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 meter scale
B. 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 cm
C. A screw gauge having 100 divisions in the circular scale and pitch as 1 mm
D. A screw gauge having 50 divisions in the circular scale and pitch as 1 mm

Answer

B

Explanation

10VSD = 9MSD `implies` 1 VSD =`9/10 ` MSD
L.C = 1 MSD - 1VSD `=(1-9/10) 1 cm ` = 0.01 cm