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

Answer

(d)

Explanation

Since, the measured value is 3.50 cm, the least count of meassuring instrument must be 0.01 cm. it is so for the given vernier calliper. Least count for vernier calliper = Vernier constant `= 1 MSD -1 VSD = 1 MSD - (9//10) MSD`
` = (1)/(10) (MSD) = (1)/(10)xx0.1 cm = 0.01 cm`