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

Options

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

Explanation

Measured length of a rod is 3.50cm, hence least count of instrument used near should be 0.01cm = 0.1mm.
Least count = 1MSD - IVSD ... (1).
In vemier scale IMSD = Imm and
9MSD = 10VSD
`:.` IVSD=0.9 MSD
`:.` From equation (i)
Least count of vernier calliper
`=1MSD=0.9MSD`
`=0.1 MSD =0.1 mm, 0.01 cm`