Question

The least count of a vernier caliper is $0.01\text{cm}$ and $10$ divisions of the vernier scale is equal to $n$ divisions of main scale. The value of $n$ will be ($1$ division of main scale is $1\text{mm}$)

• A. $6$
• B. $9$
• C. $10$
• D. $12$

Answer 1

The correct option is B $9$

The least count of Vernier caliper is given by,

$\text{L.C}=1\text{M.S.D}-1\text{V.S.D}$

Given, $n\text{M.S.D}=10\text{V.S.D}$

So, $1\text{V.S.D}=\frac{n\text{M.S.D}}{10}$

Hence , $\text{L.C}=(1-\frac{n}{10})\text{M.S.D}$

$\Rightarrow \frac{1}{10}=(1-\frac{n}{10})1$

$\Rightarrow \frac{n}{10}=\frac{9}{10}$

$n=9$