 1 M . S . D . - 1 V . S . D . \\ \Rightarrow 0.05 m m = 1 m m - x \\ \Rightarrow x = 0.95 m m \\ G i v e n : (n - 4) \times 1 m m = n \times 0.95 m m \\ \Rightarrow n = 80 \\ Length of pencil = 8.4 c m + [0.005 c m \times (\frac{80}{5})] = 8.48 cm \end{array}$

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