Question

A vernier scale has 10 divisions. It slides over the main scale,whose$1\mathrm{M}.\mathrm{S}.\mathrm{D}.\mathrm{is}1.0\mathrm{mm}$. If the number of divisions to the left side of the zero of the vernier scale on the main scale is $49\mathrm{and}\mathrm{the}8\mathrm{th}$vernier scale division coincides with the main scale, calculate the length in centimetres.

Answer 1

Step 1:- Given data

Vernier scale division $\left(\mathrm{N}\right)=10$

1 Main scale division $\left(\mathrm{MSD}\right)=1\mathrm{mm}$

Main scale reading $\left(\mathrm{MSR}\right)=49\mathrm{mm}$

Vernier caliper coincide $\left(\mathrm{VSR}\right)=8$

Step 2:- Formula for Vernier caliper:

Since least count $\raisebox{1ex}{$=\mathrm{MSD}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{N}$}\right.$

$$\begin{gathered} =\frac{1}{10} \\ =0.1\mathrm{mm} \end{gathered}$$

$\mathrm{Length}=\mathrm{MSR}+(\mathrm{VSR}\mathrm{x}\mathrm{L}.\mathrm{C})$

$$\begin{gathered} =49+8\mathrm{x}0.1 \\ =49.8\mathrm{mm} \end{gathered}$$

Hence, the length in centimeters will be $4.98\mathrm{cm}.$