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

Answer 1

The correct option is A A vermier 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 uncertianly of order $0.01\text{cm}$.

$L.C$ of $V.C$ = $1MSD-1VSD$
$=\frac{1}{10}[1-\frac{9}{10}]=\frac{1}{100}\text{cm}$

Hence, $B$ is correct option.