Question

A student measured the length of a rod and wrote it as $3.50\text{cm}$. Which instrument did he use to measure it ?

• A. A screw gauge have $50$ divisions in the circular scale and pitch as $1\text{mm}$
• B. A screw gauge having $100$ divisions in the circular scale and pitch as $1\text{mm}$.
• C. A meter scale
• D. A vernier calliper where the $10$ divisions in vernier scale matches with $9$ division in main scale and main scale has $10$ divisions in $1\text{cm}$.

Answer 1

The correct option is D A vernier calliper where the $10$ divisions in vernier scale matches with $9$ division in main scale and main scale has $10$ divisions in $1\text{cm}$.

If student measure $3.50\text{cm}$, it means that there in an uncertainity of order of $0.01\text{cm}$. So, such measurement can be done by instrument of least count $0.01\text{cm}$.

From options:
$\left(a\right)$ Least count of meter scale $=1\text{mm}$


$\left(b\right)$ Least count of $\text{V.C.}=1\text{MSD}-1\text{VSD}$

$=\frac{1}{10}[1-\frac{9}{10}]=\frac{1}{100}\text{cm}$

$\left(c\right)$ Least count of screw gauge $=\frac{p}{n}=\frac{1}{100}\text{mm}$

$\left(d\right)$ Least count of screw gauge $=\frac{p}{n}=\frac{1}{50}\text{mm}$

So, LC of VC = $0.01\text{cm}$

Hence, $\left(\text{B}\right)$ is the correct option.