Question

$1cm$ of main scale of a vernier caliper is divided into $10$ divisions. The least count of the calipers is $0.005cm$, then the vernier scale must have

• A. $10$ divisions
• B. $20$ divisions
• C. $25$ divisions
• D. $50$ divisions

Answer 1

The correct option is B $20$ divisions

$\begin{array}{cc}1MSD& =\frac{1}{10}cm\\ & =0.1cm\\ L\cdot C=& 0.005cm\\ L\cdot C=& \mathrm{1\; M}SD-1VSD⟶\left(i\right)\end{array}$

$\begin{array}{c}0.005cm=0.1-1VSD\\ \text{1 VSD}=0.095cm\\ \text{let,}\\ nVSD=m\text{MSD}\\ n\times 0.095=m\times 0.1\\ \frac{n}{m}=\frac{100}{95}\end{array}$

$\begin{array}{c}n\&m\text{are integers.}\text{Minimum values of}n\text{\&}m\\ \text{are}20\text{and}19\text{respectively}\\ \therefore n=20\end{array}$