Question

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

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

Answer 1

Answer: (B)

If student measures $3.50 \,cm$, it means that there is an uncertainly of order $0.01\, cm$.
For vernier scale with $1 MSD =\frac{1}{10} cm$ and
and $9 \,MSD =10 \,VSD$
$LC$ of vernier calliper $=1 \,MSD -1 \,VSD$
$=\frac{1}{10}\left(1-\frac{9}{10}\right)=\frac{1}{100} cm =0.01\, cm$