Question

A student performs an experiment of measuring the thickness of a slab with a vernier calliper whose $50$ divisions of the vernier scale are equal to $49$ divisions of the main scale. He noted that zero of the vernier scale is between $7.00$ $cm$ and $7.05$ $cm$ mark of the main scale and ${23}^{rd}$ division of the vernier scale exactly coincides with the main scale. The measured value of the thickness of the given slab using the calliper will be:

• A. $7.23cm$
• B. $7.023cm$
• C. $7.073cm$
• D. $7.73cm$

Answer 1

The correct option is B $7.023cm$

$\begin{array}{c}\text{Given - vernier scale with}50VSD=49MSD\\ \text{VSD= Vernier Scale div}\\ \text{MSD = Main scale div}\end{array}$

$\begin{array}{c}{23}^{rd}\text{division of the vernier scale matches}\\ \text{with main scale.}\end{array}$

$\begin{array}{cc}\text{Least count}& =\text{1MSD - 1VSD}\\ & =1MSD-\frac{49}{50}MSD=\frac{1}{50}MSD\\ & \text{Since}50VSD=49MSD\end{array}$

$\begin{array}{c}\text{we have}1MSD=0.05cm\\ \Rightarrow \text{Least count}=\frac{1}{50}\times 0.05=0.001cm\end{array}$

$\begin{array}{c}\text{We have Rading -}\\ \text{Reading}=\text{Main scale Reading}+23\times \text{least count}\\ \begin{array}{cc}& =7.00cm+23\times 0.001\\ & =7.023cm.\end{array}\end{array}$